CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Thermochemistry
Question Bank - Set 2
Liberty University
Question 1
Question
For the reaction:
2NO(g) + O2(g)→2NO2(g)
the standard enthalpy change is ∆H◦=−113.1 kJ. Calculate the standard
enthalpy change, ∆H◦
f, for formation of 1 mole of NO2(g) from its elements.
Solution
Step 1: Write the formation reaction for NO2from its elements:
2NO(g) + O2(g)→2NO2(g)
∆H◦
1=−113.1 kJ
Step 2: Write the formation reactions for NO and O2from their elements:
2N2(g) + O2(g)→2NO(g)
∆H◦
2=?
2O2(g)→2O(g)
∆H◦
3=?
Step 3: Calculate the standard enthalpy change for formation of NO2from
its elements using Hess’s Law:
∆H◦
f(NO2)=∆H◦
1−∆H◦
2−∆H◦
3
Substitute the known values to find ∆H◦
f(NO2).
Question 2
Question
Calculate the enthalpy change (∆H) for the reaction:
2C(graphite) + O2(g)→2CO(g)
given the following standard enthalpies of formation:
C(graphite) = 0 kJ/mol
CO(g) = −110.5 kJ/mol
O2(g) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation and list the standard enthalpies
of formation for the reactants and products involved in the reaction. Step 2:
Calculate the standard enthalpy change for the reaction using the formula:
∆H=XProducts −XReactants
Step 3: Substitute the values for the standard enthalpies of formation into
the formula and calculate the enthalpy change for the reaction.
Step 1: The balanced chemical equation is: 2C(graphite) + O2(g)→
2CO(g)
Given standard enthalpies of formation:
C(graphite) = 0 kJ/mol
CO(g) = −110.5 kJ/mol
O2(g) = 0 kJ/mol
Step 2: Calculate ∆Husing the formula:
∆H=XProducts −XReactants
∆H= (2 × −110.5 kJ/mol) −(2 ×0 kJ/mol + 0 kJ/mol)
∆H=−221.0 kJ/mol
Step 3: The enthalpy change for the reaction is −221.0 kJ/mol.
Question 3
Question
A reaction has a standard enthalpy change of -275 kJ/mol according to the
equation:
2A(l)+3B(g)→C(s)+2D(g)
2
If 6.50 g of A and 4.50 g of B are reacted, what is the total amount of heat
released or absorbed by the reaction?
Given: Molar mass of A = 20 g/mol Molar mass of B = 30 g/mol Molar
mass of C = 40 g/mol Molar mass of D = 10 g/mol
Solution
Step 1: Calculate the number of moles of A and B used in the reaction.
Moles of A = Mass of A
Molar mass of A
=6.50g
20 g/mol
= 0.325 mol
Moles of B = Mass of B
Molar mass of B
=4.50g
30 g/mol
= 0.150 mol
Step 2: Determine the limiting reactant. Since we need 2 moles of A for
every 3 moles of B, let’s compare the actual ratio of moles of A to B:
0.325 mol A
2:0.150 mol B
3
Simplifying, we get:
0.325 mol A : 0.150 mol B
13 : 6
Since we need a 2:3 ratio for A:B, A is the limiting reactant.
Step 3: Calculate the heat involved in the reaction. The total heat released
or absorbed is given by the standard enthalpy change.
∆H=−275 kJ/mol
Since the reaction involves 2 moles of A, the total heat will be:
2× −275 kJ = −550 kJ
Therefore, the total amount of heat released by the reaction is 550 kJ.
Question 4
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2H2(g) + O2(g)→2H2O(l)
3
Given the following standard enthalpy of formation values:
∆H◦
f(H2O(l)) = −285.8 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
Solution
Step 1: Write the given chemical equation and standard enthalpy change ex-
pression. The given equation is:
2H2(g) + O2(g)→2H2O(l)
The standard enthalpy change expression for this reaction can be written as:
∆H◦= Σn∆H◦
f(products) −Σm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients of the products and reactants,
respectively.
Step 2: Calculate the standard enthalpy change for the reaction. Substitute
the given standard enthalpy of formation values into the expression:
∆H◦= 2(−285.8) −[2(0) + 1(0)]
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is -571.6 kJ/mol.
Question 5
Question
A reaction has a standard enthalpy change of ∆H◦=−342 kJ/mol. Calculate
the wavelength of light that corresponds to this energy change.
(Note: Planck’s constant h= 6.626 ×10−34 J s, speed of light c= 3.00 ×108
m/s, and 1 J = 1 kg m2/s2)
Solution
Step 1: Convert the given enthalpy change to Joules:
∆H◦=−342 kJ/mol ×1000 J
1 kJ
=−342000 J/mol
Step 2: Use the relationship between energy, Planck’s constant, frequency,
and speed of light:
E=hν =hc
λ
4
where Eis the energy change in Joules, his Planck’s constant, νis the frequency,
cis the speed of light, and λis the wavelength.
Step 3: Solve for the wavelength λ:
∆H◦=hc/λ
λ=hc
∆H◦
=(6.626 ×10−34 J s)(3.00 ×108m/s)
−342000 J/mol
≈5.52 ×10−7m
≈552 nm
Therefore, the wavelength of light that corresponds to this energy change is
approximately 552 nm.
Question 6
Question
Calculate the standard enthalpy change for the following reaction at 298 K given
the standard enthalpies of formation:
2C(graphite)+3H2(g)→CH4(g)
Given:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(C(graphite)) = 0 kJ/mol
∆H◦
f(H2) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the given reaction:
2C(graphite) + 3H2(g)→CH4(g)
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation. The standard enthalpy change for the reaction is given
by:
∆H◦
rxn =Xν∆H◦
f(products) −Xν∆H◦
f(reactants)
where νis the stoichiometric coefficient of each substance.
Substitute the given values:
∆H◦
rxn = [1(−74.8)] −[2(0) + 3(0)]
=−74.8 kJ/mol
Therefore, the standard enthalpy change for the reaction is -74.8 kJ/mol at
298 K.
5
Question 7
Question
Calculate the enthalpy change (∆H) for the following reaction at 25
°
C:
2SO2(g)+O2(g)→2SO3(g)
Given the following bond dissociation energies:
S-O = 498 kJ/mol,O-O = 498 kJ/mol
SO2= 519 kJ/mol,SO3= 642 kJ/mol
Solution
Step 1: Calculate the ∆Hrxn using the bond dissociation energies provided.
∆Hrxn =XBDE(bonds broken) −XBDE(bonds formed)
Step 2: Determine the bonds broken and formed in the reaction.
Bonds broken: 2×[S-O] + 1×[O=O]
Bonds formed: 2×[S=O] + 3×[O-O]
Step 3: Substitute bond dissociation energies and calculate ∆Hrxn.
∆Hrxn = [2 ×(498 kJ/mol) + 1 ×(498 kJ/mol)] −[2 ×(519 kJ/mol) + 3 ×(642 kJ/mol)]
= [996 kJ/mol + 498 kJ/mol] −[1038 kJ/mol + 1926 kJ/mol]
= 1494 kJ/mol −2964 kJ/mol
=−1470 kJ/mol
Therefore, the enthalpy change for the given reaction is ∆H=−1470 kJ/mol.
Question 8
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25
°
C:
2Fe(s)+O2(g)→2FeO(s)
Given:
∆H◦
f(FeO(s)) = −827.4 kJ/mol
∆H◦
f(Fe(s)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
6
Solution
Step 1: Write the balanced chemical equation for the reaction.
2Fe(s)+O2(g)→2FeO(s)
Step 2: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the given values into the equation and calculate.
∆H◦=2×∆H◦
f(FeO(s))−2×∆H◦
f(Fe(s)) + ∆H◦
f(O2(g))
∆H◦= (2 × −827.4) −(2 ×0 + 0)
∆H◦=−1654.8 kJ/mol
Therefore, the standard enthalpy change for the reaction is −1654.8 kJ/mol .
Question 9
Question
Given the following reaction:
2H2O2(l)→2H2O(l)+O2(g) ∆H=−196.0 kJ
If 50.0 g of hydrogen peroxide decomposes in a coffee-cup calorimeter with a
heat capacity of 12.5 J/K, what is the final temperature of the system? (Assume
the specific heat of the resulting water is 4.18 J/g
°
C, the density of water is 1.00
g/mL, and there is no heat lost to the surroundings).
Solution
Step 1: Calculate the moles of H2O2that reacted:
Molar mass of H2O2= 2(1.008) + 2(15.999)
= 34.014 g/mol
Moles of H2O2=50.0 g
34.014 g/mol
≈1.47 mol
Step 2: Calculate the heat released in the reaction:
Heat released = moles of H2O2×∆H
= 1.47 mol × −196.0 kJ/mol
=−287.52 kJ
7
Step 3: Calculate the energy absorbed by the coffee-cup calorimeter:
qcalorimeter =−qreaction
=−(−287.52 kJ)
= 287.52 kJ
= 287520 J
Step 4: Calculate the final temperature of the system:
qcalorimeter =Ccalorimeter ×∆T
∆T=qcalorimeter
Ccalorimeter
∆T=287520 J
12.5 J/K
∆T≈23001.6 K
∆T≈23001.6
°
C
Therefore, the final temperature of the system is approximately 23001.6
°
C.
Question 10
Question
A reaction has a standard enthalpy change of -335 kJ/mol at 298 K. If the
standard enthalpy change of the reaction is -300 kJ/mol at 500 K, calculate the
standard entropy change of the reaction.
Solution
Step 1: Write down the relationship between standard enthalpy change (∆H◦),
standard entropy change (∆S◦), and standard Gibbs free energy change (∆G◦)
using the formula:
∆G◦= ∆H◦−T∆S◦
Step 2: We can rearrange the formula to solve for standard entropy change:
∆S◦=∆H◦−∆G◦
T
Step 3: Calculate the standard Gibbs free energy change ∆G◦at 298 K
using the formula:
∆G◦= ∆H◦−T∆S◦
∆G◦=−335 kJ/mol −(298 K)(8.314 J/mol K)(x)
∆G◦=−335 ∗103J/mol −(298 K)(8.314 J/mol K)(x)
8
Step 4: Now, calculate the standard Gibbs free energy change ∆G◦at 500
K using the formula:
∆G◦= ∆H◦−T∆S◦
∆G◦=−300 kJ/mol −(500 K)(8.314 J/mol K)(x)
∆G◦=−300 ∗103J/mol −(500 K)(8.314 J/mol K)(x)
Step 5: Plug in the calculated ∆G◦values and the corresponding tempera-
tures into the formula for standard entropy change:
x=−335 ∗103+ 298 ∗8.314 ∗x
298
x=−300 ∗103+ 500 ∗8.314 ∗x
500
Solving these two equations will give us the standard entropy change of the
reaction.
Question 11
Question
A reaction has a standard enthalpy change of -335 kJ/mol. If 2.50 mol of the
reactant is consumed in the reaction, how much heat is released or absorbed?
Solution
Step 1: Determine the heat change for 1 mol of the reaction. Given Standard
enthalpy change ∆H=−335 kJ/mol, the heat change for 1 mol of the reaction
is −335 kJ.
Step 2: Calculate the heat change for 2.50 mol of the reaction. To calculate
the heat change for 2.50 mol of the reaction, we use the following equation:
Heat change = Standard enthalpy change ×Number of moles
Substitute the values:
Heat change = −335 kJ/mol ×2.50 mol
Heat change = −335 ×2.50 kJ
Heat change = −837.5 kJ
Therefore, when 2.50 mol of the reactant is consumed in the reaction, the
heat change is -837.5 kJ.
9
Question 12
Question
In a chemical reaction, 5.00 moles of methane gas (CH4) react with excess
oxygen gas to form carbon dioxide gas (CO2) and water vapor. The reaction
releases 793.5 kJ of energy. Calculate the standard enthalpy change (∆H◦) for
the reaction at constant pressure and 298 K.
Solution
Step 1: Write the balanced chemical equation for the reaction. Step 2: Calculate
the moles of water vapor and carbon dioxide produced. Step 3: Calculate the
moles of reactant remaining after the reaction. Step 4: Determine the standard
enthalpy change (∆H◦) for the reaction.
Question 13
Question
Given the reaction, 2CO(g)+O2(g)→2CO2(g), with ∆H=−566 kJ. Calcu-
late the enthalpy change when 4.50 g of CO(g) is burnt in excess oxygen. (molar
mass of CO(g) = 28.0 g/mol)
Solution
Step 1: Calculate the number of moles of CO(g) present. Given mass of CO(g)
= 4.50 g and molar mass of CO(g) = 28.0 g/mol, we can calculate the number
of moles using the formula:
moles = mass
molar mass
moles = 4.50 g
28.0 g/mol = 0.161 mol CO
Step 2: Use the stoichiometry of the reaction to determine the enthalpy
change. From the balanced equation 2CO(g) + O2(g)→2CO2(g), we see that
the molar ratio between CO(g) and CO2(g) is 2:2, or 1:1. This means that for
each mole of CO(g) reacted, the enthalpy change is −566 kJ.
Step 3: Calculate the enthalpy change when 0.161 mol of CO(g) is burnt.
Enthalpy change = moles ×∆H
Enthalpy change = 0.161 mol × −566 kJ/mol = −91.0 kJ
Answer: The enthalpy change when 4.50 g of CO(g) is burnt in excess
oxygen is −91.0kJ.
10
Question 14
Question
Consider the following reaction:
2CH3OH(l)→2CO2(g) + 4H2O(l)
Given that the standard enthalpy of formation (∆H◦
f) for methanol (CH3OH)
is -238.6 kJ/mol, carbon dioxide (CO2) is -393.5 kJ/mol, and water (H2O) is
-285.8 kJ/mol, calculate the standard enthalpy change (∆H◦) for the reaction
above.
Solution
Step 1: Calculate the standard enthalpy change for the reaction using the stan-
dard enthalpy of formation values.
The standard enthalpy change for the reaction can be calculated using the
formula:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Where nand mare the coefficients of the products and reactants respectively.
Plugging in the given values:
∆H◦= 2(2(−393.5 kJ/mol) + 4(−285.8 kJ/mol)) −2(−238.6 kJ/mol)
Step 2: Solve for ∆H◦.
Calculating the above expression gives:
∆H◦= 2(−787.0 kJ/mol) −477.2 kJ/mol = −2061.2 kJ/mol
Therefore, the standard enthalpy change for the reaction is -2061.2 kJ/mol.
Question 15
Question
Calculate the standard enthalpy change (∆H◦) for the reaction below using the
given bond dissociation energies:
H2O2(l)→2HO·(g)
Given bond dissociation energies:
O-H = 463 kJ/mol,O=O = 495 kJ/mol,H-H = 432 kJ/mol
11
Solution
Step 1: Write out the given equation and apply Hess’s Law to find the standard
enthalpy change (∆H◦):
∆H◦=XBDE(bonds broken) −XBDE(bonds formed)
Step 2: Calculate the energy required to break the bonds in the reactants:
BDE(O-O) + BDE(H-H) = 2 ×495 kJ/mol + 432 kJ/mol
Step 3: Calculate the energy released when the new bonds are formed in the
products:
4×BDE(O-H)
Step 4: Substitute the values into the equation to find ∆H◦:
∆H◦= (2 ×495 + 432) −4×463 kJ/mol
Step 5: Perform the calculations to find the answer for ∆H◦:
∆H◦= (990 + 432) −1852 kJ/mol
∆H◦= 1142 −1852 kJ/mol
∆H◦=−710 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−710
kJ/mol.
Question 16
Question
A sample of gaseous ethanol (C2H5OH) is burned in a bomb calorimeter. The
combustion reaction is as follows:
C2H5OH(g)+3O2(g)→2CO2(g)+3H2O(l)
The standard enthalpy of formation for ethanol is -277.7 kJ/mol and for water
is -285.8 kJ/mol. Calculate the standard enthalpy change for this combustion
reaction and determine whether it is exothermic or endothermic.
12
Solution
Step 1: Calculate the standard enthalpy change for the combustion reaction
using the enthalpy of formation values given. The standard enthalpy change for
a reaction can be calculated using the formula:
∆H◦
reaction =Xνi∆H◦
f, products −Xνj∆H◦
f, reactants
where νiand νjare the stoichiometric coefficients of the products and reactants,
respectively.
Plugging in the values from the reaction and the enthalpy of formation val-
ues:
∆H◦
reaction = [2(−393.5) + 3(−285.8)] −[−277.7 + 3(0)]
∆H◦
reaction =−787.4 kJ/mol + 277.7 kJ/mol
∆H◦
reaction =−509.7 kJ/mol
Step 2: Determine whether the combustion reaction is exothermic or en-
dothermic. Since the standard enthalpy change for the reaction is negative
(-509.7 kJ/mol), the combustion reaction is exothermic. This means that heat
is released to the surroundings during the reaction.
Question 17
Question
A reaction has a standard enthalpy change of -125 kJ/mol. If 0.500 mol of the
limiting reactant is consumed, what is the heat transferred at constant pressure?
Solution
Step 1: Determine the heat transferred using the given information.
Given: Standard enthalpy change (∆H◦) = -125 kJ/mol Amount of limiting
reactant consumed = 0.500 mol
The heat transferred can be calculated using the equation:
Heat transferred = Standard enthalpy change×Amount of limiting reactant consumed
Step 2: Substitute the given values into the equation and solve for the heat
transferred.
Heat transferred = −125 kJ/mol ×0.500 mol
Heat transferred = −62.5 kJ
Therefore, the heat transferred at constant pressure when 0.500 mol of the
limiting reactant is consumed is -62.5 kJ.
13
Question 18
Question
Calculate the standard enthalpy change (∆H◦) for the reaction where 2 moles
of ammonia gas (NH3) react with 3 moles of oxygen gas (O2) to form 2 moles
of nitrogen gas (N2) and 3 moles of water vapor (H2O). Given:
4NH3(g)+5O2(g)→4NO(g)+6H2O(l) ∆H◦=−1450 kJ
N2(g) + O2(g)→2NO(g) ∆H◦= 180 kJ
Solution
Step 1: First, we will write the balanced chemical equation for the reaction given.
We need to use the provided thermochemical equations to find the enthalpy
change for the reaction. The given reaction can be split into two steps:
Step 1: 4NH3(g)+5O2(g)→4NO(g)+6H2O(l)
Step 2: N2(g) + O2(g)→2NO(g)
Step 2: Now, we will calculate the enthalpy change for the overall reaction
by adding the enthalpy changes for the individual steps. As the second step
is reversed in the given reaction, we need to change the sign for the enthalpy
change of this step.
∆H◦
overall = ∆H◦
Step 1 −∆H◦
Step 2
Step 3: Substituting the given enthalpy changes:
∆H◦
overall = (−1450 kJ) −(−180 kJ)
∆H◦
overall =−1450 kJ + 180 kJ
∆H◦
overall =−1270 kJ
Therefore, the standard enthalpy change for the reaction where 2 moles of
NH3react with 3 moles of O2to form 2 moles of N2and 3 moles of H2Ois
−1270 kJ .
Question 19
Question
Calculate the standard enthalpy change for the reaction:
2CO(g) + O2(g)→2CO2(g)
given the following standard enthalpies of formation: ∆H◦
f(CO) = −110.5
kJ/mol, ∆H◦
f(CO2) = −393.5 kJ/mol, and ∆H◦
f(O2) = 0 kJ/mol.
14
Solution
Step 1: Write the balanced chemical equation and determine the overall change
in enthalpy. The balanced chemical equation is:
2CO(g) + O2(g)→2CO2(g)
The overall change in enthalpy (∆H◦
rxn) can be calculated using the standard
enthalpies of formation of the reactants and products:
∆H◦
rxn =X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦
rxn = 2∆H◦
f(CO2)−[2∆H◦
f(CO)+∆H◦
f(O2)]
Step 2: Substitute the given values into the equation and calculate.
∆H◦
rxn = 2(−393.5) −[2(−110.5) + 0]
∆H◦
rxn =−787.0 + 221.0
∆H◦
rxn =−566.0 kJ/mol
Therefore, the standard enthalpy change for the reaction is −566.0 kJ/mol .
Question 20
Question
Consider the following reaction:
2C(s) + 3H2(g)→C2H6(g)
The standard enthalpy change for this reaction, ∆H◦, is −284 kJ/mol. Cal-
culate the standard enthalpy of formation, ∆H◦
f, for C2H6(g) given the following
standard enthalpies of formation: ∆H◦
f[H2(g)] = 0 kJ/mol and ∆H◦
f[C(s)] =
0 kJ/mol.
Solution
Step 1: Write the standard enthalpy of formation equation for the given reaction:
∆H◦=Xν∆H◦
f(products) −Xν∆H◦
f(reactants)
Given reaction:
2C(s) + 3H2(g)→C2H6(g)
Step 2: Plug in the known enthalpy of formation values and solve for
∆H◦
f[C2H6(g)]:
−284 kJ/mol = (1)∆H◦
f[C2H6(g)] −(2)(0) −(3)(0)
−284 kJ/mol = ∆H◦
f[C2H6(g)]
Therefore, the standard enthalpy of formation for C2H6(g) is ∆H◦
f[C2H6(g)] =
−284 kJ/mol.
15
Question 21
Question
Given the following reaction:
2NO(g) + O2(g)→2NO2(g)
with ∆H=−113 kJ. Calculate the enthalpy change when 2 moles of NO(g)
react with excess O(g) to form NO(g).
Solution
Step 1: Calculate the enthalpy change for the reaction as given. Given: ∆H=
−113 kJ
Step 2: Determine the moles of NO(g) reacting in the given reaction. Since
the stoichiometric coefficient of NO(g) is 2, 2 moles of NO(g) will react.
Step 3: Use the information from Step 2 and the enthalpy change in Step 1
to calculate the enthalpy change for 2 moles of NO(g) reacting. Since 2 moles
of NO(g) are reacting, the enthalpy change is doubled: 2 × −113 kJ =−226 kJ
Answer
The enthalpy change when 2 moles of NO(g) react with excess O(g) to form
NO(g) is −226 kJ.
Question 22
Question
Calculate the enthalpy change (∆H) for the reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(g)
given the following standard enthalpies of formation:
∆H◦
f(C2H6) = −84.68 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution
Step 1: Calculate the standard enthalpy of formation (∆H◦
f) for the given
reaction using the standard enthalpies of formation provided.
16
The standard enthalpy of formation for the reactants is given as:
∆H◦
f(2C2H6) = 2(−84.68 kJ/mol)
=−169.36 kJ/mol
∆H◦
f(7O2) = 7(0 kJ/mol)
= 0 kJ/mol
The standard enthalpy of formation for the products is given as:
∆H◦
f(4CO2) = 4(−393.5 kJ/mol)
=−1574 kJ/mol
∆H◦
f(6H2O) = 6(−285.8 kJ/mol)
=−1714.8 kJ/mol
Step 2: Calculate the overall ∆Hfor the reaction by summing the standard
enthalpies of formation of the products and subtracting the sum of the standard
enthalpies of formation of the reactants.
∆H=X∆H◦
f(products)−X∆H◦
f(reactants)
∆H= (−1574 kJ/mol −1714.8 kJ/mol) −(−169.36 kJ/mol + 0 kJ/mol)
∆H=−3288.8 kJ/mol + 169.36 kJ/mol
∆H=−3119.44 kJ/mol
Therefore, the enthalpy change (∆H) for the given reaction is -3119.44
kJ/mol.
Question 23
Question
For the reaction 2A(g) + B(g) →3C(g), the standard enthalpy change is -1450
kJ/mol. If 2.50 moles of A react completely with excess B, how much heat is
absorbed or released?
Solution
Step 1: Determine the moles of C produced from the reaction. Given the
stoichiometry of the reaction, for every 2 moles of A reacting, 3 moles of C are
produced. Thus, the number of moles of C produced will be:
2.50 moles A ×3 moles C
2 moles A = 3.75 moles C
17
Step 2: Calculate the heat change for the reaction. The heat change for the
reaction can be calculated using the formula:
Heat change (J) = moles of C ×∆H◦
Substitute the values:
Heat change (J) = 3.75 moles C×(−1450 kJ/mol×103J/kJ) = −5.4375×106J
Step 3: Determine if heat is absorbed or released. Since the heat change
is negative, the reaction is exothermic, and heat is released. Therefore, the
amount of heat released is 5.4375 ×106J.
Question 24
Question
During a chemical reaction, 200.0 g of water is heated from 25
°
C to 100
°
C. If
the specific heat capacity of water is 4.18 J/g
°
C, calculate the amount of heat
absorbed by the water.
Solution
Step 1: Calculate the change in temperature Given: Initial temperature, Tinitial =
25CFinal temperature, Tfinal = 100CChange in temperature, ∆T=Tfinal −
Tinitial ∆T= 100C−25C= 75C
Step 2: Calculate the heat absorbed by the water The heat absorbed by the
water can be calculated using the formula:
q=mc∆T
where: q= heat absorbed by the water m= mass of water = 200.0 g c= specific
heat capacity of water = 4.18 J/g
°
C ∆T= change in temperature = 75
°
C
Substitute the given values into the formula to find the heat absorbed by
the water:
q= (200.0 g)(4.18 J/g
°
C)(75C) = 62700 J
Therefore, the amount of heat absorbed by the water during the heating
process is 62,700 J.
Question 25
Question
A reaction has a standard enthalpy change of -350 kJ/mol. If 2.00 moles of the
reactant are consumed in the reaction, what is the heat transfer that occurs?
18
Solution
Step 1: Determine the moles of heat transferred per mole of reactant. Given:
Standard enthalpy change (∆H◦) = -350 kJ/mol
Step 2: Calculate the heat transfer for 2.00 moles of reactant. Given: Moles
of reactant = 2.00 mol
Now, we can use the following relationship to calculate the heat transfer:
Heat transfer = moles of reactant ×∆H◦
Heat transfer = 2.00 mol × −350 kJ/mol
Step 3: Calculate the heat transfer.
Heat transfer = −700 kJ
Therefore, the heat transfer that occurs when 2.00 moles of the reactant are
consumed in the reaction is -700 kJ.
Question 26
Question
Calculate the enthalpy change for the combustion of 1 mol of butane (C4H10)
at 25
°
C using the following information:
Standard enthalpy of formation of CO2(g): -393.5 kJ/mol
Standard enthalpy of formation of H2O(l): -285.8 kJ/mol
Standard enthalpy of combustion of butane: -2877 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the combustion of butane:
C4H10(g) + 13/2O2(g)→4CO2(g)+5H2O(l)
Step 2: Calculate the enthalpy change for the combustion reaction using the
given standard enthalpies of formation:
∆H=X(products) −X(reactants)
= [4(−393.5) + 5(−285.8)] −[−2877]
=−3935.8+(−1429) + 2877
=−1488.8 kJ/mol
Therefore, the enthalpy change for the combustion of 1 mol of butane is
-1488.8 kJ/mol at 25
°
C.
19
Question 27
Question
Given the following reaction at 25
°
C:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(g)
which has an enthalpy change of -3129 kJ. Calculate the standard enthalpy
of formation of ethane, C2H6, given the standard enthalpies of formation are:
∆H◦
f(CO2) = −393.5 kJ/mol and ∆H◦
f(H2O) = −285.8 kJ/mol.
Solution
Step 1: Calculate the enthalpy change for the formation of one mole of C2H6.
∆H= 4∆H◦
f(CO2) + 6∆H◦
f(H2O)−2∆H◦
f(C2H6)
−3129 = 4(−393.5) + 6(−285.8) −2∆H◦
f(C2H6)
−3129 = −1574 −1714.8−2∆H◦
f(C2H6)
−3129 = −3288.8−2∆H◦
f(C2H6)
2∆H◦
f(C2H6) = 159.8
∆H◦
f(C2H6) = 79.9 kJ/mol
Therefore, the standard enthalpy of formation of ethane, C2H6, is 79.9
kJ/mol.
Question 28
Question
Given the following reaction and enthalpy changes:
2SO2(g) + O2(g)→2SO3(g) ∆H1=−198.2 kJ
2SO3(g)→2SO2(g) + O2(g) ∆H2= 198.2 kJ
Calculate the standard enthalpy change for the reaction:
SO2(g) + 1
2O2(g)→SO3(g)
Solution
Step 1: Write out the given reactions and their enthalpy changes.
1) 2SO2(g) + O2(g)→2SO3(g) ∆H1=−198.2 kJ
2) 2SO3(g)→2SO2(g) + O2(g) ∆H2= 198.2 kJ
20
Step 2: Rearrange the given reactions to obtain the target reaction. Adding
the two given reactions together cancels out 2SO2(g) and 2SO3(g), leaving the
target reaction. Adding the two equations:
2SO2(g) + O2(g)+2SO3(g)→2SO3(g)+2SO2(g) + O2(g)
Step 3: Calculate the standard enthalpy change for the target reaction. Since
the given reactions add up to the target reaction,
∆Htarget = ∆H1+ ∆H2
∆Htarget =−198.2 kJ + 198.2 kJ
∆Htarget = 0 kJ
Therefore, the standard enthalpy change for the reaction SO2(g)+ 1
2O2(g)→
SO3(g) is 0 kJ .
Question 29
Question
Given the following thermochemical equation:
2SO2(g)+O2(g)→2SO3(g)
∆H◦=−196 kJ
Calculate the standard enthalpy change for the reaction:
SO2(g) + 1
2O2(g)→SO3(g)
Solution
Step 1: Write the given thermochemical equation:
2SO2(g)+O2(g)→2SO3(g) ∆H◦=−196 kJ
Step 2: Determine the enthalpy change for the desired reaction using the
given equation:
SO2(g) + 1
2O2(g)→SO3(g)
Step 3: Recall that reversing an equation changes the sign of ∆H◦:
2SO3(g)→2SO2(g)+O2(g) ∆H◦= 196 kJ
Step 4: Divide the above equation by 2 to obtain the desired reaction:
SO3(g)→SO2(g) + 1
2O2(g) ∆H◦= 98 kJ
Therefore, the standard enthalpy change for the reaction:
SO2(g) + 1
2O2(g)→SO3(g)
is 98 kJ .
21
Question 2
Question
Calculate the enthalpy change (∆H) for the reaction:
2C(graphite) + O2(g)→2CO(g)
given the following standard enthalpies of formation:
C(graphite) = 0 kJ/mol
CO(g) = −110.5 kJ/mol
O2(g) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation and list the standard enthalpies
of formation for the reactants and products involved in the reaction. Step 2:
Calculate the standard enthalpy change for the reaction using the formula:
∆H=XProducts −XReactants
Step 3: Substitute the values for the standard enthalpies of formation into
the formula and calculate the enthalpy change for the reaction.
Step 1: The balanced chemical equation is: 2C(graphite) + O2(g)→
2CO(g)
Given standard enthalpies of formation:
C(graphite) = 0 kJ/mol
CO(g) = −110.5 kJ/mol
O2(g) = 0 kJ/mol
Step 2: Calculate ∆Husing the formula:
∆H=XProducts −XReactants
∆H= (2 × −110.5 kJ/mol) −(2 ×0 kJ/mol + 0 kJ/mol)
∆H=−221.0 kJ/mol
Step 3: The enthalpy change for the reaction is −221.0 kJ/mol.
Question 3
Question
A reaction has a standard enthalpy change of -275 kJ/mol according to the
equation:
2A(l)+3B(g)→C(s)+2D(g)
2
If 6.50 g of A and 4.50 g of B are reacted, what is the total amount of heat
released or absorbed by the reaction?
Given: Molar mass of A = 20 g/mol Molar mass of B = 30 g/mol Molar
mass of C = 40 g/mol Molar mass of D = 10 g/mol
Solution
Step 1: Calculate the number of moles of A and B used in the reaction.
Moles of A = Mass of A
Molar mass of A
=6.50g
20 g/mol
= 0.325 mol
Moles of B = Mass of B
Molar mass of B
=4.50g
30 g/mol
= 0.150 mol
Step 2: Determine the limiting reactant. Since we need 2 moles of A for
every 3 moles of B, let’s compare the actual ratio of moles of A to B:
0.325 mol A
2:0.150 mol B
3
Simplifying, we get:
0.325 mol A : 0.150 mol B
13 : 6
Since we need a 2:3 ratio for A:B, A is the limiting reactant.
Step 3: Calculate the heat involved in the reaction. The total heat released
or absorbed is given by the standard enthalpy change.
∆H=−275 kJ/mol
Since the reaction involves 2 moles of A, the total heat will be:
2× −275 kJ = −550 kJ
Therefore, the total amount of heat released by the reaction is 550 kJ.
Question 4
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2H2(g) + O2(g)→2H2O(l)
3
Given the following standard enthalpy of formation values:
∆H◦
f(H2O(l)) = −285.8 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
Solution
Step 1: Write the given chemical equation and standard enthalpy change ex-
pression. The given equation is:
2H2(g) + O2(g)→2H2O(l)
The standard enthalpy change expression for this reaction can be written as:
∆H◦= Σn∆H◦
f(products) −Σm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients of the products and reactants,
respectively.
Step 2: Calculate the standard enthalpy change for the reaction. Substitute
the given standard enthalpy of formation values into the expression:
∆H◦= 2(−285.8) −[2(0) + 1(0)]
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is -571.6 kJ/mol.
Question 5
Question
A reaction has a standard enthalpy change of ∆H◦=−342 kJ/mol. Calculate
the wavelength of light that corresponds to this energy change.
(Note: Planck’s constant h= 6.626 ×10−34 J s, speed of light c= 3.00 ×108
m/s, and 1 J = 1 kg m2/s2)
Solution
Step 1: Convert the given enthalpy change to Joules:
∆H◦=−342 kJ/mol ×1000 J
1 kJ
=−342000 J/mol
Step 2: Use the relationship between energy, Planck’s constant, frequency,
and speed of light:
E=hν =hc
λ
4
where Eis the energy change in Joules, his Planck’s constant, νis the frequency,
cis the speed of light, and λis the wavelength.
Step 3: Solve for the wavelength λ:
∆H◦=hc/λ
λ=hc
∆H◦
=(6.626 ×10−34 J s)(3.00 ×108m/s)
−342000 J/mol
≈5.52 ×10−7m
≈552 nm
Therefore, the wavelength of light that corresponds to this energy change is
approximately 552 nm.
Question 6
Question
Calculate the standard enthalpy change for the following reaction at 298 K given
the standard enthalpies of formation:
2C(graphite)+3H2(g)→CH4(g)
Given:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(C(graphite)) = 0 kJ/mol
∆H◦
f(H2) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the given reaction:
2C(graphite) + 3H2(g)→CH4(g)
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation. The standard enthalpy change for the reaction is given
by:
∆H◦
rxn =Xν∆H◦
f(products) −Xν∆H◦
f(reactants)
where νis the stoichiometric coefficient of each substance.
Substitute the given values:
∆H◦
rxn = [1(−74.8)] −[2(0) + 3(0)]
=−74.8 kJ/mol
Therefore, the standard enthalpy change for the reaction is -74.8 kJ/mol at
298 K.
5
Question 7
Question
Calculate the enthalpy change (∆H) for the following reaction at 25
°
C:
2SO2(g)+O2(g)→2SO3(g)
Given the following bond dissociation energies:
S-O = 498 kJ/mol,O-O = 498 kJ/mol
SO2= 519 kJ/mol,SO3= 642 kJ/mol
Solution
Step 1: Calculate the ∆Hrxn using the bond dissociation energies provided.
∆Hrxn =XBDE(bonds broken) −XBDE(bonds formed)
Step 2: Determine the bonds broken and formed in the reaction.
Bonds broken: 2×[S-O] + 1×[O=O]
Bonds formed: 2×[S=O] + 3×[O-O]
Step 3: Substitute bond dissociation energies and calculate ∆Hrxn.
∆Hrxn = [2 ×(498 kJ/mol) + 1 ×(498 kJ/mol)] −[2 ×(519 kJ/mol) + 3 ×(642 kJ/mol)]
= [996 kJ/mol + 498 kJ/mol] −[1038 kJ/mol + 1926 kJ/mol]
= 1494 kJ/mol −2964 kJ/mol
=−1470 kJ/mol
Therefore, the enthalpy change for the given reaction is ∆H=−1470 kJ/mol.
Question 8
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25
°
C:
2Fe(s)+O2(g)→2FeO(s)
Given:
∆H◦
f(FeO(s)) = −827.4 kJ/mol
∆H◦
f(Fe(s)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
6
Solution
Step 1: Write the balanced chemical equation for the reaction.
2Fe(s)+O2(g)→2FeO(s)
Step 2: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the given values into the equation and calculate.
∆H◦=2×∆H◦
f(FeO(s))−2×∆H◦
f(Fe(s)) + ∆H◦
f(O2(g))
∆H◦= (2 × −827.4) −(2 ×0 + 0)
∆H◦=−1654.8 kJ/mol
Therefore, the standard enthalpy change for the reaction is −1654.8 kJ/mol .
Question 9
Question
Given the following reaction:
2H2O2(l)→2H2O(l)+O2(g) ∆H=−196.0 kJ
If 50.0 g of hydrogen peroxide decomposes in a coffee-cup calorimeter with a
heat capacity of 12.5 J/K, what is the final temperature of the system? (Assume
the specific heat of the resulting water is 4.18 J/g
°
C, the density of water is 1.00
g/mL, and there is no heat lost to the surroundings).
Solution
Step 1: Calculate the moles of H2O2that reacted:
Molar mass of H2O2= 2(1.008) + 2(15.999)
= 34.014 g/mol
Moles of H2O2=50.0 g
34.014 g/mol
≈1.47 mol
Step 2: Calculate the heat released in the reaction:
Heat released = moles of H2O2×∆H
= 1.47 mol × −196.0 kJ/mol
=−287.52 kJ
7
Step 3: Calculate the energy absorbed by the coffee-cup calorimeter:
qcalorimeter =−qreaction
=−(−287.52 kJ)
= 287.52 kJ
= 287520 J
Step 4: Calculate the final temperature of the system:
qcalorimeter =Ccalorimeter ×∆T
∆T=qcalorimeter
Ccalorimeter
∆T=287520 J
12.5 J/K
∆T≈23001.6 K
∆T≈23001.6
°
C
Therefore, the final temperature of the system is approximately 23001.6
°
C.
Question 10
Question
A reaction has a standard enthalpy change of -335 kJ/mol at 298 K. If the
standard enthalpy change of the reaction is -300 kJ/mol at 500 K, calculate the
standard entropy change of the reaction.
Solution
Step 1: Write down the relationship between standard enthalpy change (∆H◦),
standard entropy change (∆S◦), and standard Gibbs free energy change (∆G◦)
using the formula:
∆G◦= ∆H◦−T∆S◦
Step 2: We can rearrange the formula to solve for standard entropy change:
∆S◦=∆H◦−∆G◦
T
Step 3: Calculate the standard Gibbs free energy change ∆G◦at 298 K
using the formula:
∆G◦= ∆H◦−T∆S◦
∆G◦=−335 kJ/mol −(298 K)(8.314 J/mol K)(x)
∆G◦=−335 ∗103J/mol −(298 K)(8.314 J/mol K)(x)
8
Step 4: Now, calculate the standard Gibbs free energy change ∆G◦at 500
K using the formula:
∆G◦= ∆H◦−T∆S◦
∆G◦=−300 kJ/mol −(500 K)(8.314 J/mol K)(x)
∆G◦=−300 ∗103J/mol −(500 K)(8.314 J/mol K)(x)
Step 5: Plug in the calculated ∆G◦values and the corresponding tempera-
tures into the formula for standard entropy change:
x=−335 ∗103+ 298 ∗8.314 ∗x
298
x=−300 ∗103+ 500 ∗8.314 ∗x
500
Solving these two equations will give us the standard entropy change of the
reaction.
Question 11
Question
A reaction has a standard enthalpy change of -335 kJ/mol. If 2.50 mol of the
reactant is consumed in the reaction, how much heat is released or absorbed?
Solution
Step 1: Determine the heat change for 1 mol of the reaction. Given Standard
enthalpy change ∆H=−335 kJ/mol, the heat change for 1 mol of the reaction
is −335 kJ.
Step 2: Calculate the heat change for 2.50 mol of the reaction. To calculate
the heat change for 2.50 mol of the reaction, we use the following equation:
Heat change = Standard enthalpy change ×Number of moles
Substitute the values:
Heat change = −335 kJ/mol ×2.50 mol
Heat change = −335 ×2.50 kJ
Heat change = −837.5 kJ
Therefore, when 2.50 mol of the reactant is consumed in the reaction, the
heat change is -837.5 kJ.
9
Question 12
Question
In a chemical reaction, 5.00 moles of methane gas (CH4) react with excess
oxygen gas to form carbon dioxide gas (CO2) and water vapor. The reaction
releases 793.5 kJ of energy. Calculate the standard enthalpy change (∆H◦) for
the reaction at constant pressure and 298 K.
Solution
Step 1: Write the balanced chemical equation for the reaction. Step 2: Calculate
the moles of water vapor and carbon dioxide produced. Step 3: Calculate the
moles of reactant remaining after the reaction. Step 4: Determine the standard
enthalpy change (∆H◦) for the reaction.
Question 13
Question
Given the reaction, 2CO(g)+O2(g)→2CO2(g), with ∆H=−566 kJ. Calcu-
late the enthalpy change when 4.50 g of CO(g) is burnt in excess oxygen. (molar
mass of CO(g) = 28.0 g/mol)
Solution
Step 1: Calculate the number of moles of CO(g) present. Given mass of CO(g)
= 4.50 g and molar mass of CO(g) = 28.0 g/mol, we can calculate the number
of moles using the formula:
moles = mass
molar mass
moles = 4.50 g
28.0 g/mol = 0.161 mol CO
Step 2: Use the stoichiometry of the reaction to determine the enthalpy
change. From the balanced equation 2CO(g) + O2(g)→2CO2(g), we see that
the molar ratio between CO(g) and CO2(g) is 2:2, or 1:1. This means that for
each mole of CO(g) reacted, the enthalpy change is −566 kJ.
Step 3: Calculate the enthalpy change when 0.161 mol of CO(g) is burnt.
Enthalpy change = moles ×∆H
Enthalpy change = 0.161 mol × −566 kJ/mol = −91.0 kJ
Answer: The enthalpy change when 4.50 g of CO(g) is burnt in excess
oxygen is −91.0kJ.
10
Question 14
Question
Consider the following reaction:
2CH3OH(l)→2CO2(g) + 4H2O(l)
Given that the standard enthalpy of formation (∆H◦
f) for methanol (CH3OH)
is -238.6 kJ/mol, carbon dioxide (CO2) is -393.5 kJ/mol, and water (H2O) is
-285.8 kJ/mol, calculate the standard enthalpy change (∆H◦) for the reaction
above.
Solution
Step 1: Calculate the standard enthalpy change for the reaction using the stan-
dard enthalpy of formation values.
The standard enthalpy change for the reaction can be calculated using the
formula:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Where nand mare the coefficients of the products and reactants respectively.
Plugging in the given values:
∆H◦= 2(2(−393.5 kJ/mol) + 4(−285.8 kJ/mol)) −2(−238.6 kJ/mol)
Step 2: Solve for ∆H◦.
Calculating the above expression gives:
∆H◦= 2(−787.0 kJ/mol) −477.2 kJ/mol = −2061.2 kJ/mol
Therefore, the standard enthalpy change for the reaction is -2061.2 kJ/mol.
Question 15
Question
Calculate the standard enthalpy change (∆H◦) for the reaction below using the
given bond dissociation energies:
H2O2(l)→2HO·(g)
Given bond dissociation energies:
O-H = 463 kJ/mol,O=O = 495 kJ/mol,H-H = 432 kJ/mol
11
Solution
Step 1: Write out the given equation and apply Hess’s Law to find the standard
enthalpy change (∆H◦):
∆H◦=XBDE(bonds broken) −XBDE(bonds formed)
Step 2: Calculate the energy required to break the bonds in the reactants:
BDE(O-O) + BDE(H-H) = 2 ×495 kJ/mol + 432 kJ/mol
Step 3: Calculate the energy released when the new bonds are formed in the
products:
4×BDE(O-H)
Step 4: Substitute the values into the equation to find ∆H◦:
∆H◦= (2 ×495 + 432) −4×463 kJ/mol
Step 5: Perform the calculations to find the answer for ∆H◦:
∆H◦= (990 + 432) −1852 kJ/mol
∆H◦= 1142 −1852 kJ/mol
∆H◦=−710 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−710
kJ/mol.
Question 16
Question
A sample of gaseous ethanol (C2H5OH) is burned in a bomb calorimeter. The
combustion reaction is as follows:
C2H5OH(g)+3O2(g)→2CO2(g)+3H2O(l)
The standard enthalpy of formation for ethanol is -277.7 kJ/mol and for water
is -285.8 kJ/mol. Calculate the standard enthalpy change for this combustion
reaction and determine whether it is exothermic or endothermic.
12
Solution
Step 1: Calculate the standard enthalpy change for the combustion reaction
using the enthalpy of formation values given. The standard enthalpy change for
a reaction can be calculated using the formula:
∆H◦
reaction =Xνi∆H◦
f, products −Xνj∆H◦
f, reactants
where νiand νjare the stoichiometric coefficients of the products and reactants,
respectively.
Plugging in the values from the reaction and the enthalpy of formation val-
ues:
∆H◦
reaction = [2(−393.5) + 3(−285.8)] −[−277.7 + 3(0)]
∆H◦
reaction =−787.4 kJ/mol + 277.7 kJ/mol
∆H◦
reaction =−509.7 kJ/mol
Step 2: Determine whether the combustion reaction is exothermic or en-
dothermic. Since the standard enthalpy change for the reaction is negative
(-509.7 kJ/mol), the combustion reaction is exothermic. This means that heat
is released to the surroundings during the reaction.
Question 17
Question
A reaction has a standard enthalpy change of -125 kJ/mol. If 0.500 mol of the
limiting reactant is consumed, what is the heat transferred at constant pressure?
Solution
Step 1: Determine the heat transferred using the given information.
Given: Standard enthalpy change (∆H◦) = -125 kJ/mol Amount of limiting
reactant consumed = 0.500 mol
The heat transferred can be calculated using the equation:
Heat transferred = Standard enthalpy change×Amount of limiting reactant consumed
Step 2: Substitute the given values into the equation and solve for the heat
transferred.
Heat transferred = −125 kJ/mol ×0.500 mol
Heat transferred = −62.5 kJ
Therefore, the heat transferred at constant pressure when 0.500 mol of the
limiting reactant is consumed is -62.5 kJ.
13
Question 18
Question
Calculate the standard enthalpy change (∆H◦) for the reaction where 2 moles
of ammonia gas (NH3) react with 3 moles of oxygen gas (O2) to form 2 moles
of nitrogen gas (N2) and 3 moles of water vapor (H2O). Given:
4NH3(g)+5O2(g)→4NO(g)+6H2O(l) ∆H◦=−1450 kJ
N2(g) + O2(g)→2NO(g) ∆H◦= 180 kJ
Solution
Step 1: First, we will write the balanced chemical equation for the reaction given.
We need to use the provided thermochemical equations to find the enthalpy
change for the reaction. The given reaction can be split into two steps:
Step 1: 4NH3(g)+5O2(g)→4NO(g)+6H2O(l)
Step 2: N2(g) + O2(g)→2NO(g)
Step 2: Now, we will calculate the enthalpy change for the overall reaction
by adding the enthalpy changes for the individual steps. As the second step
is reversed in the given reaction, we need to change the sign for the enthalpy
change of this step.
∆H◦
overall = ∆H◦
Step 1 −∆H◦
Step 2
Step 3: Substituting the given enthalpy changes:
∆H◦
overall = (−1450 kJ) −(−180 kJ)
∆H◦
overall =−1450 kJ + 180 kJ
∆H◦
overall =−1270 kJ
Therefore, the standard enthalpy change for the reaction where 2 moles of
NH3react with 3 moles of O2to form 2 moles of N2and 3 moles of H2Ois
−1270 kJ .
Question 19
Question
Calculate the standard enthalpy change for the reaction:
2CO(g) + O2(g)→2CO2(g)
given the following standard enthalpies of formation: ∆H◦
f(CO) = −110.5
kJ/mol, ∆H◦
f(CO2) = −393.5 kJ/mol, and ∆H◦
f(O2) = 0 kJ/mol.
14
Solution
Step 1: Write the balanced chemical equation and determine the overall change
in enthalpy. The balanced chemical equation is:
2CO(g) + O2(g)→2CO2(g)
The overall change in enthalpy (∆H◦
rxn) can be calculated using the standard
enthalpies of formation of the reactants and products:
∆H◦
rxn =X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦
rxn = 2∆H◦
f(CO2)−[2∆H◦
f(CO)+∆H◦
f(O2)]
Step 2: Substitute the given values into the equation and calculate.
∆H◦
rxn = 2(−393.5) −[2(−110.5) + 0]
∆H◦
rxn =−787.0 + 221.0
∆H◦
rxn =−566.0 kJ/mol
Therefore, the standard enthalpy change for the reaction is −566.0 kJ/mol .
Question 20
Question
Consider the following reaction:
2C(s) + 3H2(g)→C2H6(g)
The standard enthalpy change for this reaction, ∆H◦, is −284 kJ/mol. Cal-
culate the standard enthalpy of formation, ∆H◦
f, for C2H6(g) given the following
standard enthalpies of formation: ∆H◦
f[H2(g)] = 0 kJ/mol and ∆H◦
f[C(s)] =
0 kJ/mol.
Solution
Step 1: Write the standard enthalpy of formation equation for the given reaction:
∆H◦=Xν∆H◦
f(products) −Xν∆H◦
f(reactants)
Given reaction:
2C(s) + 3H2(g)→C2H6(g)
Step 2: Plug in the known enthalpy of formation values and solve for
∆H◦
f[C2H6(g)]:
−284 kJ/mol = (1)∆H◦
f[C2H6(g)] −(2)(0) −(3)(0)
−284 kJ/mol = ∆H◦
f[C2H6(g)]
Therefore, the standard enthalpy of formation for C2H6(g) is ∆H◦
f[C2H6(g)] =
−284 kJ/mol.
15
Question 21
Question
Given the following reaction:
2NO(g) + O2(g)→2NO2(g)
with ∆H=−113 kJ. Calculate the enthalpy change when 2 moles of NO(g)
react with excess O(g) to form NO(g).
Solution
Step 1: Calculate the enthalpy change for the reaction as given. Given: ∆H=
−113 kJ
Step 2: Determine the moles of NO(g) reacting in the given reaction. Since
the stoichiometric coefficient of NO(g) is 2, 2 moles of NO(g) will react.
Step 3: Use the information from Step 2 and the enthalpy change in Step 1
to calculate the enthalpy change for 2 moles of NO(g) reacting. Since 2 moles
of NO(g) are reacting, the enthalpy change is doubled: 2 × −113 kJ =−226 kJ
Answer
The enthalpy change when 2 moles of NO(g) react with excess O(g) to form
NO(g) is −226 kJ.
Question 22
Question
Calculate the enthalpy change (∆H) for the reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(g)
given the following standard enthalpies of formation:
∆H◦
f(C2H6) = −84.68 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution
Step 1: Calculate the standard enthalpy of formation (∆H◦
f) for the given
reaction using the standard enthalpies of formation provided.
16
The standard enthalpy of formation for the reactants is given as:
∆H◦
f(2C2H6) = 2(−84.68 kJ/mol)
=−169.36 kJ/mol
∆H◦
f(7O2) = 7(0 kJ/mol)
= 0 kJ/mol
The standard enthalpy of formation for the products is given as:
∆H◦
f(4CO2) = 4(−393.5 kJ/mol)
=−1574 kJ/mol
∆H◦
f(6H2O) = 6(−285.8 kJ/mol)
=−1714.8 kJ/mol
Step 2: Calculate the overall ∆Hfor the reaction by summing the standard
enthalpies of formation of the products and subtracting the sum of the standard
enthalpies of formation of the reactants.
∆H=X∆H◦
f(products)−X∆H◦
f(reactants)
∆H= (−1574 kJ/mol −1714.8 kJ/mol) −(−169.36 kJ/mol + 0 kJ/mol)
∆H=−3288.8 kJ/mol + 169.36 kJ/mol
∆H=−3119.44 kJ/mol
Therefore, the enthalpy change (∆H) for the given reaction is -3119.44
kJ/mol.
Question 23
Question
For the reaction 2A(g) + B(g) →3C(g), the standard enthalpy change is -1450
kJ/mol. If 2.50 moles of A react completely with excess B, how much heat is
absorbed or released?
Solution
Step 1: Determine the moles of C produced from the reaction. Given the
stoichiometry of the reaction, for every 2 moles of A reacting, 3 moles of C are
produced. Thus, the number of moles of C produced will be:
2.50 moles A ×3 moles C
2 moles A = 3.75 moles C
17
Step 2: Calculate the heat change for the reaction. The heat change for the
reaction can be calculated using the formula:
Heat change (J) = moles of C ×∆H◦
Substitute the values:
Heat change (J) = 3.75 moles C×(−1450 kJ/mol×103J/kJ) = −5.4375×106J
Step 3: Determine if heat is absorbed or released. Since the heat change
is negative, the reaction is exothermic, and heat is released. Therefore, the
amount of heat released is 5.4375 ×106J.
Question 24
Question
During a chemical reaction, 200.0 g of water is heated from 25
°
C to 100
°
C. If
the specific heat capacity of water is 4.18 J/g
°
C, calculate the amount of heat
absorbed by the water.
Solution
Step 1: Calculate the change in temperature Given: Initial temperature, Tinitial =
25CFinal temperature, Tfinal = 100CChange in temperature, ∆T=Tfinal −
Tinitial ∆T= 100C−25C= 75C
Step 2: Calculate the heat absorbed by the water The heat absorbed by the
water can be calculated using the formula:
q=mc∆T
where: q= heat absorbed by the water m= mass of water = 200.0 g c= specific
heat capacity of water = 4.18 J/g
°
C ∆T= change in temperature = 75
°
C
Substitute the given values into the formula to find the heat absorbed by
the water:
q= (200.0 g)(4.18 J/g
°
C)(75C) = 62700 J
Therefore, the amount of heat absorbed by the water during the heating
process is 62,700 J.
Question 25
Question
A reaction has a standard enthalpy change of -350 kJ/mol. If 2.00 moles of the
reactant are consumed in the reaction, what is the heat transfer that occurs?
18
Solution
Step 1: Determine the moles of heat transferred per mole of reactant. Given:
Standard enthalpy change (∆H◦) = -350 kJ/mol
Step 2: Calculate the heat transfer for 2.00 moles of reactant. Given: Moles
of reactant = 2.00 mol
Now, we can use the following relationship to calculate the heat transfer:
Heat transfer = moles of reactant ×∆H◦
Heat transfer = 2.00 mol × −350 kJ/mol
Step 3: Calculate the heat transfer.
Heat transfer = −700 kJ
Therefore, the heat transfer that occurs when 2.00 moles of the reactant are
consumed in the reaction is -700 kJ.
Question 26
Question
Calculate the enthalpy change for the combustion of 1 mol of butane (C4H10)
at 25
°
C using the following information:
Standard enthalpy of formation of CO2(g): -393.5 kJ/mol
Standard enthalpy of formation of H2O(l): -285.8 kJ/mol
Standard enthalpy of combustion of butane: -2877 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the combustion of butane:
C4H10(g) + 13/2O2(g)→4CO2(g)+5H2O(l)
Step 2: Calculate the enthalpy change for the combustion reaction using the
given standard enthalpies of formation:
∆H=X(products) −X(reactants)
= [4(−393.5) + 5(−285.8)] −[−2877]
=−3935.8+(−1429) + 2877
=−1488.8 kJ/mol
Therefore, the enthalpy change for the combustion of 1 mol of butane is
-1488.8 kJ/mol at 25
°
C.
19
Question 27
Question
Given the following reaction at 25
°
C:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(g)
which has an enthalpy change of -3129 kJ. Calculate the standard enthalpy
of formation of ethane, C2H6, given the standard enthalpies of formation are:
∆H◦
f(CO2) = −393.5 kJ/mol and ∆H◦
f(H2O) = −285.8 kJ/mol.
Solution
Step 1: Calculate the enthalpy change for the formation of one mole of C2H6.
∆H= 4∆H◦
f(CO2) + 6∆H◦
f(H2O)−2∆H◦
f(C2H6)
−3129 = 4(−393.5) + 6(−285.8) −2∆H◦
f(C2H6)
−3129 = −1574 −1714.8−2∆H◦
f(C2H6)
−3129 = −3288.8−2∆H◦
f(C2H6)
2∆H◦
f(C2H6) = 159.8
∆H◦
f(C2H6) = 79.9 kJ/mol
Therefore, the standard enthalpy of formation of ethane, C2H6, is 79.9
kJ/mol.
Question 28
Question
Given the following reaction and enthalpy changes:
2SO2(g) + O2(g)→2SO3(g) ∆H1=−198.2 kJ
2SO3(g)→2SO2(g) + O2(g) ∆H2= 198.2 kJ
Calculate the standard enthalpy change for the reaction:
SO2(g) + 1
2O2(g)→SO3(g)
Solution
Step 1: Write out the given reactions and their enthalpy changes.
1) 2SO2(g) + O2(g)→2SO3(g) ∆H1=−198.2 kJ
2) 2SO3(g)→2SO2(g) + O2(g) ∆H2= 198.2 kJ
20
Step 2: Rearrange the given reactions to obtain the target reaction. Adding
the two given reactions together cancels out 2SO2(g) and 2SO3(g), leaving the
target reaction. Adding the two equations:
2SO2(g) + O2(g)+2SO3(g)→2SO3(g)+2SO2(g) + O2(g)
Step 3: Calculate the standard enthalpy change for the target reaction. Since
the given reactions add up to the target reaction,
∆Htarget = ∆H1+ ∆H2
∆Htarget =−198.2 kJ + 198.2 kJ
∆Htarget = 0 kJ
Therefore, the standard enthalpy change for the reaction SO2(g)+ 1
2O2(g)→
SO3(g) is 0 kJ .
Question 29
Question
Given the following thermochemical equation:
2SO2(g)+O2(g)→2SO3(g)
∆H◦=−196 kJ
Calculate the standard enthalpy change for the reaction:
SO2(g) + 1
2O2(g)→SO3(g)
Solution
Step 1: Write the given thermochemical equation:
2SO2(g)+O2(g)→2SO3(g) ∆H◦=−196 kJ
Step 2: Determine the enthalpy change for the desired reaction using the
given equation:
SO2(g) + 1
2O2(g)→SO3(g)
Step 3: Recall that reversing an equation changes the sign of ∆H◦:
2SO3(g)→2SO2(g)+O2(g) ∆H◦= 196 kJ
Step 4: Divide the above equation by 2 to obtain the desired reaction:
SO3(g)→SO2(g) + 1
2O2(g) ∆H◦= 98 kJ
Therefore, the standard enthalpy change for the reaction:
SO2(g) + 1
2O2(g)→SO3(g)
is 98 kJ .
21
Question 2
Question
Calculate the enthalpy change (∆H) for the reaction:
2C(graphite) + O2(g)→2CO(g)
given the following standard enthalpies of formation:
C(graphite) = 0 kJ/mol
CO(g) = −110.5 kJ/mol
O2(g) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation and list the standard enthalpies
of formation for the reactants and products involved in the reaction. Step 2:
Calculate the standard enthalpy change for the reaction using the formula:
∆H=XProducts −XReactants
Step 3: Substitute the values for the standard enthalpies of formation into
the formula and calculate the enthalpy change for the reaction.
Step 1: The balanced chemical equation is: 2C(graphite) + O2(g)→
2CO(g)
Given standard enthalpies of formation:
C(graphite) = 0 kJ/mol
CO(g) = −110.5 kJ/mol
O2(g) = 0 kJ/mol
Step 2: Calculate ∆Husing the formula:
∆H=XProducts −XReactants
∆H= (2 × −110.5 kJ/mol) −(2 ×0 kJ/mol + 0 kJ/mol)
∆H=−221.0 kJ/mol
Step 3: The enthalpy change for the reaction is −221.0 kJ/mol.
Question 3
Question
A reaction has a standard enthalpy change of -275 kJ/mol according to the
equation:
2A(l)+3B(g)→C(s)+2D(g)
2
If 6.50 g of A and 4.50 g of B are reacted, what is the total amount of heat
released or absorbed by the reaction?
Given: Molar mass of A = 20 g/mol Molar mass of B = 30 g/mol Molar
mass of C = 40 g/mol Molar mass of D = 10 g/mol
Solution
Step 1: Calculate the number of moles of A and B used in the reaction.
Moles of A = Mass of A
Molar mass of A
=6.50g
20 g/mol
= 0.325 mol
Moles of B = Mass of B
Molar mass of B
=4.50g
30 g/mol
= 0.150 mol
Step 2: Determine the limiting reactant. Since we need 2 moles of A for
every 3 moles of B, let’s compare the actual ratio of moles of A to B:
0.325 mol A
2:0.150 mol B
3
Simplifying, we get:
0.325 mol A : 0.150 mol B
13 : 6
Since we need a 2:3 ratio for A:B, A is the limiting reactant.
Step 3: Calculate the heat involved in the reaction. The total heat released
or absorbed is given by the standard enthalpy change.
∆H=−275 kJ/mol
Since the reaction involves 2 moles of A, the total heat will be:
2× −275 kJ = −550 kJ
Therefore, the total amount of heat released by the reaction is 550 kJ.
Question 4
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2H2(g) + O2(g)→2H2O(l)
3
Given the following standard enthalpy of formation values:
∆H◦
f(H2O(l)) = −285.8 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
Solution
Step 1: Write the given chemical equation and standard enthalpy change ex-
pression. The given equation is:
2H2(g) + O2(g)→2H2O(l)
The standard enthalpy change expression for this reaction can be written as:
∆H◦= Σn∆H◦
f(products) −Σm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients of the products and reactants,
respectively.
Step 2: Calculate the standard enthalpy change for the reaction. Substitute
the given standard enthalpy of formation values into the expression:
∆H◦= 2(−285.8) −[2(0) + 1(0)]
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is -571.6 kJ/mol.
Question 5
Question
A reaction has a standard enthalpy change of ∆H◦=−342 kJ/mol. Calculate
the wavelength of light that corresponds to this energy change.
(Note: Planck’s constant h= 6.626 ×10−34 J s, speed of light c= 3.00 ×108
m/s, and 1 J = 1 kg m2/s2)
Solution
Step 1: Convert the given enthalpy change to Joules:
∆H◦=−342 kJ/mol ×1000 J
1 kJ
=−342000 J/mol
Step 2: Use the relationship between energy, Planck’s constant, frequency,
and speed of light:
E=hν =hc
λ
4
where Eis the energy change in Joules, his Planck’s constant, νis the frequency,
cis the speed of light, and λis the wavelength.
Step 3: Solve for the wavelength λ:
∆H◦=hc/λ
λ=hc
∆H◦
=(6.626 ×10−34 J s)(3.00 ×108m/s)
−342000 J/mol
≈5.52 ×10−7m
≈552 nm
Therefore, the wavelength of light that corresponds to this energy change is
approximately 552 nm.
Question 6
Question
Calculate the standard enthalpy change for the following reaction at 298 K given
the standard enthalpies of formation:
2C(graphite)+3H2(g)→CH4(g)
Given:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(C(graphite)) = 0 kJ/mol
∆H◦
f(H2) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the given reaction:
2C(graphite) + 3H2(g)→CH4(g)
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation. The standard enthalpy change for the reaction is given
by:
∆H◦
rxn =Xν∆H◦
f(products) −Xν∆H◦
f(reactants)
where νis the stoichiometric coefficient of each substance.
Substitute the given values:
∆H◦
rxn = [1(−74.8)] −[2(0) + 3(0)]
=−74.8 kJ/mol
Therefore, the standard enthalpy change for the reaction is -74.8 kJ/mol at
298 K.
5
Question 7
Question
Calculate the enthalpy change (∆H) for the following reaction at 25
°
C:
2SO2(g)+O2(g)→2SO3(g)
Given the following bond dissociation energies:
S-O = 498 kJ/mol,O-O = 498 kJ/mol
SO2= 519 kJ/mol,SO3= 642 kJ/mol
Solution
Step 1: Calculate the ∆Hrxn using the bond dissociation energies provided.
∆Hrxn =XBDE(bonds broken) −XBDE(bonds formed)
Step 2: Determine the bonds broken and formed in the reaction.
Bonds broken: 2×[S-O] + 1×[O=O]
Bonds formed: 2×[S=O] + 3×[O-O]
Step 3: Substitute bond dissociation energies and calculate ∆Hrxn.
∆Hrxn = [2 ×(498 kJ/mol) + 1 ×(498 kJ/mol)] −[2 ×(519 kJ/mol) + 3 ×(642 kJ/mol)]
= [996 kJ/mol + 498 kJ/mol] −[1038 kJ/mol + 1926 kJ/mol]
= 1494 kJ/mol −2964 kJ/mol
=−1470 kJ/mol
Therefore, the enthalpy change for the given reaction is ∆H=−1470 kJ/mol.
Question 8
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25
°
C:
2Fe(s)+O2(g)→2FeO(s)
Given:
∆H◦
f(FeO(s)) = −827.4 kJ/mol
∆H◦
f(Fe(s)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
6
Solution
Step 1: Write the balanced chemical equation for the reaction.
2Fe(s)+O2(g)→2FeO(s)
Step 2: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the given values into the equation and calculate.
∆H◦=2×∆H◦
f(FeO(s))−2×∆H◦
f(Fe(s)) + ∆H◦
f(O2(g))
∆H◦= (2 × −827.4) −(2 ×0 + 0)
∆H◦=−1654.8 kJ/mol
Therefore, the standard enthalpy change for the reaction is −1654.8 kJ/mol .
Question 9
Question
Given the following reaction:
2H2O2(l)→2H2O(l)+O2(g) ∆H=−196.0 kJ
If 50.0 g of hydrogen peroxide decomposes in a coffee-cup calorimeter with a
heat capacity of 12.5 J/K, what is the final temperature of the system? (Assume
the specific heat of the resulting water is 4.18 J/g
°
C, the density of water is 1.00
g/mL, and there is no heat lost to the surroundings).
Solution
Step 1: Calculate the moles of H2O2that reacted:
Molar mass of H2O2= 2(1.008) + 2(15.999)
= 34.014 g/mol
Moles of H2O2=50.0 g
34.014 g/mol
≈1.47 mol
Step 2: Calculate the heat released in the reaction:
Heat released = moles of H2O2×∆H
= 1.47 mol × −196.0 kJ/mol
=−287.52 kJ
7
Step 3: Calculate the energy absorbed by the coffee-cup calorimeter:
qcalorimeter =−qreaction
=−(−287.52 kJ)
= 287.52 kJ
= 287520 J
Step 4: Calculate the final temperature of the system:
qcalorimeter =Ccalorimeter ×∆T
∆T=qcalorimeter
Ccalorimeter
∆T=287520 J
12.5 J/K
∆T≈23001.6 K
∆T≈23001.6
°
C
Therefore, the final temperature of the system is approximately 23001.6
°
C.
Question 10
Question
A reaction has a standard enthalpy change of -335 kJ/mol at 298 K. If the
standard enthalpy change of the reaction is -300 kJ/mol at 500 K, calculate the
standard entropy change of the reaction.
Solution
Step 1: Write down the relationship between standard enthalpy change (∆H◦),
standard entropy change (∆S◦), and standard Gibbs free energy change (∆G◦)
using the formula:
∆G◦= ∆H◦−T∆S◦
Step 2: We can rearrange the formula to solve for standard entropy change:
∆S◦=∆H◦−∆G◦
T
Step 3: Calculate the standard Gibbs free energy change ∆G◦at 298 K
using the formula:
∆G◦= ∆H◦−T∆S◦
∆G◦=−335 kJ/mol −(298 K)(8.314 J/mol K)(x)
∆G◦=−335 ∗103J/mol −(298 K)(8.314 J/mol K)(x)
8
Step 4: Now, calculate the standard Gibbs free energy change ∆G◦at 500
K using the formula:
∆G◦= ∆H◦−T∆S◦
∆G◦=−300 kJ/mol −(500 K)(8.314 J/mol K)(x)
∆G◦=−300 ∗103J/mol −(500 K)(8.314 J/mol K)(x)
Step 5: Plug in the calculated ∆G◦values and the corresponding tempera-
tures into the formula for standard entropy change:
x=−335 ∗103+ 298 ∗8.314 ∗x
298
x=−300 ∗103+ 500 ∗8.314 ∗x
500
Solving these two equations will give us the standard entropy change of the
reaction.
Question 11
Question
A reaction has a standard enthalpy change of -335 kJ/mol. If 2.50 mol of the
reactant is consumed in the reaction, how much heat is released or absorbed?
Solution
Step 1: Determine the heat change for 1 mol of the reaction. Given Standard
enthalpy change ∆H=−335 kJ/mol, the heat change for 1 mol of the reaction
is −335 kJ.
Step 2: Calculate the heat change for 2.50 mol of the reaction. To calculate
the heat change for 2.50 mol of the reaction, we use the following equation:
Heat change = Standard enthalpy change ×Number of moles
Substitute the values:
Heat change = −335 kJ/mol ×2.50 mol
Heat change = −335 ×2.50 kJ
Heat change = −837.5 kJ
Therefore, when 2.50 mol of the reactant is consumed in the reaction, the
heat change is -837.5 kJ.
9
Question 12
Question
In a chemical reaction, 5.00 moles of methane gas (CH4) react with excess
oxygen gas to form carbon dioxide gas (CO2) and water vapor. The reaction
releases 793.5 kJ of energy. Calculate the standard enthalpy change (∆H◦) for
the reaction at constant pressure and 298 K.
Solution
Step 1: Write the balanced chemical equation for the reaction. Step 2: Calculate
the moles of water vapor and carbon dioxide produced. Step 3: Calculate the
moles of reactant remaining after the reaction. Step 4: Determine the standard
enthalpy change (∆H◦) for the reaction.
Question 13
Question
Given the reaction, 2CO(g)+O2(g)→2CO2(g), with ∆H=−566 kJ. Calcu-
late the enthalpy change when 4.50 g of CO(g) is burnt in excess oxygen. (molar
mass of CO(g) = 28.0 g/mol)
Solution
Step 1: Calculate the number of moles of CO(g) present. Given mass of CO(g)
= 4.50 g and molar mass of CO(g) = 28.0 g/mol, we can calculate the number
of moles using the formula:
moles = mass
molar mass
moles = 4.50 g
28.0 g/mol = 0.161 mol CO
Step 2: Use the stoichiometry of the reaction to determine the enthalpy
change. From the balanced equation 2CO(g) + O2(g)→2CO2(g), we see that
the molar ratio between CO(g) and CO2(g) is 2:2, or 1:1. This means that for
each mole of CO(g) reacted, the enthalpy change is −566 kJ.
Step 3: Calculate the enthalpy change when 0.161 mol of CO(g) is burnt.
Enthalpy change = moles ×∆H
Enthalpy change = 0.161 mol × −566 kJ/mol = −91.0 kJ
Answer: The enthalpy change when 4.50 g of CO(g) is burnt in excess
oxygen is −91.0kJ.
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Question 14
Question
Consider the following reaction:
2CH3OH(l)→2CO2(g) + 4H2O(l)
Given that the standard enthalpy of formation (∆H◦
f) for methanol (CH3OH)
is -238.6 kJ/mol, carbon dioxide (CO2) is -393.5 kJ/mol, and water (H2O) is
-285.8 kJ/mol, calculate the standard enthalpy change (∆H◦) for the reaction
above.
Solution
Step 1: Calculate the standard enthalpy change for the reaction using the stan-
dard enthalpy of formation values.
The standard enthalpy change for the reaction can be calculated using the
formula:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Where nand mare the coefficients of the products and reactants respectively.
Plugging in the given values:
∆H◦= 2(2(−393.5 kJ/mol) + 4(−285.8 kJ/mol)) −2(−238.6 kJ/mol)
Step 2: Solve for ∆H◦.
Calculating the above expression gives:
∆H◦= 2(−787.0 kJ/mol) −477.2 kJ/mol = −2061.2 kJ/mol
Therefore, the standard enthalpy change for the reaction is -2061.2 kJ/mol.
Question 15
Question
Calculate the standard enthalpy change (∆H◦) for the reaction below using the
given bond dissociation energies:
H2O2(l)→2HO·(g)
Given bond dissociation energies:
O-H = 463 kJ/mol,O=O = 495 kJ/mol,H-H = 432 kJ/mol
11
Solution
Step 1: Write out the given equation and apply Hess’s Law to find the standard
enthalpy change (∆H◦):
∆H◦=XBDE(bonds broken) −XBDE(bonds formed)
Step 2: Calculate the energy required to break the bonds in the reactants:
BDE(O-O) + BDE(H-H) = 2 ×495 kJ/mol + 432 kJ/mol
Step 3: Calculate the energy released when the new bonds are formed in the
products:
4×BDE(O-H)
Step 4: Substitute the values into the equation to find ∆H◦:
∆H◦= (2 ×495 + 432) −4×463 kJ/mol
Step 5: Perform the calculations to find the answer for ∆H◦:
∆H◦= (990 + 432) −1852 kJ/mol
∆H◦= 1142 −1852 kJ/mol
∆H◦=−710 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−710
kJ/mol.
Question 16
Question
A sample of gaseous ethanol (C2H5OH) is burned in a bomb calorimeter. The
combustion reaction is as follows:
C2H5OH(g)+3O2(g)→2CO2(g)+3H2O(l)
The standard enthalpy of formation for ethanol is -277.7 kJ/mol and for water
is -285.8 kJ/mol. Calculate the standard enthalpy change for this combustion
reaction and determine whether it is exothermic or endothermic.
12
Solution
Step 1: Calculate the standard enthalpy change for the combustion reaction
using the enthalpy of formation values given. The standard enthalpy change for
a reaction can be calculated using the formula:
∆H◦
reaction =Xνi∆H◦
f, products −Xνj∆H◦
f, reactants
where νiand νjare the stoichiometric coefficients of the products and reactants,
respectively.
Plugging in the values from the reaction and the enthalpy of formation val-
ues:
∆H◦
reaction = [2(−393.5) + 3(−285.8)] −[−277.7 + 3(0)]
∆H◦
reaction =−787.4 kJ/mol + 277.7 kJ/mol
∆H◦
reaction =−509.7 kJ/mol
Step 2: Determine whether the combustion reaction is exothermic or en-
dothermic. Since the standard enthalpy change for the reaction is negative
(-509.7 kJ/mol), the combustion reaction is exothermic. This means that heat
is released to the surroundings during the reaction.
Question 17
Question
A reaction has a standard enthalpy change of -125 kJ/mol. If 0.500 mol of the
limiting reactant is consumed, what is the heat transferred at constant pressure?
Solution
Step 1: Determine the heat transferred using the given information.
Given: Standard enthalpy change (∆H◦) = -125 kJ/mol Amount of limiting
reactant consumed = 0.500 mol
The heat transferred can be calculated using the equation:
Heat transferred = Standard enthalpy change×Amount of limiting reactant consumed
Step 2: Substitute the given values into the equation and solve for the heat
transferred.
Heat transferred = −125 kJ/mol ×0.500 mol
Heat transferred = −62.5 kJ
Therefore, the heat transferred at constant pressure when 0.500 mol of the
limiting reactant is consumed is -62.5 kJ.
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Question 18
Question
Calculate the standard enthalpy change (∆H◦) for the reaction where 2 moles
of ammonia gas (NH3) react with 3 moles of oxygen gas (O2) to form 2 moles
of nitrogen gas (N2) and 3 moles of water vapor (H2O). Given:
4NH3(g)+5O2(g)→4NO(g)+6H2O(l) ∆H◦=−1450 kJ
N2(g) + O2(g)→2NO(g) ∆H◦= 180 kJ
Solution
Step 1: First, we will write the balanced chemical equation for the reaction given.
We need to use the provided thermochemical equations to find the enthalpy
change for the reaction. The given reaction can be split into two steps:
Step 1: 4NH3(g)+5O2(g)→4NO(g)+6H2O(l)
Step 2: N2(g) + O2(g)→2NO(g)
Step 2: Now, we will calculate the enthalpy change for the overall reaction
by adding the enthalpy changes for the individual steps. As the second step
is reversed in the given reaction, we need to change the sign for the enthalpy
change of this step.
∆H◦
overall = ∆H◦
Step 1 −∆H◦
Step 2
Step 3: Substituting the given enthalpy changes:
∆H◦
overall = (−1450 kJ) −(−180 kJ)
∆H◦
overall =−1450 kJ + 180 kJ
∆H◦
overall =−1270 kJ
Therefore, the standard enthalpy change for the reaction where 2 moles of
NH3react with 3 moles of O2to form 2 moles of N2and 3 moles of H2Ois
−1270 kJ .
Question 19
Question
Calculate the standard enthalpy change for the reaction:
2CO(g) + O2(g)→2CO2(g)
given the following standard enthalpies of formation: ∆H◦
f(CO) = −110.5
kJ/mol, ∆H◦
f(CO2) = −393.5 kJ/mol, and ∆H◦
f(O2) = 0 kJ/mol.
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Solution
Step 1: Write the balanced chemical equation and determine the overall change
in enthalpy. The balanced chemical equation is:
2CO(g) + O2(g)→2CO2(g)
The overall change in enthalpy (∆H◦
rxn) can be calculated using the standard
enthalpies of formation of the reactants and products:
∆H◦
rxn =X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦
rxn = 2∆H◦
f(CO2)−[2∆H◦
f(CO)+∆H◦
f(O2)]
Step 2: Substitute the given values into the equation and calculate.
∆H◦
rxn = 2(−393.5) −[2(−110.5) + 0]
∆H◦
rxn =−787.0 + 221.0
∆H◦
rxn =−566.0 kJ/mol
Therefore, the standard enthalpy change for the reaction is −566.0 kJ/mol .
Question 20
Question
Consider the following reaction:
2C(s) + 3H2(g)→C2H6(g)
The standard enthalpy change for this reaction, ∆H◦, is −284 kJ/mol. Cal-
culate the standard enthalpy of formation, ∆H◦
f, for C2H6(g) given the following
standard enthalpies of formation: ∆H◦
f[H2(g)] = 0 kJ/mol and ∆H◦
f[C(s)] =
0 kJ/mol.
Solution
Step 1: Write the standard enthalpy of formation equation for the given reaction:
∆H◦=Xν∆H◦
f(products) −Xν∆H◦
f(reactants)
Given reaction:
2C(s) + 3H2(g)→C2H6(g)
Step 2: Plug in the known enthalpy of formation values and solve for
∆H◦
f[C2H6(g)]:
−284 kJ/mol = (1)∆H◦
f[C2H6(g)] −(2)(0) −(3)(0)
−284 kJ/mol = ∆H◦
f[C2H6(g)]
Therefore, the standard enthalpy of formation for C2H6(g) is ∆H◦
f[C2H6(g)] =
−284 kJ/mol.
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Question 21
Question
Given the following reaction:
2NO(g) + O2(g)→2NO2(g)
with ∆H=−113 kJ. Calculate the enthalpy change when 2 moles of NO(g)
react with excess O(g) to form NO(g).
Solution
Step 1: Calculate the enthalpy change for the reaction as given. Given: ∆H=
−113 kJ
Step 2: Determine the moles of NO(g) reacting in the given reaction. Since
the stoichiometric coefficient of NO(g) is 2, 2 moles of NO(g) will react.
Step 3: Use the information from Step 2 and the enthalpy change in Step 1
to calculate the enthalpy change for 2 moles of NO(g) reacting. Since 2 moles
of NO(g) are reacting, the enthalpy change is doubled: 2 × −113 kJ =−226 kJ
Answer
The enthalpy change when 2 moles of NO(g) react with excess O(g) to form
NO(g) is −226 kJ.
Question 22
Question
Calculate the enthalpy change (∆H) for the reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(g)
given the following standard enthalpies of formation:
∆H◦
f(C2H6) = −84.68 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution
Step 1: Calculate the standard enthalpy of formation (∆H◦
f) for the given
reaction using the standard enthalpies of formation provided.
16
The standard enthalpy of formation for the reactants is given as:
∆H◦
f(2C2H6) = 2(−84.68 kJ/mol)
=−169.36 kJ/mol
∆H◦
f(7O2) = 7(0 kJ/mol)
= 0 kJ/mol
The standard enthalpy of formation for the products is given as:
∆H◦
f(4CO2) = 4(−393.5 kJ/mol)
=−1574 kJ/mol
∆H◦
f(6H2O) = 6(−285.8 kJ/mol)
=−1714.8 kJ/mol
Step 2: Calculate the overall ∆Hfor the reaction by summing the standard
enthalpies of formation of the products and subtracting the sum of the standard
enthalpies of formation of the reactants.
∆H=X∆H◦
f(products)−X∆H◦
f(reactants)
∆H= (−1574 kJ/mol −1714.8 kJ/mol) −(−169.36 kJ/mol + 0 kJ/mol)
∆H=−3288.8 kJ/mol + 169.36 kJ/mol
∆H=−3119.44 kJ/mol
Therefore, the enthalpy change (∆H) for the given reaction is -3119.44
kJ/mol.
Question 23
Question
For the reaction 2A(g) + B(g) →3C(g), the standard enthalpy change is -1450
kJ/mol. If 2.50 moles of A react completely with excess B, how much heat is
absorbed or released?
Solution
Step 1: Determine the moles of C produced from the reaction. Given the
stoichiometry of the reaction, for every 2 moles of A reacting, 3 moles of C are
produced. Thus, the number of moles of C produced will be:
2.50 moles A ×3 moles C
2 moles A = 3.75 moles C
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Step 2: Calculate the heat change for the reaction. The heat change for the
reaction can be calculated using the formula:
Heat change (J) = moles of C ×∆H◦
Substitute the values:
Heat change (J) = 3.75 moles C×(−1450 kJ/mol×103J/kJ) = −5.4375×106J
Step 3: Determine if heat is absorbed or released. Since the heat change
is negative, the reaction is exothermic, and heat is released. Therefore, the
amount of heat released is 5.4375 ×106J.
Question 24
Question
During a chemical reaction, 200.0 g of water is heated from 25
°
C to 100
°
C. If
the specific heat capacity of water is 4.18 J/g
°
C, calculate the amount of heat
absorbed by the water.
Solution
Step 1: Calculate the change in temperature Given: Initial temperature, Tinitial =
25CFinal temperature, Tfinal = 100CChange in temperature, ∆T=Tfinal −
Tinitial ∆T= 100C−25C= 75C
Step 2: Calculate the heat absorbed by the water The heat absorbed by the
water can be calculated using the formula:
q=mc∆T
where: q= heat absorbed by the water m= mass of water = 200.0 g c= specific
heat capacity of water = 4.18 J/g
°
C ∆T= change in temperature = 75
°
C
Substitute the given values into the formula to find the heat absorbed by
the water:
q= (200.0 g)(4.18 J/g
°
C)(75C) = 62700 J
Therefore, the amount of heat absorbed by the water during the heating
process is 62,700 J.
Question 25
Question
A reaction has a standard enthalpy change of -350 kJ/mol. If 2.00 moles of the
reactant are consumed in the reaction, what is the heat transfer that occurs?
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Solution
Step 1: Determine the moles of heat transferred per mole of reactant. Given:
Standard enthalpy change (∆H◦) = -350 kJ/mol
Step 2: Calculate the heat transfer for 2.00 moles of reactant. Given: Moles
of reactant = 2.00 mol
Now, we can use the following relationship to calculate the heat transfer:
Heat transfer = moles of reactant ×∆H◦
Heat transfer = 2.00 mol × −350 kJ/mol
Step 3: Calculate the heat transfer.
Heat transfer = −700 kJ
Therefore, the heat transfer that occurs when 2.00 moles of the reactant are
consumed in the reaction is -700 kJ.
Question 26
Question
Calculate the enthalpy change for the combustion of 1 mol of butane (C4H10)
at 25
°
C using the following information:
Standard enthalpy of formation of CO2(g): -393.5 kJ/mol
Standard enthalpy of formation of H2O(l): -285.8 kJ/mol
Standard enthalpy of combustion of butane: -2877 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the combustion of butane:
C4H10(g) + 13/2O2(g)→4CO2(g)+5H2O(l)
Step 2: Calculate the enthalpy change for the combustion reaction using the
given standard enthalpies of formation:
∆H=X(products) −X(reactants)
= [4(−393.5) + 5(−285.8)] −[−2877]
=−3935.8+(−1429) + 2877
=−1488.8 kJ/mol
Therefore, the enthalpy change for the combustion of 1 mol of butane is
-1488.8 kJ/mol at 25
°
C.
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Question 27
Question
Given the following reaction at 25
°
C:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(g)
which has an enthalpy change of -3129 kJ. Calculate the standard enthalpy
of formation of ethane, C2H6, given the standard enthalpies of formation are:
∆H◦
f(CO2) = −393.5 kJ/mol and ∆H◦
f(H2O) = −285.8 kJ/mol.
Solution
Step 1: Calculate the enthalpy change for the formation of one mole of C2H6.
∆H= 4∆H◦
f(CO2) + 6∆H◦
f(H2O)−2∆H◦
f(C2H6)
−3129 = 4(−393.5) + 6(−285.8) −2∆H◦
f(C2H6)
−3129 = −1574 −1714.8−2∆H◦
f(C2H6)
−3129 = −3288.8−2∆H◦
f(C2H6)
2∆H◦
f(C2H6) = 159.8
∆H◦
f(C2H6) = 79.9 kJ/mol
Therefore, the standard enthalpy of formation of ethane, C2H6, is 79.9
kJ/mol.
Question 28
Question
Given the following reaction and enthalpy changes:
2SO2(g) + O2(g)→2SO3(g) ∆H1=−198.2 kJ
2SO3(g)→2SO2(g) + O2(g) ∆H2= 198.2 kJ
Calculate the standard enthalpy change for the reaction:
SO2(g) + 1
2O2(g)→SO3(g)
Solution
Step 1: Write out the given reactions and their enthalpy changes.
1) 2SO2(g) + O2(g)→2SO3(g) ∆H1=−198.2 kJ
2) 2SO3(g)→2SO2(g) + O2(g) ∆H2= 198.2 kJ
20
Step 2: Rearrange the given reactions to obtain the target reaction. Adding
the two given reactions together cancels out 2SO2(g) and 2SO3(g), leaving the
target reaction. Adding the two equations:
2SO2(g) + O2(g)+2SO3(g)→2SO3(g)+2SO2(g) + O2(g)
Step 3: Calculate the standard enthalpy change for the target reaction. Since
the given reactions add up to the target reaction,
∆Htarget = ∆H1+ ∆H2
∆Htarget =−198.2 kJ + 198.2 kJ
∆Htarget = 0 kJ
Therefore, the standard enthalpy change for the reaction SO2(g)+ 1
2O2(g)→
SO3(g) is 0 kJ .
Question 29
Question
Given the following thermochemical equation:
2SO2(g)+O2(g)→2SO3(g)
∆H◦=−196 kJ
Calculate the standard enthalpy change for the reaction:
SO2(g) + 1
2O2(g)→SO3(g)
Solution
Step 1: Write the given thermochemical equation:
2SO2(g)+O2(g)→2SO3(g) ∆H◦=−196 kJ
Step 2: Determine the enthalpy change for the desired reaction using the
given equation:
SO2(g) + 1
2O2(g)→SO3(g)
Step 3: Recall that reversing an equation changes the sign of ∆H◦:
2SO3(g)→2SO2(g)+O2(g) ∆H◦= 196 kJ
Step 4: Divide the above equation by 2 to obtain the desired reaction:
SO3(g)→SO2(g) + 1
2O2(g) ∆H◦= 98 kJ
Therefore, the standard enthalpy change for the reaction:
SO2(g) + 1
2O2(g)→SO3(g)
is 98 kJ .
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Question 30
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25◦C given the following bond dissociation energies (all values in kJ/mol):
CH4(g)→C(g)+4H(g): 1660
C(g)+2Cl(g)→CCl4(g): 905
H2(g) + Cl2(g)→2HCl(g): 677
CH4(g)+2Cl2(g)→CCl4(g)+2H2(g)
Solution
Step 1: Calculate the total bond dissociation energy of the reactants.
CH4: (1 ×1660 kJ/mol) = 1660 kJ/mol
Cl2: (2 ×242 kJ/mol) = 484 kJ/mol (using the average bond dissociation energy for Cl-Cl bond)
Total energy of reactants = 1660 kJ/mol + 484 kJ/mol = 2144 kJ/mol
Step 2: Calculate the total bond dissociation energy of the products.
CCl4: (1 ×905 kJ/mol) = 905 kJ/mol
H2: (2 ×436 kJ/mol) = 872 kJ/mol (using the average bond dissociation energy for H-H bond)
Total energy of products = 905 kJ/mol + 872 kJ/mol = 1777 kJ/mol
Step 3: Calculate the change in enthalpy.
∆H◦= Energy of products −Energy of reactants
= 1777 kJ/mol −2144 kJ/mol
=−367 kJ/mol
Therefore, the standard enthalpy change for the given reaction at 25◦C is
−367 kJ/mol.
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