CHEM 105 - ELEMENTS OF GENERAL
CHEMISTRY - Thermochemistry Question Bank
Question 1
Solution: Given: Initial temperature of water 1, T1= 25.0C
Initial temperature of water 2, T2= 75.0C
Mass of water 1, m1= 150.0g
Mass of water 2, m2= 100.0g
Specific heat capacity of water, c= 4.18J/gC
Step 1: Calculate the heat lost by water 1 and the heat gained by water 2:
Q=mcT
For water 1:
Q1=m1cT1= 150.0g×4.18J/gC ×(Tf−25.0C)
For water 2:
Q2=m2cT2= 100.0g×4.18J/gC ×(75.0C−Tf)
Step 2: Since the heat lost by water 1 is equal to the heat gained by water
2, we have:
Q1=Q2
150.0g×4.18J/gC ×(Tf−25.0C) = 100.0g×4.18J/gC ×(75.0C−Tf)
Solving the above equation will give the final temperature, Tf.Question 1:
Calculate the final temperature after 150.0 g of water at 25.0
°
C is
mixed with 100.0 g of water at 75.0
°
C. Assume no heat is lost to the
surroundings. The specific heat capacity of water is 4.18 J/g
°
C.
Solution: Given: Initial temperature of water 1, T1= 25.0C
Initial temperature of water 2, T2= 75.0C
Mass of water 1, m1= 150.0g
Mass of water 2, m2= 100.0g
Specific heat capacity of water, c= 4.18J/gC
Step 1: Calculate the heat lost by water 1 and the heat gained by
water 2:
Q=mcT
1
For water 1:
Q1=m1cT1= 150.0g×4.18J/gC ×(Tf−25.0C)
For water 2:
Q2=m2cT2= 100.0g×4.18J/gC ×(75.0C−Tf)
Step 2: Since the heat lost by water 1 is equal to the heat gained
by water 2, we have:
Q1=Q2
150.0g×4.18J/gC ×(Tf−25.0C) = 100.0g×4.18J/gC ×(75.0C−Tf)
Solving the above equation will give the final temperature, Tf.
Question 2
Calculate the enthalpy change for the reaction:
2H2(g)+O2(g)→2H2O(l)
Given the following bond dissociation energies:
H−H: 436 kJ/mol
O=O: 498 kJ/mol
H−O: 464 kJ/mol
Solution:
The enthalpy change for the reaction can be calculated using the
bond energies of the reactants and products. The enthalpy change
(∆H) for a reaction is given by the formula:
∆H=X(bond energies of reactants)−X(bond energies of products)
1. Calculate the total bond energy of the reactants:
2(H−H) + 1(O=O) = 2(436) + 1(498) = 872 + 498 = 1370 kJ/mol
2. Calculate the total bond energy of the products:
2(H−O) = 2(464) = 928 kJ/mol
3. Calculate the enthalpy change:
∆H=Total bond energy of reactants−Total bond energy of products = 1370−928 = 442kJ/mol
2
Therefore, the enthalpy change for the reaction is ∆H= 442kJ/mol.Question
2:
Calculate the enthalpy change for the reaction:
2H2(g)+O2(g)→2H2O(l)
Given the following bond dissociation energies:
H−H: 436 kJ/mol
O=O: 498 kJ/mol
H−O: 464 kJ/mol
Solution:
The enthalpy change for the reaction can be calculated using the
bond energies of the reactants and products. The enthalpy change
(∆H) for a reaction is given by the formula:
∆H=X(bond energies of reactants)−X(bond energies of products)
1. Calculate the total bond energy of the reactants:
2(H−H) + 1(O=O) = 2(436) + 1(498) = 872 + 498 = 1370 kJ/mol
2. Calculate the total bond energy of the products:
2(H−O) = 2(464) = 928 kJ/mol
3. Calculate the enthalpy change:
∆H=Total bond energy of reactants−Total bond energy of products = 1370−928 = 442kJ/mol
Therefore, the enthalpy change for the reaction is ∆H= 442kJ/mol.
Question 3
The combustion of propane (C3H8) releases 2220 kJ/mol of heat.
Calculate the heat released when 5.00 g of propane burns.
Solution:
Given: Heat released per mole of propane = 2220 kJ/mol
Molar mass of C3H8= 3(12.01 g/mol) + 8(1.008 g/mol) = 44.11
g/mol
Number of moles of propane burned = 5.00 g
44.11 g/mol
To calculate the heat released when 5.00 g of propane burns, we
can use the formula:
3
Heat released = (Number of moles of propane burned) * (Heat
released per mole of propane)
Plugging in the values,
Heat released = 5.00 g
44.11 g/mol ∗2220 kJ/mol
Heat released 0.1133 mol
1∗2220 kJ/mol
Heat released 251.2 kJ
Therefore, when 5.00 g of propane burns, approximately 251.2 kJ
of heat is released.Question 3:
The combustion of propane (C3H8) releases 2220 kJ/mol of heat.
Calculate the heat released when 5.00 g of propane burns.
Solution:
Given: Heat released per mole of propane = 2220 kJ/mol
Molar mass of C3H8= 3(12.01 g/mol) + 8(1.008 g/mol) = 44.11
g/mol
Number of moles of propane burned = 5.00 g
44.11 g/mol
To calculate the heat released when 5.00 g of propane burns, we
can use the formula:
Heat released = (Number of moles of propane burned) * (Heat
released per mole of propane)
Plugging in the values,
Heat released = 5.00 g
44.11 g/mol ∗2220 kJ/mol
Heat released 0.1133 mol
1∗2220 kJ/mol
Heat released 251.2 kJ
Therefore, when 5.00 g of propane burns, approximately 251.2 kJ
of heat is released.
Question 4
Standard enthalpy of formation for C(s)= 0 kJ/mol
Standard enthalpy of formation for H2(g)= 0 kJ/mol
Standard enthalpy of formation for C2H6(g)= -84.68 kJ/mol
Step-by-step Solution: The enthalpy change for the reaction can
be calculated using the standard enthalpy of formation values:
1. Write the balanced chemical equation for the reaction: 2C(s) +
3H2(g)→C2H6(g)
2. Calculate the standard enthalpy of formation for the reaction
using the given values: The standard enthalpy change (∆H) for the
reaction can be calculated by taking the sum of the standard enthalpy
of formation of the products minus the sum of the standard enthalpy
of formation of the reactants:
∆H= (PProducts)−(PReactants)
4
∆H= (1 × −84.68 kJ/mol)−(2 ×0kJ/mol + 3 ×0kJ/mol)
∆H=−84.68 kJ/mol
3. Therefore, the enthalpy change for the reaction is -84.68 kJ/mol.Question
4: Calculate the enthalpy change for the reaction:
2C(s)+3H2(g)→C2H6(g)
given the following information:
Standard enthalpy of formation for C(s)= 0 kJ/mol
Standard enthalpy of formation for H2(g)= 0 kJ/mol
Standard enthalpy of formation for C2H6(g)= -84.68 kJ/mol
Step-by-step Solution: The enthalpy change for the reaction can
be calculated using the standard enthalpy of formation values:
1. Write the balanced chemical equation for the reaction: 2C(s) +
3H2(g)→C2H6(g)
2. Calculate the standard enthalpy of formation for the reaction
using the given values: The standard enthalpy change (∆H) for the
reaction can be calculated by taking the sum of the standard enthalpy
of formation of the products minus the sum of the standard enthalpy
of formation of the reactants:
∆H= (PProducts)−(PReactants)
∆H= (1 × −84.68 kJ/mol)−(2 ×0kJ/mol + 3 ×0kJ/mol)
∆H=−84.68 kJ/mol
3. Therefore, the enthalpy change for the reaction is -84.68 kJ/mol.
Question 5
Calculate the standard enthalpy change for the following reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
Given the following standard enthalpy of formations:
∆Hf(C2H6) = −84.68 kJ/mol
∆Hf(CO2) = −393.51 kJ/mol
∆Hf(H2O) = −285.83 kJ/mol
∆Hf(O2)=0kJ/mol
Solution: To find the standard enthalpy change for the reaction,
we can use Hess’s Law.
5
1. Calculate the sum of the standard enthalpy of formations of the
products:
∆Hproducts = 4(∆Hf(CO2)) + 6(∆Hf(H2O))
= 4(−393.51 kJ/mol) + 6(−285.83 kJ/mol)
=−1574.04 kJ/mol −1714.98 kJ/mol
=−3288.02 kJ/mol
2. Calculate the sum of the standard enthalpy of formations of the
reactants:
∆Hreactants = 2(∆Hf(C2H6)) + 7(∆Hf(O2))
= 2(−84.68 kJ/mol) + 7(0 kJ/mol)
=−169.36 kJ/mol
3. Calculate the standard enthalpy change for the reaction:
∆Hreaction = ∆Hproducts −∆Hreactants
=−3288.02 kJ/mol −(−169.36 kJ/mol)
=−3288.02 kJ/mol + 169.36 kJ/mol
=−3118.66 kJ/mol
Therefore, the standard enthalpy change for the reaction is -3118.66
kJ/mol.Question 5:
Calculate the standard enthalpy change for the following reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
Given the following standard enthalpy of formations:
∆Hf(C2H6) = −84.68 kJ/mol
∆Hf(CO2) = −393.51 kJ/mol
∆Hf(H2O) = −285.83 kJ/mol
∆Hf(O2)=0kJ/mol
Solution: To find the standard enthalpy change for the reaction,
we can use Hess’s Law.
1. Calculate the sum of the standard enthalpy of formations of the
products:
∆Hproducts = 4(∆Hf(CO2)) + 6(∆Hf(H2O))
= 4(−393.51 kJ/mol) + 6(−285.83 kJ/mol)
=−1574.04 kJ/mol −1714.98 kJ/mol
=−3288.02 kJ/mol
6
2. Calculate the sum of the standard enthalpy of formations of the
reactants:
∆Hreactants = 2(∆Hf(C2H6)) + 7(∆Hf(O2))
= 2(−84.68 kJ/mol) + 7(0 kJ/mol)
=−169.36 kJ/mol
3. Calculate the standard enthalpy change for the reaction:
∆Hreaction = ∆Hproducts −∆Hreactants
=−3288.02 kJ/mol −(−169.36 kJ/mol)
=−3288.02 kJ/mol + 169.36 kJ/mol
=−3118.66 kJ/mol
Therefore, the standard enthalpy change for the reaction is -3118.66
kJ/mol.
Question 6
4F e(s)+3O2(g)→2F e2O3(s)
given the following enthalpy changes:
∆H◦
f(F e2O3) = −824.2kJ/mol
∆H◦
f(F e) = 0 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution:
1. Write the chemical equation for the reaction:
4F e(s)+3O2(g)→2F e2O3(s)
2. Calculate the enthalpy change using the given enthalpy changes:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
Substitute the values:
∆H◦= [2 ·∆H◦
f(F e2O3)] −[4 ·∆H◦
f(F e)+3·∆H◦
f(O2)]
∆H◦= [2 ·(−824.2)] −[4 ·0+3·0]
∆H◦=−1648.4kJ/mol
Therefore, the enthalpy change for the reaction is −1648.4kJ/mol.
“‘“‘markdown 6. Calculate the enthalpy change for the reaction:
7
4F e(s)+3O2(g)→2F e2O3(s)
given the following enthalpy changes:
∆H◦
f(F e2O3) = −824.2kJ/mol
∆H◦
f(F e) = 0 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution:
1. Write the chemical equation for the reaction:
4F e(s)+3O2(g)→2F e2O3(s)
2. Calculate the enthalpy change using the given enthalpy changes:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
Substitute the values:
∆H◦= [2 ·∆H◦
f(F e2O3)] −[4 ·∆H◦
f(F e)+3·∆H◦
f(O2)]
∆H◦= [2 ·(−824.2)] −[4 ·0+3·0]
∆H◦=−1648.4kJ/mol
Therefore, the enthalpy change for the reaction is −1648.4kJ/mol.
“‘
Question 7
Calculate the enthalpy change for the reaction:
2H2(g) + O2(g)→2H2O(g)
given the following standard enthalpies of formation:
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2O(g)) = −286 kJ/mol
Step-by-step Solution:
1. Write the balanced chemical equation for the reaction.
2H2(g) + O2(g)→2H2O(g)
2. Calculate the change in enthalpy using the formula:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
8
3. Substitute the values of standard enthalpies of formation into
the formula.
∆H◦= 2∆H◦
f(H2O(g)) −[2∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
4. Calculate the enthalpy change for the reaction.
∆H◦= 2(−286) −[2(0) + 0]
∆H◦=−572 kJ/mol
Therefore, the enthalpy change for the reaction is -572 kJ/mol.Question
7:
Calculate the enthalpy change for the reaction:
2H2(g) + O2(g)→2H2O(g)
given the following standard enthalpies of formation:
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2O(g)) = −286 kJ/mol
Step-by-step Solution:
1. Write the balanced chemical equation for the reaction.
2H2(g) + O2(g)→2H2O(g)
2. Calculate the change in enthalpy using the formula:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
3. Substitute the values of standard enthalpies of formation into
the formula.
∆H◦= 2∆H◦
f(H2O(g)) −[2∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
4. Calculate the enthalpy change for the reaction.
∆H◦= 2(−286) −[2(0) + 0]
∆H◦=−572 kJ/mol
Therefore, the enthalpy change for the reaction is -572 kJ/mol.
9
Question 8
Question 8: Calculate the enthalpy change for the reaction below
using the given bond energies: 2C2H2(g) + 5O2(g)→4CO2(g) + 2H2O(g)
Given bond energies: C−C: 347 kJ/mol,C=C: 611 kJ/mol,C=O:
799 kJ/mol,O=O: 498 kJ/mol,O−H: 463 kJ/mol
Step-by-step Solution: 1. Calculate the total energy required to
break the bonds in the reactants: 2×(4×C−C)+2×(2×C=C)+5×O=
O= 2 ×4×347 + 2 ×2×611 + 5 ×498 = 6634 kJ
2. Calculate the total energy released in forming the bonds in the
products: 4×(2 ×C=O)+2×O−H= 4 ×2×799 + 2 ×463 = 6878 kJ
3. Calculate the overall enthalpy change: ∆H=Total energy in bonds broken−
Total energy in bonds formed = 6634 −6878 = −244 kJ
Therefore, the enthalpy change for the reaction is −244 kJ.
“‘latex article amsmath
Question 8: Calculate the enthalpy change for the reaction below
using the given bond energies: 2C2H2(g) + 5O2(g)→4CO2(g) + 2H2O(g)
Given bond energies: C−C: 347 kJ/mol,C=C: 611 kJ/mol,C=O:
799 kJ/mol,O=O: 498 kJ/mol,O−H: 463 kJ/mol
Step-by-step Solution:
1. Calculate the total energy required to break the bonds in the
reactants: 2×(4 ×C−C) + 2 ×(2 ×C=C) + 5 ×O=O= 2 ×4×
347 + 2 ×2×611 + 5 ×498 = 6634 kJ
2. Calculate the total energy released in forming the bonds in the
products: 4×(2 ×C=O)+2×O−H= 4 ×2×799 + 2 ×463 = 6878 kJ
3. Calculate the overall enthalpy change: ∆H=Total energy in bonds broken−
Total energy in bonds formed = 6634 −6878 = −244 kJ
Therefore, the enthalpy change for the reaction is −244 kJ.
10
For water 1:
Q1=m1cT1= 150.0g×4.18J/gC ×(Tf−25.0C)
For water 2:
Q2=m2cT2= 100.0g×4.18J/gC ×(75.0C−Tf)
Step 2: Since the heat lost by water 1 is equal to the heat gained
by water 2, we have:
Q1=Q2
150.0g×4.18J/gC ×(Tf−25.0C) = 100.0g×4.18J/gC ×(75.0C−Tf)
Solving the above equation will give the final temperature, Tf.
Question 2
Calculate the enthalpy change for the reaction:
2H2(g)+O2(g)→2H2O(l)
Given the following bond dissociation energies:
H−H: 436 kJ/mol
O=O: 498 kJ/mol
H−O: 464 kJ/mol
Solution:
The enthalpy change for the reaction can be calculated using the
bond energies of the reactants and products. The enthalpy change
(∆H) for a reaction is given by the formula:
∆H=X(bond energies of reactants)−X(bond energies of products)
1. Calculate the total bond energy of the reactants:
2(H−H) + 1(O=O) = 2(436) + 1(498) = 872 + 498 = 1370 kJ/mol
2. Calculate the total bond energy of the products:
2(H−O) = 2(464) = 928 kJ/mol
3. Calculate the enthalpy change:
∆H=Total bond energy of reactants−Total bond energy of products = 1370−928 = 442kJ/mol
2
Therefore, the enthalpy change for the reaction is ∆H= 442kJ/mol.Question
2:
Calculate the enthalpy change for the reaction:
2H2(g)+O2(g)→2H2O(l)
Given the following bond dissociation energies:
H−H: 436 kJ/mol
O=O: 498 kJ/mol
H−O: 464 kJ/mol
Solution:
The enthalpy change for the reaction can be calculated using the
bond energies of the reactants and products. The enthalpy change
(∆H) for a reaction is given by the formula:
∆H=X(bond energies of reactants)−X(bond energies of products)
1. Calculate the total bond energy of the reactants:
2(H−H) + 1(O=O) = 2(436) + 1(498) = 872 + 498 = 1370 kJ/mol
2. Calculate the total bond energy of the products:
2(H−O) = 2(464) = 928 kJ/mol
3. Calculate the enthalpy change:
∆H=Total bond energy of reactants−Total bond energy of products = 1370−928 = 442kJ/mol
Therefore, the enthalpy change for the reaction is ∆H= 442kJ/mol.
Question 3
The combustion of propane (C3H8) releases 2220 kJ/mol of heat.
Calculate the heat released when 5.00 g of propane burns.
Solution:
Given: Heat released per mole of propane = 2220 kJ/mol
Molar mass of C3H8= 3(12.01 g/mol) + 8(1.008 g/mol) = 44.11
g/mol
Number of moles of propane burned = 5.00 g
44.11 g/mol
To calculate the heat released when 5.00 g of propane burns, we
can use the formula:
3
Heat released = (Number of moles of propane burned) * (Heat
released per mole of propane)
Plugging in the values,
Heat released = 5.00 g
44.11 g/mol ∗2220 kJ/mol
Heat released 0.1133 mol
1∗2220 kJ/mol
Heat released 251.2 kJ
Therefore, when 5.00 g of propane burns, approximately 251.2 kJ
of heat is released.Question 3:
The combustion of propane (C3H8) releases 2220 kJ/mol of heat.
Calculate the heat released when 5.00 g of propane burns.
Solution:
Given: Heat released per mole of propane = 2220 kJ/mol
Molar mass of C3H8= 3(12.01 g/mol) + 8(1.008 g/mol) = 44.11
g/mol
Number of moles of propane burned = 5.00 g
44.11 g/mol
To calculate the heat released when 5.00 g of propane burns, we
can use the formula:
Heat released = (Number of moles of propane burned) * (Heat
released per mole of propane)
Plugging in the values,
Heat released = 5.00 g
44.11 g/mol ∗2220 kJ/mol
Heat released 0.1133 mol
1∗2220 kJ/mol
Heat released 251.2 kJ
Therefore, when 5.00 g of propane burns, approximately 251.2 kJ
of heat is released.
Question 4
Standard enthalpy of formation for C(s)= 0 kJ/mol
Standard enthalpy of formation for H2(g)= 0 kJ/mol
Standard enthalpy of formation for C2H6(g)= -84.68 kJ/mol
Step-by-step Solution: The enthalpy change for the reaction can
be calculated using the standard enthalpy of formation values:
1. Write the balanced chemical equation for the reaction: 2C(s) +
3H2(g)→C2H6(g)
2. Calculate the standard enthalpy of formation for the reaction
using the given values: The standard enthalpy change (∆H) for the
reaction can be calculated by taking the sum of the standard enthalpy
of formation of the products minus the sum of the standard enthalpy
of formation of the reactants:
∆H= (PProducts)−(PReactants)
4
∆H= (1 × −84.68 kJ/mol)−(2 ×0kJ/mol + 3 ×0kJ/mol)
∆H=−84.68 kJ/mol
3. Therefore, the enthalpy change for the reaction is -84.68 kJ/mol.Question
4: Calculate the enthalpy change for the reaction:
2C(s)+3H2(g)→C2H6(g)
given the following information:
Standard enthalpy of formation for C(s)= 0 kJ/mol
Standard enthalpy of formation for H2(g)= 0 kJ/mol
Standard enthalpy of formation for C2H6(g)= -84.68 kJ/mol
Step-by-step Solution: The enthalpy change for the reaction can
be calculated using the standard enthalpy of formation values:
1. Write the balanced chemical equation for the reaction: 2C(s) +
3H2(g)→C2H6(g)
2. Calculate the standard enthalpy of formation for the reaction
using the given values: The standard enthalpy change (∆H) for the
reaction can be calculated by taking the sum of the standard enthalpy
of formation of the products minus the sum of the standard enthalpy
of formation of the reactants:
∆H= (PProducts)−(PReactants)
∆H= (1 × −84.68 kJ/mol)−(2 ×0kJ/mol + 3 ×0kJ/mol)
∆H=−84.68 kJ/mol
3. Therefore, the enthalpy change for the reaction is -84.68 kJ/mol.
Question 5
Calculate the standard enthalpy change for the following reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
Given the following standard enthalpy of formations:
∆Hf(C2H6) = −84.68 kJ/mol
∆Hf(CO2) = −393.51 kJ/mol
∆Hf(H2O) = −285.83 kJ/mol
∆Hf(O2)=0kJ/mol
Solution: To find the standard enthalpy change for the reaction,
we can use Hess’s Law.
5
1. Calculate the sum of the standard enthalpy of formations of the
products:
∆Hproducts = 4(∆Hf(CO2)) + 6(∆Hf(H2O))
= 4(−393.51 kJ/mol) + 6(−285.83 kJ/mol)
=−1574.04 kJ/mol −1714.98 kJ/mol
=−3288.02 kJ/mol
2. Calculate the sum of the standard enthalpy of formations of the
reactants:
∆Hreactants = 2(∆Hf(C2H6)) + 7(∆Hf(O2))
= 2(−84.68 kJ/mol) + 7(0 kJ/mol)
=−169.36 kJ/mol
3. Calculate the standard enthalpy change for the reaction:
∆Hreaction = ∆Hproducts −∆Hreactants
=−3288.02 kJ/mol −(−169.36 kJ/mol)
=−3288.02 kJ/mol + 169.36 kJ/mol
=−3118.66 kJ/mol
Therefore, the standard enthalpy change for the reaction is -3118.66
kJ/mol.Question 5:
Calculate the standard enthalpy change for the following reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
Given the following standard enthalpy of formations:
∆Hf(C2H6) = −84.68 kJ/mol
∆Hf(CO2) = −393.51 kJ/mol
∆Hf(H2O) = −285.83 kJ/mol
∆Hf(O2)=0kJ/mol
Solution: To find the standard enthalpy change for the reaction,
we can use Hess’s Law.
1. Calculate the sum of the standard enthalpy of formations of the
products:
∆Hproducts = 4(∆Hf(CO2)) + 6(∆Hf(H2O))
= 4(−393.51 kJ/mol) + 6(−285.83 kJ/mol)
=−1574.04 kJ/mol −1714.98 kJ/mol
=−3288.02 kJ/mol
6
2. Calculate the sum of the standard enthalpy of formations of the
reactants:
∆Hreactants = 2(∆Hf(C2H6)) + 7(∆Hf(O2))
= 2(−84.68 kJ/mol) + 7(0 kJ/mol)
=−169.36 kJ/mol
3. Calculate the standard enthalpy change for the reaction:
∆Hreaction = ∆Hproducts −∆Hreactants
=−3288.02 kJ/mol −(−169.36 kJ/mol)
=−3288.02 kJ/mol + 169.36 kJ/mol
=−3118.66 kJ/mol
Therefore, the standard enthalpy change for the reaction is -3118.66
kJ/mol.
Question 6
4F e(s)+3O2(g)→2F e2O3(s)
given the following enthalpy changes:
∆H◦
f(F e2O3) = −824.2kJ/mol
∆H◦
f(F e) = 0 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution:
1. Write the chemical equation for the reaction:
4F e(s)+3O2(g)→2F e2O3(s)
2. Calculate the enthalpy change using the given enthalpy changes:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
Substitute the values:
∆H◦= [2 ·∆H◦
f(F e2O3)] −[4 ·∆H◦
f(F e)+3·∆H◦
f(O2)]
∆H◦= [2 ·(−824.2)] −[4 ·0+3·0]
∆H◦=−1648.4kJ/mol
Therefore, the enthalpy change for the reaction is −1648.4kJ/mol.
“‘“‘markdown 6. Calculate the enthalpy change for the reaction:
7
4F e(s)+3O2(g)→2F e2O3(s)
given the following enthalpy changes:
∆H◦
f(F e2O3) = −824.2kJ/mol
∆H◦
f(F e) = 0 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution:
1. Write the chemical equation for the reaction:
4F e(s)+3O2(g)→2F e2O3(s)
2. Calculate the enthalpy change using the given enthalpy changes:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
Substitute the values:
∆H◦= [2 ·∆H◦
f(F e2O3)] −[4 ·∆H◦
f(F e)+3·∆H◦
f(O2)]
∆H◦= [2 ·(−824.2)] −[4 ·0+3·0]
∆H◦=−1648.4kJ/mol
Therefore, the enthalpy change for the reaction is −1648.4kJ/mol.
“‘
Question 7
Calculate the enthalpy change for the reaction:
2H2(g) + O2(g)→2H2O(g)
given the following standard enthalpies of formation:
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2O(g)) = −286 kJ/mol
Step-by-step Solution:
1. Write the balanced chemical equation for the reaction.
2H2(g) + O2(g)→2H2O(g)
2. Calculate the change in enthalpy using the formula:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
8
3. Substitute the values of standard enthalpies of formation into
the formula.
∆H◦= 2∆H◦
f(H2O(g)) −[2∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
4. Calculate the enthalpy change for the reaction.
∆H◦= 2(−286) −[2(0) + 0]
∆H◦=−572 kJ/mol
Therefore, the enthalpy change for the reaction is -572 kJ/mol.Question
7:
Calculate the enthalpy change for the reaction:
2H2(g) + O2(g)→2H2O(g)
given the following standard enthalpies of formation:
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2O(g)) = −286 kJ/mol
Step-by-step Solution:
1. Write the balanced chemical equation for the reaction.
2H2(g) + O2(g)→2H2O(g)
2. Calculate the change in enthalpy using the formula:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
3. Substitute the values of standard enthalpies of formation into
the formula.
∆H◦= 2∆H◦
f(H2O(g)) −[2∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
4. Calculate the enthalpy change for the reaction.
∆H◦= 2(−286) −[2(0) + 0]
∆H◦=−572 kJ/mol
Therefore, the enthalpy change for the reaction is -572 kJ/mol.
9
Question 8
Question 8: Calculate the enthalpy change for the reaction below
using the given bond energies: 2C2H2(g) + 5O2(g)→4CO2(g) + 2H2O(g)
Given bond energies: C−C: 347 kJ/mol,C=C: 611 kJ/mol,C=O:
799 kJ/mol,O=O: 498 kJ/mol,O−H: 463 kJ/mol
Step-by-step Solution: 1. Calculate the total energy required to
break the bonds in the reactants: 2×(4×C−C)+2×(2×C=C)+5×O=
O= 2 ×4×347 + 2 ×2×611 + 5 ×498 = 6634 kJ
2. Calculate the total energy released in forming the bonds in the
products: 4×(2 ×C=O)+2×O−H= 4 ×2×799 + 2 ×463 = 6878 kJ
3. Calculate the overall enthalpy change: ∆H=Total energy in bonds broken−
Total energy in bonds formed = 6634 −6878 = −244 kJ
Therefore, the enthalpy change for the reaction is −244 kJ.
“‘latex article amsmath
Question 8: Calculate the enthalpy change for the reaction below
using the given bond energies: 2C2H2(g) + 5O2(g)→4CO2(g) + 2H2O(g)
Given bond energies: C−C: 347 kJ/mol,C=C: 611 kJ/mol,C=O:
799 kJ/mol,O=O: 498 kJ/mol,O−H: 463 kJ/mol
Step-by-step Solution:
1. Calculate the total energy required to break the bonds in the
reactants: 2×(4 ×C−C) + 2 ×(2 ×C=C) + 5 ×O=O= 2 ×4×
347 + 2 ×2×611 + 5 ×498 = 6634 kJ
2. Calculate the total energy released in forming the bonds in the
products: 4×(2 ×C=O)+2×O−H= 4 ×2×799 + 2 ×463 = 6878 kJ
3. Calculate the overall enthalpy change: ∆H=Total energy in bonds broken−
Total energy in bonds formed = 6634 −6878 = −244 kJ
Therefore, the enthalpy change for the reaction is −244 kJ.
10
For water 1:
Q1=m1cT1= 150.0g×4.18J/gC ×(Tf−25.0C)
For water 2:
Q2=m2cT2= 100.0g×4.18J/gC ×(75.0C−Tf)
Step 2: Since the heat lost by water 1 is equal to the heat gained
by water 2, we have:
Q1=Q2
150.0g×4.18J/gC ×(Tf−25.0C) = 100.0g×4.18J/gC ×(75.0C−Tf)
Solving the above equation will give the final temperature, Tf.
Question 2
Calculate the enthalpy change for the reaction:
2H2(g)+O2(g)→2H2O(l)
Given the following bond dissociation energies:
H−H: 436 kJ/mol
O=O: 498 kJ/mol
H−O: 464 kJ/mol
Solution:
The enthalpy change for the reaction can be calculated using the
bond energies of the reactants and products. The enthalpy change
(∆H) for a reaction is given by the formula:
∆H=X(bond energies of reactants)−X(bond energies of products)
1. Calculate the total bond energy of the reactants:
2(H−H) + 1(O=O) = 2(436) + 1(498) = 872 + 498 = 1370 kJ/mol
2. Calculate the total bond energy of the products:
2(H−O) = 2(464) = 928 kJ/mol
3. Calculate the enthalpy change:
∆H=Total bond energy of reactants−Total bond energy of products = 1370−928 = 442kJ/mol
2
Therefore, the enthalpy change for the reaction is ∆H= 442kJ/mol.Question
2:
Calculate the enthalpy change for the reaction:
2H2(g)+O2(g)→2H2O(l)
Given the following bond dissociation energies:
H−H: 436 kJ/mol
O=O: 498 kJ/mol
H−O: 464 kJ/mol
Solution:
The enthalpy change for the reaction can be calculated using the
bond energies of the reactants and products. The enthalpy change
(∆H) for a reaction is given by the formula:
∆H=X(bond energies of reactants)−X(bond energies of products)
1. Calculate the total bond energy of the reactants:
2(H−H) + 1(O=O) = 2(436) + 1(498) = 872 + 498 = 1370 kJ/mol
2. Calculate the total bond energy of the products:
2(H−O) = 2(464) = 928 kJ/mol
3. Calculate the enthalpy change:
∆H=Total bond energy of reactants−Total bond energy of products = 1370−928 = 442kJ/mol
Therefore, the enthalpy change for the reaction is ∆H= 442kJ/mol.
Question 3
The combustion of propane (C3H8) releases 2220 kJ/mol of heat.
Calculate the heat released when 5.00 g of propane burns.
Solution:
Given: Heat released per mole of propane = 2220 kJ/mol
Molar mass of C3H8= 3(12.01 g/mol) + 8(1.008 g/mol) = 44.11
g/mol
Number of moles of propane burned = 5.00 g
44.11 g/mol
To calculate the heat released when 5.00 g of propane burns, we
can use the formula:
3
Heat released = (Number of moles of propane burned) * (Heat
released per mole of propane)
Plugging in the values,
Heat released = 5.00 g
44.11 g/mol ∗2220 kJ/mol
Heat released 0.1133 mol
1∗2220 kJ/mol
Heat released 251.2 kJ
Therefore, when 5.00 g of propane burns, approximately 251.2 kJ
of heat is released.Question 3:
The combustion of propane (C3H8) releases 2220 kJ/mol of heat.
Calculate the heat released when 5.00 g of propane burns.
Solution:
Given: Heat released per mole of propane = 2220 kJ/mol
Molar mass of C3H8= 3(12.01 g/mol) + 8(1.008 g/mol) = 44.11
g/mol
Number of moles of propane burned = 5.00 g
44.11 g/mol
To calculate the heat released when 5.00 g of propane burns, we
can use the formula:
Heat released = (Number of moles of propane burned) * (Heat
released per mole of propane)
Plugging in the values,
Heat released = 5.00 g
44.11 g/mol ∗2220 kJ/mol
Heat released 0.1133 mol
1∗2220 kJ/mol
Heat released 251.2 kJ
Therefore, when 5.00 g of propane burns, approximately 251.2 kJ
of heat is released.
Question 4
Standard enthalpy of formation for C(s)= 0 kJ/mol
Standard enthalpy of formation for H2(g)= 0 kJ/mol
Standard enthalpy of formation for C2H6(g)= -84.68 kJ/mol
Step-by-step Solution: The enthalpy change for the reaction can
be calculated using the standard enthalpy of formation values:
1. Write the balanced chemical equation for the reaction: 2C(s) +
3H2(g)→C2H6(g)
2. Calculate the standard enthalpy of formation for the reaction
using the given values: The standard enthalpy change (∆H) for the
reaction can be calculated by taking the sum of the standard enthalpy
of formation of the products minus the sum of the standard enthalpy
of formation of the reactants:
∆H= (PProducts)−(PReactants)
4
∆H= (1 × −84.68 kJ/mol)−(2 ×0kJ/mol + 3 ×0kJ/mol)
∆H=−84.68 kJ/mol
3. Therefore, the enthalpy change for the reaction is -84.68 kJ/mol.Question
4: Calculate the enthalpy change for the reaction:
2C(s)+3H2(g)→C2H6(g)
given the following information:
Standard enthalpy of formation for C(s)= 0 kJ/mol
Standard enthalpy of formation for H2(g)= 0 kJ/mol
Standard enthalpy of formation for C2H6(g)= -84.68 kJ/mol
Step-by-step Solution: The enthalpy change for the reaction can
be calculated using the standard enthalpy of formation values:
1. Write the balanced chemical equation for the reaction: 2C(s) +
3H2(g)→C2H6(g)
2. Calculate the standard enthalpy of formation for the reaction
using the given values: The standard enthalpy change (∆H) for the
reaction can be calculated by taking the sum of the standard enthalpy
of formation of the products minus the sum of the standard enthalpy
of formation of the reactants:
∆H= (PProducts)−(PReactants)
∆H= (1 × −84.68 kJ/mol)−(2 ×0kJ/mol + 3 ×0kJ/mol)
∆H=−84.68 kJ/mol
3. Therefore, the enthalpy change for the reaction is -84.68 kJ/mol.
Question 5
Calculate the standard enthalpy change for the following reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
Given the following standard enthalpy of formations:
∆Hf(C2H6) = −84.68 kJ/mol
∆Hf(CO2) = −393.51 kJ/mol
∆Hf(H2O) = −285.83 kJ/mol
∆Hf(O2)=0kJ/mol
Solution: To find the standard enthalpy change for the reaction,
we can use Hess’s Law.
5
1. Calculate the sum of the standard enthalpy of formations of the
products:
∆Hproducts = 4(∆Hf(CO2)) + 6(∆Hf(H2O))
= 4(−393.51 kJ/mol) + 6(−285.83 kJ/mol)
=−1574.04 kJ/mol −1714.98 kJ/mol
=−3288.02 kJ/mol
2. Calculate the sum of the standard enthalpy of formations of the
reactants:
∆Hreactants = 2(∆Hf(C2H6)) + 7(∆Hf(O2))
= 2(−84.68 kJ/mol) + 7(0 kJ/mol)
=−169.36 kJ/mol
3. Calculate the standard enthalpy change for the reaction:
∆Hreaction = ∆Hproducts −∆Hreactants
=−3288.02 kJ/mol −(−169.36 kJ/mol)
=−3288.02 kJ/mol + 169.36 kJ/mol
=−3118.66 kJ/mol
Therefore, the standard enthalpy change for the reaction is -3118.66
kJ/mol.Question 5:
Calculate the standard enthalpy change for the following reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
Given the following standard enthalpy of formations:
∆Hf(C2H6) = −84.68 kJ/mol
∆Hf(CO2) = −393.51 kJ/mol
∆Hf(H2O) = −285.83 kJ/mol
∆Hf(O2)=0kJ/mol
Solution: To find the standard enthalpy change for the reaction,
we can use Hess’s Law.
1. Calculate the sum of the standard enthalpy of formations of the
products:
∆Hproducts = 4(∆Hf(CO2)) + 6(∆Hf(H2O))
= 4(−393.51 kJ/mol) + 6(−285.83 kJ/mol)
=−1574.04 kJ/mol −1714.98 kJ/mol
=−3288.02 kJ/mol
6
2. Calculate the sum of the standard enthalpy of formations of the
reactants:
∆Hreactants = 2(∆Hf(C2H6)) + 7(∆Hf(O2))
= 2(−84.68 kJ/mol) + 7(0 kJ/mol)
=−169.36 kJ/mol
3. Calculate the standard enthalpy change for the reaction:
∆Hreaction = ∆Hproducts −∆Hreactants
=−3288.02 kJ/mol −(−169.36 kJ/mol)
=−3288.02 kJ/mol + 169.36 kJ/mol
=−3118.66 kJ/mol
Therefore, the standard enthalpy change for the reaction is -3118.66
kJ/mol.
Question 6
4F e(s)+3O2(g)→2F e2O3(s)
given the following enthalpy changes:
∆H◦
f(F e2O3) = −824.2kJ/mol
∆H◦
f(F e) = 0 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution:
1. Write the chemical equation for the reaction:
4F e(s)+3O2(g)→2F e2O3(s)
2. Calculate the enthalpy change using the given enthalpy changes:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
Substitute the values:
∆H◦= [2 ·∆H◦
f(F e2O3)] −[4 ·∆H◦
f(F e)+3·∆H◦
f(O2)]
∆H◦= [2 ·(−824.2)] −[4 ·0+3·0]
∆H◦=−1648.4kJ/mol
Therefore, the enthalpy change for the reaction is −1648.4kJ/mol.
“‘“‘markdown 6. Calculate the enthalpy change for the reaction:
7
4F e(s)+3O2(g)→2F e2O3(s)
given the following enthalpy changes:
∆H◦
f(F e2O3) = −824.2kJ/mol
∆H◦
f(F e) = 0 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution:
1. Write the chemical equation for the reaction:
4F e(s)+3O2(g)→2F e2O3(s)
2. Calculate the enthalpy change using the given enthalpy changes:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
Substitute the values:
∆H◦= [2 ·∆H◦
f(F e2O3)] −[4 ·∆H◦
f(F e)+3·∆H◦
f(O2)]
∆H◦= [2 ·(−824.2)] −[4 ·0+3·0]
∆H◦=−1648.4kJ/mol
Therefore, the enthalpy change for the reaction is −1648.4kJ/mol.
“‘
Question 7
Calculate the enthalpy change for the reaction:
2H2(g) + O2(g)→2H2O(g)
given the following standard enthalpies of formation:
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2O(g)) = −286 kJ/mol
Step-by-step Solution:
1. Write the balanced chemical equation for the reaction.
2H2(g) + O2(g)→2H2O(g)
2. Calculate the change in enthalpy using the formula:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
8
3. Substitute the values of standard enthalpies of formation into
the formula.
∆H◦= 2∆H◦
f(H2O(g)) −[2∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
4. Calculate the enthalpy change for the reaction.
∆H◦= 2(−286) −[2(0) + 0]
∆H◦=−572 kJ/mol
Therefore, the enthalpy change for the reaction is -572 kJ/mol.Question
7:
Calculate the enthalpy change for the reaction:
2H2(g) + O2(g)→2H2O(g)
given the following standard enthalpies of formation:
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2O(g)) = −286 kJ/mol
Step-by-step Solution:
1. Write the balanced chemical equation for the reaction.
2H2(g) + O2(g)→2H2O(g)
2. Calculate the change in enthalpy using the formula:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
3. Substitute the values of standard enthalpies of formation into
the formula.
∆H◦= 2∆H◦
f(H2O(g)) −[2∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
4. Calculate the enthalpy change for the reaction.
∆H◦= 2(−286) −[2(0) + 0]
∆H◦=−572 kJ/mol
Therefore, the enthalpy change for the reaction is -572 kJ/mol.
9
Question 8
Question 8: Calculate the enthalpy change for the reaction below
using the given bond energies: 2C2H2(g) + 5O2(g)→4CO2(g) + 2H2O(g)
Given bond energies: C−C: 347 kJ/mol,C=C: 611 kJ/mol,C=O:
799 kJ/mol,O=O: 498 kJ/mol,O−H: 463 kJ/mol
Step-by-step Solution: 1. Calculate the total energy required to
break the bonds in the reactants: 2×(4×C−C)+2×(2×C=C)+5×O=
O= 2 ×4×347 + 2 ×2×611 + 5 ×498 = 6634 kJ
2. Calculate the total energy released in forming the bonds in the
products: 4×(2 ×C=O)+2×O−H= 4 ×2×799 + 2 ×463 = 6878 kJ
3. Calculate the overall enthalpy change: ∆H=Total energy in bonds broken−
Total energy in bonds formed = 6634 −6878 = −244 kJ
Therefore, the enthalpy change for the reaction is −244 kJ.
“‘latex article amsmath
Question 8: Calculate the enthalpy change for the reaction below
using the given bond energies: 2C2H2(g) + 5O2(g)→4CO2(g) + 2H2O(g)
Given bond energies: C−C: 347 kJ/mol,C=C: 611 kJ/mol,C=O:
799 kJ/mol,O=O: 498 kJ/mol,O−H: 463 kJ/mol
Step-by-step Solution:
1. Calculate the total energy required to break the bonds in the
reactants: 2×(4 ×C−C) + 2 ×(2 ×C=C) + 5 ×O=O= 2 ×4×
347 + 2 ×2×611 + 5 ×498 = 6634 kJ
2. Calculate the total energy released in forming the bonds in the
products: 4×(2 ×C=O)+2×O−H= 4 ×2×799 + 2 ×463 = 6878 kJ
3. Calculate the overall enthalpy change: ∆H=Total energy in bonds broken−
Total energy in bonds formed = 6634 −6878 = −244 kJ
Therefore, the enthalpy change for the reaction is −244 kJ.
10
For water 1:
Q1=m1cT1= 150.0g×4.18J/gC ×(Tf−25.0C)
For water 2:
Q2=m2cT2= 100.0g×4.18J/gC ×(75.0C−Tf)
Step 2: Since the heat lost by water 1 is equal to the heat gained
by water 2, we have:
Q1=Q2
150.0g×4.18J/gC ×(Tf−25.0C) = 100.0g×4.18J/gC ×(75.0C−Tf)
Solving the above equation will give the final temperature, Tf.
Question 2
Calculate the enthalpy change for the reaction:
2H2(g)+O2(g)→2H2O(l)
Given the following bond dissociation energies:
H−H: 436 kJ/mol
O=O: 498 kJ/mol
H−O: 464 kJ/mol
Solution:
The enthalpy change for the reaction can be calculated using the
bond energies of the reactants and products. The enthalpy change
(∆H) for a reaction is given by the formula:
∆H=X(bond energies of reactants)−X(bond energies of products)
1. Calculate the total bond energy of the reactants:
2(H−H) + 1(O=O) = 2(436) + 1(498) = 872 + 498 = 1370 kJ/mol
2. Calculate the total bond energy of the products:
2(H−O) = 2(464) = 928 kJ/mol
3. Calculate the enthalpy change:
∆H=Total bond energy of reactants−Total bond energy of products = 1370−928 = 442kJ/mol
2
Therefore, the enthalpy change for the reaction is ∆H= 442kJ/mol.Question
2:
Calculate the enthalpy change for the reaction:
2H2(g)+O2(g)→2H2O(l)
Given the following bond dissociation energies:
H−H: 436 kJ/mol
O=O: 498 kJ/mol
H−O: 464 kJ/mol
Solution:
The enthalpy change for the reaction can be calculated using the
bond energies of the reactants and products. The enthalpy change
(∆H) for a reaction is given by the formula:
∆H=X(bond energies of reactants)−X(bond energies of products)
1. Calculate the total bond energy of the reactants:
2(H−H) + 1(O=O) = 2(436) + 1(498) = 872 + 498 = 1370 kJ/mol
2. Calculate the total bond energy of the products:
2(H−O) = 2(464) = 928 kJ/mol
3. Calculate the enthalpy change:
∆H=Total bond energy of reactants−Total bond energy of products = 1370−928 = 442kJ/mol
Therefore, the enthalpy change for the reaction is ∆H= 442kJ/mol.
Question 3
The combustion of propane (C3H8) releases 2220 kJ/mol of heat.
Calculate the heat released when 5.00 g of propane burns.
Solution:
Given: Heat released per mole of propane = 2220 kJ/mol
Molar mass of C3H8= 3(12.01 g/mol) + 8(1.008 g/mol) = 44.11
g/mol
Number of moles of propane burned = 5.00 g
44.11 g/mol
To calculate the heat released when 5.00 g of propane burns, we
can use the formula:
3
Heat released = (Number of moles of propane burned) * (Heat
released per mole of propane)
Plugging in the values,
Heat released = 5.00 g
44.11 g/mol ∗2220 kJ/mol
Heat released 0.1133 mol
1∗2220 kJ/mol
Heat released 251.2 kJ
Therefore, when 5.00 g of propane burns, approximately 251.2 kJ
of heat is released.Question 3:
The combustion of propane (C3H8) releases 2220 kJ/mol of heat.
Calculate the heat released when 5.00 g of propane burns.
Solution:
Given: Heat released per mole of propane = 2220 kJ/mol
Molar mass of C3H8= 3(12.01 g/mol) + 8(1.008 g/mol) = 44.11
g/mol
Number of moles of propane burned = 5.00 g
44.11 g/mol
To calculate the heat released when 5.00 g of propane burns, we
can use the formula:
Heat released = (Number of moles of propane burned) * (Heat
released per mole of propane)
Plugging in the values,
Heat released = 5.00 g
44.11 g/mol ∗2220 kJ/mol
Heat released 0.1133 mol
1∗2220 kJ/mol
Heat released 251.2 kJ
Therefore, when 5.00 g of propane burns, approximately 251.2 kJ
of heat is released.
Question 4
Standard enthalpy of formation for C(s)= 0 kJ/mol
Standard enthalpy of formation for H2(g)= 0 kJ/mol
Standard enthalpy of formation for C2H6(g)= -84.68 kJ/mol
Step-by-step Solution: The enthalpy change for the reaction can
be calculated using the standard enthalpy of formation values:
1. Write the balanced chemical equation for the reaction: 2C(s) +
3H2(g)→C2H6(g)
2. Calculate the standard enthalpy of formation for the reaction
using the given values: The standard enthalpy change (∆H) for the
reaction can be calculated by taking the sum of the standard enthalpy
of formation of the products minus the sum of the standard enthalpy
of formation of the reactants:
∆H= (PProducts)−(PReactants)
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∆H= (1 × −84.68 kJ/mol)−(2 ×0kJ/mol + 3 ×0kJ/mol)
∆H=−84.68 kJ/mol
3. Therefore, the enthalpy change for the reaction is -84.68 kJ/mol.Question
4: Calculate the enthalpy change for the reaction:
2C(s)+3H2(g)→C2H6(g)
given the following information:
Standard enthalpy of formation for C(s)= 0 kJ/mol
Standard enthalpy of formation for H2(g)= 0 kJ/mol
Standard enthalpy of formation for C2H6(g)= -84.68 kJ/mol
Step-by-step Solution: The enthalpy change for the reaction can
be calculated using the standard enthalpy of formation values:
1. Write the balanced chemical equation for the reaction: 2C(s) +
3H2(g)→C2H6(g)
2. Calculate the standard enthalpy of formation for the reaction
using the given values: The standard enthalpy change (∆H) for the
reaction can be calculated by taking the sum of the standard enthalpy
of formation of the products minus the sum of the standard enthalpy
of formation of the reactants:
∆H= (PProducts)−(PReactants)
∆H= (1 × −84.68 kJ/mol)−(2 ×0kJ/mol + 3 ×0kJ/mol)
∆H=−84.68 kJ/mol
3. Therefore, the enthalpy change for the reaction is -84.68 kJ/mol.
Question 5
Calculate the standard enthalpy change for the following reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
Given the following standard enthalpy of formations:
∆Hf(C2H6) = −84.68 kJ/mol
∆Hf(CO2) = −393.51 kJ/mol
∆Hf(H2O) = −285.83 kJ/mol
∆Hf(O2)=0kJ/mol
Solution: To find the standard enthalpy change for the reaction,
we can use Hess’s Law.
5
1. Calculate the sum of the standard enthalpy of formations of the
products:
∆Hproducts = 4(∆Hf(CO2)) + 6(∆Hf(H2O))
= 4(−393.51 kJ/mol) + 6(−285.83 kJ/mol)
=−1574.04 kJ/mol −1714.98 kJ/mol
=−3288.02 kJ/mol
2. Calculate the sum of the standard enthalpy of formations of the
reactants:
∆Hreactants = 2(∆Hf(C2H6)) + 7(∆Hf(O2))
= 2(−84.68 kJ/mol) + 7(0 kJ/mol)
=−169.36 kJ/mol
3. Calculate the standard enthalpy change for the reaction:
∆Hreaction = ∆Hproducts −∆Hreactants
=−3288.02 kJ/mol −(−169.36 kJ/mol)
=−3288.02 kJ/mol + 169.36 kJ/mol
=−3118.66 kJ/mol
Therefore, the standard enthalpy change for the reaction is -3118.66
kJ/mol.Question 5:
Calculate the standard enthalpy change for the following reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
Given the following standard enthalpy of formations:
∆Hf(C2H6) = −84.68 kJ/mol
∆Hf(CO2) = −393.51 kJ/mol
∆Hf(H2O) = −285.83 kJ/mol
∆Hf(O2)=0kJ/mol
Solution: To find the standard enthalpy change for the reaction,
we can use Hess’s Law.
1. Calculate the sum of the standard enthalpy of formations of the
products:
∆Hproducts = 4(∆Hf(CO2)) + 6(∆Hf(H2O))
= 4(−393.51 kJ/mol) + 6(−285.83 kJ/mol)
=−1574.04 kJ/mol −1714.98 kJ/mol
=−3288.02 kJ/mol
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2. Calculate the sum of the standard enthalpy of formations of the
reactants:
∆Hreactants = 2(∆Hf(C2H6)) + 7(∆Hf(O2))
= 2(−84.68 kJ/mol) + 7(0 kJ/mol)
=−169.36 kJ/mol
3. Calculate the standard enthalpy change for the reaction:
∆Hreaction = ∆Hproducts −∆Hreactants
=−3288.02 kJ/mol −(−169.36 kJ/mol)
=−3288.02 kJ/mol + 169.36 kJ/mol
=−3118.66 kJ/mol
Therefore, the standard enthalpy change for the reaction is -3118.66
kJ/mol.
Question 6
4F e(s)+3O2(g)→2F e2O3(s)
given the following enthalpy changes:
∆H◦
f(F e2O3) = −824.2kJ/mol
∆H◦
f(F e) = 0 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution:
1. Write the chemical equation for the reaction:
4F e(s)+3O2(g)→2F e2O3(s)
2. Calculate the enthalpy change using the given enthalpy changes:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
Substitute the values:
∆H◦= [2 ·∆H◦
f(F e2O3)] −[4 ·∆H◦
f(F e)+3·∆H◦
f(O2)]
∆H◦= [2 ·(−824.2)] −[4 ·0+3·0]
∆H◦=−1648.4kJ/mol
Therefore, the enthalpy change for the reaction is −1648.4kJ/mol.
“‘“‘markdown 6. Calculate the enthalpy change for the reaction:
7
4F e(s)+3O2(g)→2F e2O3(s)
given the following enthalpy changes:
∆H◦
f(F e2O3) = −824.2kJ/mol
∆H◦
f(F e) = 0 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution:
1. Write the chemical equation for the reaction:
4F e(s)+3O2(g)→2F e2O3(s)
2. Calculate the enthalpy change using the given enthalpy changes:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
Substitute the values:
∆H◦= [2 ·∆H◦
f(F e2O3)] −[4 ·∆H◦
f(F e)+3·∆H◦
f(O2)]
∆H◦= [2 ·(−824.2)] −[4 ·0+3·0]
∆H◦=−1648.4kJ/mol
Therefore, the enthalpy change for the reaction is −1648.4kJ/mol.
“‘
Question 7
Calculate the enthalpy change for the reaction:
2H2(g) + O2(g)→2H2O(g)
given the following standard enthalpies of formation:
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2O(g)) = −286 kJ/mol
Step-by-step Solution:
1. Write the balanced chemical equation for the reaction.
2H2(g) + O2(g)→2H2O(g)
2. Calculate the change in enthalpy using the formula:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
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3. Substitute the values of standard enthalpies of formation into
the formula.
∆H◦= 2∆H◦
f(H2O(g)) −[2∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
4. Calculate the enthalpy change for the reaction.
∆H◦= 2(−286) −[2(0) + 0]
∆H◦=−572 kJ/mol
Therefore, the enthalpy change for the reaction is -572 kJ/mol.Question
7:
Calculate the enthalpy change for the reaction:
2H2(g) + O2(g)→2H2O(g)
given the following standard enthalpies of formation:
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2O(g)) = −286 kJ/mol
Step-by-step Solution:
1. Write the balanced chemical equation for the reaction.
2H2(g) + O2(g)→2H2O(g)
2. Calculate the change in enthalpy using the formula:
∆H◦=X∆H◦
f(products)−X∆H◦
f(reactants)
3. Substitute the values of standard enthalpies of formation into
the formula.
∆H◦= 2∆H◦
f(H2O(g)) −[2∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
4. Calculate the enthalpy change for the reaction.
∆H◦= 2(−286) −[2(0) + 0]
∆H◦=−572 kJ/mol
Therefore, the enthalpy change for the reaction is -572 kJ/mol.
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Question 8
Question 8: Calculate the enthalpy change for the reaction below
using the given bond energies: 2C2H2(g) + 5O2(g)→4CO2(g) + 2H2O(g)
Given bond energies: C−C: 347 kJ/mol,C=C: 611 kJ/mol,C=O:
799 kJ/mol,O=O: 498 kJ/mol,O−H: 463 kJ/mol
Step-by-step Solution: 1. Calculate the total energy required to
break the bonds in the reactants: 2×(4×C−C)+2×(2×C=C)+5×O=
O= 2 ×4×347 + 2 ×2×611 + 5 ×498 = 6634 kJ
2. Calculate the total energy released in forming the bonds in the
products: 4×(2 ×C=O)+2×O−H= 4 ×2×799 + 2 ×463 = 6878 kJ
3. Calculate the overall enthalpy change: ∆H=Total energy in bonds broken−
Total energy in bonds formed = 6634 −6878 = −244 kJ
Therefore, the enthalpy change for the reaction is −244 kJ.
“‘latex article amsmath
Question 8: Calculate the enthalpy change for the reaction below
using the given bond energies: 2C2H2(g) + 5O2(g)→4CO2(g) + 2H2O(g)
Given bond energies: C−C: 347 kJ/mol,C=C: 611 kJ/mol,C=O:
799 kJ/mol,O=O: 498 kJ/mol,O−H: 463 kJ/mol
Step-by-step Solution:
1. Calculate the total energy required to break the bonds in the
reactants: 2×(4 ×C−C) + 2 ×(2 ×C=C) + 5 ×O=O= 2 ×4×
347 + 2 ×2×611 + 5 ×498 = 6634 kJ
2. Calculate the total energy released in forming the bonds in the
products: 4×(2 ×C=O)+2×O−H= 4 ×2×799 + 2 ×463 = 6878 kJ
3. Calculate the overall enthalpy change: ∆H=Total energy in bonds broken−
Total energy in bonds formed = 6634 −6878 = −244 kJ
Therefore, the enthalpy change for the reaction is −244 kJ.
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