To solve each of these problems, we use the
standard enthalpy of reaction formula:
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ΔH°ᵣₓₙ = Σ n·ΔH°f(products) – Σ m·ΔH°f(reactants)
Where:
- ΔH°f = standard enthalpy of formation (from the table provided)
- n and m are the stoichiometric coefficients from the balanced equation.
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Let’s solve each part one by one using the given table.
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a) NaOH(s) + HCl(g) → NaCl(s) + H₂O(g)
Products:
- NaCl(s):
–411.0 kJ/mol
- H₂O(g):
–241.8 kJ/mol
Sum of products = (–411.0) + (–241.8) =
–652.8 kJ
Reactants:
- NaOH(s):
–426.7 kJ/mol
- HCl(g):
–92.3 kJ/mol
Sum of reactants = (–426.7) + (–92.3) =
–519.0 kJ
ΔH°ᵣₓₙ = –652.8 – (–519.0) = –652.8 + 519.0 = –133.8 kJ
✔ Answer a) –133.8 kJ
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b) 2 CO(g) + O₂(g) → 2 CO₂(g)
Note:
O₂(g) is an element in its standard state → ΔH°f = 0
Products:
- 2 × CO₂(g): 2 × (–393.5) =
–787.0 kJ
Reactants:
- 2 × CO(g): 2 × (–110.5) =
–221.0 kJ
- O₂(g): 0
Sum of reactants = –221.0 + 0 =
–221.0 kJ
ΔH°ᵣₓₙ = –787.0 – (–221.0) = –787.0 + 221.0 = –566.0 kJ
✔ Answer b) –566.0 kJ
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c) CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l)
Products:
- CO₂(g):
–393.5 kJ/mol
- 2 × H₂O(l): 2 × (–285.8) =
–571.6 kJ
Sum of products = –393.5 + (–571.6) =
–965.1 kJ
Reactants:
- CH₄(g):
–74.8 kJ/mol
- 2 × O₂(g): 2 × 0 =
0
Sum of reactants = –74.8 + 0 =
–74.8 kJ
ΔH°ᵣₓₙ = –965.1 – (–74.8) = –965.1 + 74.8 = –890.3 kJ
✔ Answer c) –890.3 kJ
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d) 2 H₂S(g) + 3 O₂(g) → 2 H₂O(l) + 2 SO₂(g)
Products:
- 2 × H₂O(l): 2 × (–285.8) =
–571.6 kJ
- 2 × SO₂(g): 2 × (–296.1) =
–592.2 kJ
Sum of products = –571.6 + (–592.2) =
–1163.8 kJ
Reactants:
- 2 × H₂S(g): 2 × (–20.1) =
–40.2 kJ
- 3 × O₂(g): 3 × 0 =
0
Sum of reactants = –40.2 + 0 =
–40.2 kJ
ΔH°ᵣₓₙ = –1163.8 – (–40.2) = –1163.8 + 40.2 = –1123.6 kJ
✔ Answer d) –1123.6 kJ
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e) 2 NO(g) + O₂(g) → 2 NO₂(g)
Products:
- 2 × NO₂(g): 2 × (33.9) =
+67.8 kJ
Reactants:
- 2 × NO(g): 2 × (90.4) =
+180.8 kJ
- O₂(g): 0
Sum of reactants = 180.8 + 0 =
+180.8 kJ
ΔH°ᵣₓₙ = 67.8 – 180.8 = –113.0 kJ
✔ Answer e) –113.0 kJ
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✔ Final Answers:
a) –133.8 kJ
b) –566.0 kJ
c) –890.3 kJ
d) –1123.6 kJ
e) –113.0 kJ
These values represent the
change in enthalpy for each reaction under standard conditions. Negative values indicate exothermic reactions (heat released), which is typical for combustion and acid-base reactions.
Parent Tip: Review the logic above to help your child master the concept of enthalpy worksheet.
