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Difficulty: MediumStandard Enthalpy Changes and Hess's Law

Consider the following thermochemical equations at 298 K298\text{ K}:

1. CH3OH(l)+32O2(g)→CO2(g)+2H2O(l)ΔH∘=−726 kJ mol−1\text{CH}_3\text{OH}(l) + \frac{3}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H^\circ = -726\text{ kJ mol}^{-1}
2. CO(g)+12O2(g)→CO2(g)ΔH∘=−283 kJ mol−1\text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H^\circ = -283\text{ kJ mol}^{-1}
3. H2(g)+12O2(g)→H2O(l)ΔH∘=−286 kJ mol−1\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H^\circ = -286\text{ kJ mol}^{-1}

What is the standard enthalpy change for the synthesis of liquid methanol from carbon monoxide and hydrogen gas according to the equation:
CO(g)+2H2(g)→CH3OH(l)\text{CO}(g) + 2\text{H}_2(g) \rightarrow \text{CH}_3\text{OH}(l)
  1. A
    −1581 kJ mol−1-1581\text{ kJ mol}^{-1}
  2. −129 kJ mol−1-129\text{ kJ mol}^{-1}Answer
  3. C
    +129 kJ mol−1+129\text{ kJ mol}^{-1}
  4. D
    +157 kJ mol−1+157\text{ kJ mol}^{-1}

Answer

−129 kJ mol−1-129\text{ kJ mol}^{-1}
Applying Hess's Law requires manipulating the given equations to yield the target reaction CO(g)+2H2(g)→CH3OH(l)\text{CO}(g) + 2\text{H}_2(g) \rightarrow \text{CH}_3\text{OH}(l). Reversing equation 1 gives ΔH=+726 kJ mol−1\Delta H = +726\text{ kJ mol}^{-1}. Keeping equation 2 gives ΔH=−283 kJ mol−1\Delta H = -283\text{ kJ mol}^{-1}. Doubling equation 3 gives ΔH=−572 kJ mol−1\Delta H = -572\text{ kJ mol}^{-1}. Summing these values gives +726−283−572=−129 kJ mol−1+726 - 283 - 572 = -129\text{ kJ mol}^{-1}.

Step-by-Step Solution

1
Reverse Equation (1) so that CH3OH(l)\text{CH}_3\text{OH}(l) is on the product side.
CO2(g)+2H2O(l)→CH3OH(l)+32O2(g)ΔH1=+726 kJ mol−1\text{CO}_2(g) + 2\text{H}_2\text{O}(l) \rightarrow \text{CH}_3\text{OH}(l) + \frac{3}{2}\text{O}_2(g) \quad \Delta H_1 = +726\text{ kJ mol}^{-1}
Reversing a reaction changes the sign of its enthalpy change according to Hess's Law.
2
Use Equation (2) as written to supply CO(g)\text{CO}(g) on the reactant side.
CO(g)+12O2(g)→CO2(g)ΔH2=−283 kJ mol−1\text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_2 = -283\text{ kJ mol}^{-1}
CO(g)\text{CO}(g) is required as a reactant with a coefficient of 1.
3
Multiply Equation (3) by 2 to supply 2H2(g)2\text{H}_2(g) on the reactant side.
2H2(g)+O2(g)→2H2O(l)ΔH3=2×(−286 kJ mol−1)=−572 kJ mol−12\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l) \quad \Delta H_3 = 2 \times (-286\text{ kJ mol}^{-1}) = -572\text{ kJ mol}^{-1}
The target reaction requires 2 moles of H2(g)\text{H}_2(g), so the enthalpy change must be multiplied by 2.
4
Sum the modified equations and their respective ΔH\Delta H values.
ΔH∘=+726+(−283)+(−572)=−129 kJ mol−1\Delta H^\circ = +726 + (-283) + (-572) = -129\text{ kJ mol}^{-1}
Hess's Law states that the overall enthalpy change is the sum of the enthalpy changes of individual steps.

Key Concept

Hess's Law of Constant Heat Summation
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Topics

Standard Enthalpy Changes and Hess's Law