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Difficulty: MediumStandard Enthalpy Changes and Hess's Law
Consider the thermochemical equations below:
I. C(s)+O2(g)→CO2(g)ΔH=−393.5 kJ mol−1\text{I. } \text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -393.5\text{ kJ mol}^{-1}
II. CO(g)+12O2(g)→CO2(g)ΔH=−283.0 kJ mol−1\text{II. } \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -283.0\text{ kJ mol}^{-1}

What is the standard enthalpy of formation of carbon(II) oxide, CO(g)\text{CO}(g)?

  1. −110.5 kJ mol−1-110.5\text{ kJ mol}^{-1}Answer
  2. B
    +110.5 kJ mol−1+110.5\text{ kJ mol}^{-1}
  3. C
    −676.5 kJ mol−1-676.5\text{ kJ mol}^{-1}
  4. D
    +676.5 kJ mol−1+676.5\text{ kJ mol}^{-1}

Answer

−110.5 kJ mol−1-110.5\text{ kJ mol}^{-1}
To find the enthalpy of formation of CO(g)\text{CO}(g) from C(s)\text{C}(s) and O2(g)\text{O}_2(g), Equation I is kept as written (ΔH1=−393.5 kJ mol−1\Delta H_1 = -393.5\text{ kJ mol}^{-1}) while Equation II is reversed so that CO(g)\text{CO}(g) appears on the product side (changing ΔH2\Delta H_2 from −283.0 kJ mol−1-283.0\text{ kJ mol}^{-1} to +283.0 kJ mol−1+283.0\text{ kJ mol}^{-1}). Summing both steps gives −393.5+283.0=−110.5 kJ mol−1-393.5 + 283.0 = -110.5\text{ kJ mol}^{-1}.

Step-by-Step Solution

1
Write the target equation for the standard enthalpy of formation of CO(g)\text{CO}(g)
C(s)+12O2(g)→CO(g)ΔHf∘=?\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g) \quad \Delta H_f^\circ = ?
The standard enthalpy of formation is the heat change when 1 mole of a substance is formed from its constituent elements in their standard states.
2
Manipulate the given equations so their sum yields the target equation
Keep Equation I as written:
C(s)+O2(g)→CO2(g)ΔH1=−393.5 kJ mol−1\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_1 = -393.5\text{ kJ mol}^{-1}
Reverse Equation II:
CO2(g)→CO(g)+12O2(g)ΔH2′=+283.0 kJ mol−1\text{CO}_2(g) \rightarrow \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \quad \Delta H_2' = +283.0\text{ kJ mol}^{-1}
Reversing a reaction changes the sign of its enthalpy change according to Hess's Law.
3
Sum the manipulated equations and their corresponding ΔH\Delta H values
C(s)+12O2(g)→CO(g)\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)
ΔHf∘=−393.5 kJ mol−1+283.0 kJ mol−1=−110.5 kJ mol−1\Delta H_f^\circ = -393.5\text{ kJ mol}^{-1} + 283.0\text{ kJ mol}^{-1} = -110.5\text{ kJ mol}^{-1}
Hess's Law states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes for each step.

Key Concept

Hess's Law and Standard Enthalpy of Formation
Estimated Time:1m 30s
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Topics

Standard Enthalpy Changes and Hess's Law