Examkin

Question

Difficulty: EasyStandard Enthalpy Changes and Hess's Law
Consider the following thermochemical reactions:
S(s)+O2(g)→SO2(g)ΔH=−297 kJ mol−1\text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g) \quad \Delta H = -297\text{ kJ mol}^{-1}
2SO2(g)+O2(g)→2SO3(g)ΔH=−198 kJ mol−12\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g) \quad \Delta H = -198\text{ kJ mol}^{-1}
What is the standard enthalpy change for the reaction represented by the equation below?
2S(s)+3O2(g)→2SO3(g)2\text{S}(s) + 3\text{O}_2(g) \rightarrow 2\text{SO}_3(g)
  1. −792 kJ mol−1-792\text{ kJ mol}^{-1}Answer
  2. B
    −495 kJ mol−1-495\text{ kJ mol}^{-1}
  3. C
    −396 kJ mol−1-396\text{ kJ mol}^{-1}
  4. D
    +792 kJ mol−1+792\text{ kJ mol}^{-1}

Answer

−792 kJ mol−1-792\text{ kJ mol}^{-1}
According to Hess's law, the total enthalpy change is the sum of the enthalpy changes for each step. Multiplying the first equation by 2 yields an enthalpy change of −594 kJ mol−1-594\text{ kJ mol}^{-1}. Adding the second equation directly (ΔH=−198 kJ mol−1\Delta H = -198\text{ kJ mol}^{-1}) gives a combined enthalpy change of −594+(−198)=−792 kJ mol−1-594 + (-198) = -792\text{ kJ mol}^{-1}.

Step-by-Step Solution

1
Multiply the first thermochemical equation by 2 to match the 2 moles of sulfur in the target equation.
2S(s)+2O2(g)→2SO2(g)ΔH1=2×(−297 kJ mol−1)=−594 kJ mol−12\text{S}(s) + 2\text{O}_2(g) \rightarrow 2\text{SO}_2(g) \quad \Delta H_1 = 2 \times (-297\text{ kJ mol}^{-1}) = -594\text{ kJ mol}^{-1}
The stoichiometric coefficient of solid sulfur in the target reaction is 2.
2
Add the modified first equation to the second equation.
2S(s)+3O2(g)→2SO3(g)2\text{S}(s) + 3\text{O}_2(g) \rightarrow 2\text{SO}_3(g)
Intermediate sulfur dioxide (2SO22\text{SO}_2) on opposite sides of the combined reaction cancels out.
3
Sum the enthalpy changes for both steps according to Hess's law.
ΔH=−594 kJ mol−1+(−198 kJ mol−1)=−792 kJ mol−1\Delta H = -594\text{ kJ mol}^{-1} + (-198\text{ kJ mol}^{-1}) = -792\text{ kJ mol}^{-1}
Hess's law states that the net enthalpy change of a reaction is the sum of the enthalpy changes of its constituent steps.

Key Concept

Hess's Law of Constant Heat Summation
Rate this question

Topics

Standard Enthalpy Changes and Hess's Law

More questions from this topic