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Question 8.7: Using Hess’s Law to Calculate ΔH° Water gas is the name for ......

Using Hess’s Law to Calculate ΔH°

Water gas is the name for the mixture of CO and H_2 prepared by reaction of steam with carbon at 1000 °C: \mathrm{C}(s)+\mathrm{H}_2 \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)\underset{\text{"Water gas"}}{+}\mathrm{H}_2(g)

The hydrogen is then purified and used as a starting material for preparing ammonia. Use the following information to calculate ΔH° in kilojoules for the water-gas reaction:

\begin{array}{rll}\mathrm{C}(s)+\mathrm{O}_2(g) & \longrightarrow \mathrm{CO}_2(g) & \Delta H^{\circ}=-393.5 \mathrm{~kJ} \\2 \mathrm{CO}(g)+\mathrm{O}_2(g) & \longrightarrow 2 \mathrm{CO}_2(g) & \Delta H^{\circ}=-566.0 \mathrm{~kJ} \\2 \mathrm{H}_2(g)+\mathrm{O}_2(g) & \longrightarrow 2 \mathrm{H}_2 \mathrm{O}(g) & \Delta H^{\circ}=-483.6 \mathrm{~kJ}\end{array}

STRATEGY

As in Worked Example 8.6, the idea is to find a combination of the individual reactions whose sum is the desired reaction. In this instance, it’s necessary to reverse the second and third steps and to multiply both by 1/2 to make the overall equation balance. In so doing, the signs of the enthalpy changes for those steps must be changed and multiplied by 1/2. Note that CO_2(g) and O_2(g) cancel because they appear on both the right and left sides of equations.

Step-by-Step
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\begin{array}{ll}\mathrm{C}(\mathrm{s})+\cancel{O_2 (g)} \longrightarrow \mathrm{CO}_2(g) & \Delta H^{\circ}=-393.5 \mathrm{~kJ} \\1 / 2\left[\cancel{2 \ CO_2 (g)} \longrightarrow 2 \mathrm{CO}(g)+\cancel{O_2 (g)}\right] & 1 / 2\left[\Delta H^{\circ}=566.0 \mathrm{~kJ}\right]=+283.0 \mathrm{~kJ} \\1 / 2\left[2 \mathrm{H}_2 \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_2(g)+\cancel{O_2 (g)} \right] & 1 / 2\left[\Delta H^{\circ}=483.6 \mathrm{~kJ}\right]=+241.8 \mathrm{~kJ}\end{array}

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\mathrm{C}(s)+\mathrm{H}_2 \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_2(g) \quad \Delta H^{\circ}=+131.3 \mathrm{~kJ}

The water-gas reaction is endothermic by 131.3 kJ.

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