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:
C(s) + H_{2}O(g) → \underset{“Water gas”}{CO(g) + 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:
C(s) + O_{2}(g) → CO_{2}(g) ΔH° = -393.5 kJ
2 CO(g) + O_{2}(g) → 2 CO_{2}(g) ΔH° = -566.0 kJ
2 H_{2}(g) + O_{2}(g) → 2 H_{2}O(g) ΔH° = -483.6 kJ
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.
C(s) + \cancel{O_{2}(g)} → \cancel{CO_{2}(g)} ΔH° = -393.5 kJ
1/2 [\cancel{2 CO_{2}(g)} → 2 CO(g) + \cancel{O_{2}(g)}] 1/2 [ΔH° = 566.0 kJ] = 283.0 kJ
\underline{1/2 [2 H_{2}O(g) → 2 H_{2}(g) + \cancel{O_{2}(g)} 1/2 [ΔH° = 483.6 kJ] = 241.8 kJ}C(s) + H_{2}O(g) → CO(g) + H_{2}(g) ΔH° = 131.3 kJ
The water-gas reaction is endothermic by 131.3 kJ.