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Given C(graphite) + O2(g) → CO2(g); ΔrH° = -393.5 kJ mol^-1 H2(g) + 1/2 O2(g) → H2O(l); ΔrH° = -285.8 kJ mol^-1 CO2(g) + 2H2O(l) → CH4(g) + 2O2(g); ΔrH° = +890.3 kJ mol^-1 Based on the above chemical equations, the value of ΔrH° at 298 K for the reaction C(graphite) + 2H2(g) → CH4(g) will be:
- -74.8 kJ mol^-1
- -144.0 kJ mol^-1
- +74.8 kJ mol^-1
- +144.0 kJ mol^-1
Correct answer: -74.8 kJ mol^-1
Solution
The correct option is derived by applying Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for individual steps. By rearranging and combining the given reactions, the enthalpy change for the formation of CH4 from graphite and hydrogen can be calculated, resulting in a value of -74.8 kJ mol^-1.
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