Determine the enthalpy change for the reaction 2C(s) + 2H2O(g) → CH4(g) + CO2(g) using the following:

Question

Determine the enthalpy change for the reaction 2C(s) + 2H2O(g) → CH4(g) + CO2(g) using the following:

C(s) + H2O(g) → CO(g) + H2(g) ΔH° = 131.3 kJ

CO(g) + H2O(g) → CO2(g) + H2(g) ΔH° = 41.2 kJ

CH4(g) + H2O(g) → 3H2(g) + CO(g) ΔH° = 206.1 kJ

I solved this by rearranging the equations to get to the target equation. I did this and got +97.7 kJ. However, it also tells me to solve this problem another way by looking up the standard molar enthalpies of formation of each reactant and product and insert those values into the following equation:

ΔH°reaction = ∑ {(moles of product) x (ΔH°f of product)} — ∑ {(moles of reactant) x (ΔH°f of reactant)}

I am getting a different result when I use this equation and look up the values. When I solved it the first time by rearranging the three equations to get to the target equation, I got +97.7 kJ, but when I do it by plugging the looked up values in the equation I end up getting 15.27kJ. Can someone please tell me what I am doing wrong and/or show the steps on how to get to the right answer using the method in which I look up the values and plug it into the equation?

Answer 1

I too get + 15.3 kJ using the second method. Are you sure you did it correctly using Hess' Law (the first method)? I currently don't have the time to check it, but will do so later.