Question

A scientist measures the standard enthalpy change for the following reaction to be -209.3 kJ:

CO (g) + 3 H2 (g) --> CH4 (g) + H2O (g)

based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CH4 (g) is what? (kJ/mol)

Answer 1

Given:

ΔHo rxn = -209.3 KJ/mol

Hof(CO(g)) = -110.525 KJ/mol

Hof(H2(g)) = 0.0 KJ/mol

Hof(H2O(g)) = -241.818 KJ/mol

Balanced chemical equation is:

CO(g) + 3 H2(g) ---> CH4(g) + H2O(g)

ΔHo rxn = 1*Hof(CH4(g)) + 1*Hof(H2O(g)) - 1*Hof( CO(g)) - 3*Hof(H2(g))

-209.3 = 1*Hof(CH4(g)) + 1*(-241.818) - 1*(-110.525) - 3*(0.0)

Hof(CH4(g)) = -78.007 KJ/mol

Answer: -78.01 KJ/mol