the standard enthalpy of formation of CH4(g)

Question

A scientist measures the standard enthalpy change for the following reaction to be -208.7 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

Answer 1

The standard enthalpy of this reaction is ∑H^o_rxn=∑nH^o_f(products) - ∑mH^o_f(reactants). n and m are the stoichiometric coefficients. The total H of the reaction is the given -208.7 kJ. Consulting a table on the standard enthalpies of formation for common substances, H_CO=-110.5 kJ/mol (m=1 mol), H_H2=0 (m=3 mol), H_CH4=? (n=1 mol), H_{H2O (g)}=-241.8 kJ/mol (n=1 mol).

Plugging everything in, -208.7 kJ = -241.8 kJ/mol * 1 mol + H_CH4 ( 1 mol) - (-110.5 kJ/mol (1 mol) +0)

We can then solve for the unknown variable, H_CH4 (in kJ/mol)