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QUESTION
Given the following heats of combustion. CH 3 OH(l) + 3/2 O 2 (g) CO 2 (g) + 2 H 2 O(l) H rxn = -726.4 kJ C(graphite) + O 2 (g) CO 2 (g) H rxn = -393....
Given the following heats of combustion.
CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l)ΔH°rxn = -726.4 kJC(graphite) + O2(g) → CO2(g)ΔH°rxn = -393.5 kJH2(g) + 1/2 O2(g) → H2O(l)ΔH°rxn = -285.8 kJCalculate the enthalpy of formation of methanol (CH3OH) from its elements.
C(graphite) + 2 H2(g) + 1/2 O2(g)→CH3OH(l)
Answer must be in kJ
