Question

Methanol is manufactured by the partial combustion of methane under pressure, using a copper catalyst:
CH₄(g) + ½ O₂(g) → CH₃OH(l)
Enthalpies of combustion are (in kJ mol^-1):

CH₄(g), -881
CH₃OH(l), -726

Calculate the standard enthalpy change for the manufacture of 1 mole methanol by the reaction
CH₄(g) + ½ O₂(g) → CH₃OH(l)

Follow the procedures based on Hess’s Law:
First write down the reactions corresponding to the enthalpies of combustion you have been given, reverse one of the equations remembering to change the sign of ΔH^o, then combine these equations to give the required process, and evaluate the associated ΔH^o
State the answer to three significant figures without decimal point, ±xxx

Answer 1

CH4(g) + 2O2(g) ------------------> CO2(g) + 2H2O(g)   ΔH   = -881KJ

CH3OH(l)   + 3/2O2(g) ------------> CO2(g) + 2H2O        ΔH    = -726KJ    substract two equations

(-)                  (-)                            (-)              (-)                        (+)

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CH4(g) + 1/2O2(g)   ----------------> CH3OH(l)             ΔH    = -155KJ