The standard heats of formation of CH4 H2O and CH3OH are −76,−242 and −266klmol− respectively. The enthalpy change for the following reaction is:CH3OH(l)+H2(g)⟶CH4(g)+H2O(?).

(i) C+2H2(g) ⟶CH4(g); ΔH⊥=−76kJmol−  (iii) H2(g)+12O2(g) ⟶H2O(η); ΔH2=−242kJmol−  (iii) C+2H2(g)+12O2(g)⟶CH3OH(l) ΔH3=−266kJmol−Req uired equation is:CH3OH(l)+H2(g)⟶CH4(g)+H2O(l);ΔH4=?In order to get equation (A), we have, equation (i) + equation (ii) -equation (iii). Thus:C+2H2(g)+H2(g)+12O2(g)−C−2H2(g)−12O2⟶CH4(g)+H2O(l)→CH3OH(l);ΔH1(=−76−242+266)kJmol−∴CH3OH(l)+H2(g)⟶CH4(g)+H2O;ΔH4=−52kJ mol −.So, the correct answer is (d)