Consider the reactions C2H4(g) + 2O2(g) → 2CO(g) + 2H2O(g) ∆H = −758 kJ mol−1 2C(s) + 2H2(g) → C2H4(g) ∆H = +52 kJ mol−1 H2(g) + O2(g) → H2O(g) ∆H = −242 kJ mol−1 The enthalpy of formation of carbon monoxide is A −111 kJ mol−1 B −163 kJ mol−1 C −222 kJ mol−1 D -464 kJ mol−1

C2H4(g) + 2O2(g) → 2CO(g) + 2H2O(g)       ∆H1 = −758 kJ mol−1       1)

2C(s) + 2H2(g) → C2H4(g)                           ∆H2 = +52 kJ mol−1        2)

H2(g) + O2(g) → H2O(g)                               ∆H3 = −242 kJ mol−1     3)

Now, enthalpy of formation of carbon monoxide is given by :

∆H = ∆H1 + ∆H2 - ∆H3

∆H = ( -758 + 52 - ( -242 ) ) kJ mol−1  

∆H = −464 kJ mol−1

Therefore, the enthalpy of formation of carbon monoxide is -464 kJ mol−1.

Hence, this is the required solution.