Question

Given,
 $C(s) + {O_2}(g) \to C{O_2}(g)\Delta H = + 94.2kcal$
 ${H_2}(g) + \dfrac{1}{2}{O_2}(g) \to {H_2}O(l) + 68.3\Delta H = kcal$
 $C{H_4}(g) + 2{O_2}(g) \to C{O_2}(g) + 2{H_2}O(l)\Delta H = + 210.8kcal$
The heat of formation of methane in Kcal will be:
A.45.9
B.47.8
C.20.2
D.47.3

Answer 1

Hint: Since the heat of formation reaction $(\Delta H)$ of Carbon dioxide $(C{O_2})$ is 94.2 kcal, the heat of formation of water $({H_2}O)$ is 68.3 kcal is given in the question, And now we have to find out the heat of formation of methane $(C{H_4})$, so to find that out we will calculate
 $\Delta H = $ Heat of formation of Carbon dioxide \[ + 2 \times \] Heat of formation of Water – Heat of formation of Methane.

Complete step by step answer:
 $C(s) + {O_2}(g) \to C{O_2}(g)\Delta H = + 94.2kcal$
Carbon in the form of graphite, the most stable allotropic form of carbon, is oxidized to $C{O_2}$ liberating 94.05 kcal of heat. This reaction is the formation of carbon dioxide. Here, the heat of formation reaction of Carbon dioxide is 94.2 kcal
 ${H_2}(g) + \dfrac{1}{2}{O_2}(g) \to {H_2}O(l) + 68.3\Delta H = kcal$
The oxidation i.e. combustion of hydrogen forms waters exothermically. This reaction is the formation reaction of water. Here, the heat of formation of water is 68.3 kcal
 $C{H_4}(g) + 2{O_2}(g) \to C{O_2}(g) + 2{H_2}O(l)\Delta H = + 210.8kcal$
For this reaction, $\Delta H$ is 210.8 kcal
 $\Delta H = $ Heat of formation of Carbon dioxide \[ + 2 \times \] Heat of formation of Water – Heat of formation of Methane
 \[\; \Rightarrow - 210.8 = - 94.2 + \left( {2 \times \left( { - 68.3} \right)} \right)\] – Heat of formation of $C{H_4}$
 $ \Rightarrow $ Heat of formation of $C{H_4} = - 94.2 + 136.6 - \left( { - 210.8} \right) = 20.8kcal$

Therefore, the correct answer is option (C).

Note: The enthalpy of the reaction is taken negative because the heats are added in the product side which means the reaction is exothermic and the enthalpy of the reaction is taken positive because the heats are eliminated in the product side which means the reaction is endothermic.