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Question: Given, \(C(s) + {O_2}(g) \to C{O_2}(g)\Delta H = + 94.2kcal\) \({H_2}(g) + \dfrac{1}{2}{O_2}(...

Given,
C(s)+O2(g)→CO2(g)ΔH=+94.2kcalC(s) + {O_2}(g) \to C{O_2}(g)\Delta H = + 94.2kcal
H2(g)+12O2(g)→H2O(l)+68.3ΔH=kcal{H_2}(g) + \dfrac{1}{2}{O_2}(g) \to {H_2}O(l) + 68.3\Delta H = kcal
CH4(g)+2O2(g)→CO2(g)+2H2O(l)ΔH=+210.8kcalC{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

Explanation

Solution

Since the heat of formation reaction (ΔH)(\Delta H) of Carbon dioxide (CO2)(C{O_2}) is 94.2 kcal, the heat of formation of water (H2O)({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 (CH4)(C{H_4}), so to find that out we will calculate
ΔH=\Delta H = Heat of formation of Carbon dioxide +2×+ 2 \times Heat of formation of Water – Heat of formation of Methane.

Complete step by step answer:
C(s)+O2(g)→CO2(g)ΔH=+94.2kcalC(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 CO2C{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
H2(g)+12O2(g)→H2O(l)+68.3ΔH=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
CH4(g)+2O2(g)→CO2(g)+2H2O(l)ΔH=+210.8kcalC{H_4}(g) + 2{O_2}(g) \to C{O_2}(g) + 2{H_2}O(l)\Delta H = + 210.8kcal
For this reaction, ΔH\Delta H is 210.8 kcal
ΔH=\Delta H = Heat of formation of Carbon dioxide +2×+ 2 \times Heat of formation of Water – Heat of formation of Methane
  ⇒−210.8=−94.2+(2×(−68.3))\; \Rightarrow - 210.8 = - 94.2 + \left( {2 \times \left( { - 68.3} \right)} \right) – Heat of formation of CH4C{H_4}
⇒\Rightarrow Heat of formation of CH4=−94.2+136.6−(−210.8)=20.8kcalC{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.