Question

Standard heat of formation of $C{H_4}(g),C{O_2}(g)$ and water at ${25^ \circ }C$ are $ - 17.9, - 94.1$ and $ - 68.3kcalmo{l^{ - 1}}$ respectively. Calculate the heat change ( in $kcal$) in the following reaction at ${25^ \circ }C$
$C{H_4}(g) + 2{O_2}(g) = C{O_2}(g) + 2{H_2}O(l)$
(a) $ - 144.5$
(b) $ - 180.3$
(c) $ - 248.6$
(d) $ - 212.8$

Answer 1

Hint: As we know that the standard heat of formation of a compound is the change in enthalpy during the formation of one mole of the compound from its constituent elements. Standard heat of formation of compounds is always negative that means they are exothermic reactions.

Complete step by step solution:
As we are given the standard heat of formation of different compounds, we will write their respective equations-
$
  C(s) + 2{H_2}(g) \to C{H_4}(g) \\
  C(g) + {O_2}(g) \to C{O_2}(g) \\
  {H_2}(g) + \dfrac{1}{2}{O_2}(g) \to {H_2}O(l) \\
 $ $
  \Delta {H_1} = - 17.9 \\
  \Delta {H_2} = - 94.1 \\
  \Delta {H_3} = - 68.3 \\
 $ $
  .......(1) \\
  .......(2) \\
  .......(3) \\
 $

Now if we multiply equation $(1)$ with $ - 1$ , multiply equation $(2)$ with $1$ and multiply equation $(3)$ with $2$ and add them together we will get
$C{H_4}(g) + C(g) + {O_2}(g) + 2{H_2}(g) + {O_2}(g) \to 2{H_2}(g) + C(s) + C{O_2}(g) + 2{H_2}O(l)$
$ \Rightarrow C{H_4}(g) + 2{O_2}(g) \to C{O_2}(g) + 2{H_2}O(l)$
The heat of formation of this chemical equation will be $\Delta H = ( - 1)( - 17.9) + (1)( - 94.1) + (2)( - 68.3) = - 212.8$
So from the above explanation and calculation it is clear to us that the answer of the given question is $ - 212.8$ .

So, the correct option of this question is: $(d) - 212.8$

Additional information:
The enthalpy of formation of pure elements is always zero because they are already elements and don’t need to be formed . The heat reaction of an endothermic reaction is positive . The heat of an exothermic reaction is always negative. Hess’s law is very useful to determine heat of reaction. The definition of Hess’s law is that the change of enthalpy in a chemical reaction in constant pressure is independent of the pathway between the initial and final states. The energy change of the overall reaction is equal to the summation of the energy changes of the individual reactions.

Note: Always remember that the heat of formation is the heat evolved when one mole the compound is formed from its constituent elements. The heat of formation of pure elements is always zero. Exothermic reactions have negative heat of reaction and endothermic reactions have positive heat of reaction.