Solveeit Logo
Login

Question

Question: Standard heat of formation of \(C{H_4}(g),C{O_2}(g)\) and water at \({25^ \circ }C\) are \( - 17.9, ...

Standard heat of formation of CH4(g),CO2(g)C{H_4}(g),C{O_2}(g) and water at 25∘C{25^ \circ }C are −17.9,−94.1 - 17.9, - 94.1 and −68.3kcalmol−1 - 68.3kcalmo{l^{ - 1}} respectively. Calculate the heat change ( in kcalkcal) in the following reaction at 25∘C{25^ \circ }C
CH4(g)+2O2(g)=CO2(g)+2H2O(l)C{H_4}(g) + 2{O_2}(g) = C{O_2}(g) + 2{H_2}O(l)
(a) −144.5 - 144.5
(b) −180.3 - 180.3
(c) −248.6 - 248.6
(d) −212.8 - 212.8

Explanation

Solution

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)+2H2(g)→CH4(g) C(g)+O2(g)→CO2(g) H2(g)+12O2(g)→H2O(l)  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) \\\ ΔH1=−17.9 ΔH2=−94.1 ΔH3=−68.3  \Delta {H_1} = - 17.9 \\\ \Delta {H_2} = - 94.1 \\\ \Delta {H_3} = - 68.3 \\\ .......(1) .......(2) .......(3)  .......(1) \\\ .......(2) \\\ .......(3) \\\

Now if we multiply equation (1)(1) with −1 - 1 , multiply equation (2)(2) with 11 and multiply equation (3)(3) with 22 and add them together we will get
CH4(g)+C(g)+O2(g)+2H2(g)+O2(g)→2H2(g)+C(s)+CO2(g)+2H2O(l)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)
⇒CH4(g)+2O2(g)→CO2(g)+2H2O(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 ΔH=(−1)(−17.9)+(1)(−94.1)+(2)(−68.3)=−212.8\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 - 212.8 .

So, the correct option of this question is: (d)−212.8(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.