Question

${C_3}{H_8} + {O_2} \to 3C{O_2} + 4{H_2}O{\text{ }}{\text{ }}\Delta {\text{H = - 2220kJmo}}{{\text{l}}^{ - 1}}$
${C_3}{H_6} + {H_2} \to {C_3}{H_8}{\text{ }}\Delta {\text{H = - 124kJmo}}{{\text{l}}^{ - 1}}$
${H_2} + \dfrac{1}{2}{O_2} \to {H_2}O{\text{ }}\Delta {\text{H = + 285kJmo}}{{\text{l}}^{ - 1}}$
Then find the heat of combustion of ${C_3}{H_6}$:
(A) -2059$kJmo{l^{ - 1}}$
(B) 2059$kJmo{l^{ - 1}}$
(C) -4118$kJmo{l^{ - 1}}$
(D) +4118$kJmo{l^{ - 1}}$

Answer 1

Hint: We can use Hess's law of constant heat summation in order to solve this problem. Hess’ law states that if a reaction can be divided into intermediate reactions, then the enthalpy of the reaction is equal to the sum of the enthalpies of intermediate reactions occurring at same temperature.

Complete Step-by-Step Solution:
We will find the heat of combustion of ${C_3}{H_6}$ using the Hess law of constant heat summation.
- Hess’ law states that if a reaction can be divided into intermediate reactions, then the enthalpy of the reaction is equal to the sum of the enthalpies of intermediate reactions occurring at same temperature.
Now, we can see that propene (${C_3}{H_6}$) is first hydrogenated to obtain propane and then propane is combusted according to the reactions given to us.
- We know that the given reaction is combustion of propene. We know that combustion of any compound is its reaction with oxygen. We also know that its products will be carbon dioxide and water. So, we can write the reaction as
\[{C_3}{H_6} + {O_2} \to C{O_2} + {H_2}O\]
But the above given reaction is not a balanced reaction. So, we can write the balanced reaction as
\[{C_3}{H_6} + \dfrac{9}{2}{O_2} \to 3C{O_2} + 3{H_2}O\]
Now, we need to find the enthalpy of the given reaction which is the enthalpy of combustion of propene.
- We are given the question that enthalpy of hydrogenation of propene is -124$ kJmo{l^{ - 1}}$. The enthalpy of combustion of propane is given as -2220$ kJmo{l^{ - 1}}$. The enthalpy of formation of water is $ - 285 kJmo{l^{ - 1}}$ as given.
So, we can write the reverse reaction that
\[{H_2}O \to {H_2} + \dfrac{1}{2}{O_2}\]
and the $\Delta H$ for this reaction will be +285$ kJmo{l^{ - 1}}$.
Now, we can say that, the combustion enthalpy of ${C_3}{H_6}$ = -124 + (-2220) + (+285) = -2059$ kJmo{l^{ - 1}}$

Thus, we can conclude that the correct answer is (A).

Note: Note that we need the enthalpy of formation of water as water is one of the products in the combustion of propene. We are not given the enthalpy of formation of water, so we wrote the reaction in reverse direction and took the enthalpy value having the positive sign.