Question

The difference between heat of reaction at constant pressure and constant volume for the reaction $2{C_6}{H_6}\left( l \right) + 15{O_2}\left( g \right) \to 12C{O_2}\left( g \right) + 6{H_2}O\left( l \right) at\; {25^\circ }C$ in KJ is:
(A) + 7.43
(B) + 3.72
(C) - 7.43
(D) - 3.72

Answer 1

First, we need to know what the heat of the reaction is:
Heat of reaction: The amount of heat absorbed or evolved in a reaction when the stoichiometric numbers of moles of reactants, indicated by the balanced chemical equation of the reaction, are completely converted into products at a given condition is called the heat of reaction at that condition.

Complete step by step answer:
Explanation: let us consider the balanced equation for the following reaction: $aA + bB \to cC + dD$
As per the definition, the amount of heat absorbed or evolved when a mole A reacts with b mole B and forms c mole C and d mole D is called the heat of the reaction.
Now, there are two types of reactions –

Heat of reaction constant volume$\left( {q_v} \right)$: The heat of reaction measured when the reaction is carried out at a fixed temperature & fixed volume.
Heat of reaction at constant pressure$\left( {q_p} \right)$: The heat of reaction measured when the reaction is carried out at a fixed temperature & fixed pressure.
Now, heat of reaction at constant pressure $\left( {q_p} \right)$ is always equal to the change of enthalpy of the system $\left( {\Delta H} \right)$. Thus, $\left( {q_p = \Delta H} \right)$
And, heat of reaction at constant pressure $\left( {q_v} \right)$ is always equal to the change of internal energy of the system $\left( {\Delta U} \right)$. Thus, $\left( {q_v = \Delta U} \right)$
Now, according to the definition of enthalpy change –
$\Delta H = \Delta U + P\Delta V$
or,$\Delta H - \Delta U = P\Delta V$
or,$q_p - q_v = P\Delta V$
Thus, the difference between $q_p$ & $q_v$ is equal to $P\Delta V$.
Therefore, now we have to calculate the $P\Delta V$ value of the given reaction. It is easy.
We know from the ideal gas equation –
PV = nRT
So, let at constant pressure P and temperature $T,\;{V_1}$ volume of a gas contains ${n_1}$ mole of its atoms/molecules. And ${V_2}$ volume of the gas contains ${n_2}$ moles of the atoms/molecules.

Thus, $P{V_1} = {n_1}RT.........\left( 1 \right)$ & $P{V_2} = {n_2}RT.........\left( 2 \right)$
Subtracting (1) from (2): $P({V_2} - {V_1}) = ({n_2} - {n_1})RT$
Or, $P\Delta V = \Delta nRT$

Since we know the value of $R\left( {8.314Jmo{l^{ - 1}}{K^{ - 1}}} \right)$ & the T is given $\left( {{{25}^\circ }C = 298k} \right)$, we have to calculate only $\Delta n$ from the reaction.
$\Delta n$= sum of the number of moles of gaseous product - sum of the number of moles of gaseous reactants
Thus,

$2{C_6}{H_6}\left( l \right) + \underline{\underline {15}} {O_2}\left( g \right) \to \underline{\underline {12}} C{O_2}\left( g \right) + 6{H_2}O\left( l \right) at\; {25^\circ }C$

$\Delta n = 12 - 15 = - 3$
$\therefore P\Delta V = \Delta nRT = - 3 \times 8.314 \times 298\times 10^{-3} = - 7.43KJ$

The correct answer is $\left( C \right){\text{ }} - 7.43$KJ.

Note: It is to be noted that we should only consider the number of moles of gaseous products & reactants since, the change in volume of solid & liquid is negligible.