
The difference between heats of reaction at constant pressure and constant volume for the reaction:
$2{C_6}{H_6}(l) + 15{O_2}(g) \to 12C{O_2}(g) + 6{H_2}O(l)$ at ${25^ \circ }C$ in $KJ$ is:
A: $ - 7.43$
B: $ + 3.72$
C: $ - 3.72$
D: $ + 7.43$
Answer
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Hint: In this question, we have to calculate the difference between heat of reaction at constant pressures and at constant volume. So, the difference between the sum of enthalpies of a product and the sum of enthalpies of reactants at a given temperature and constant pressure is called the heat of reaction. It is denoted by $\Delta H$.
Complete step by step answer:
Basically, the heat of reaction at constant pressure is $\Delta H$ and at constant volume is $\Delta U$.
Since, for any reaction the relation between this heat is given by the following equation.
$\Delta H = \Delta U + 9(\Delta {n_g})RT$
As we have to calculate the difference,
Therefore,
$\Delta H - \Delta U = 9(\Delta {n_g})RT$
Where $(\Delta {n_g})$= change in number of gaseous moles in any reaction.
While as the given reaction is:
$2{C_6}{H_6}(l) + 15{O_2}(g) \to 12C{O_2}(g) + 6{H_2}O(l)$
Since in the above reaction, the only gaseous species are $C{O_2}$ and ${O_2}$. So
$(\Delta {n_g}) = 12 - 15$ which is equal to $ - 3.$
Hence,
By using $\Delta H - \Delta U = \left( {\Delta {n_g}} \right)RT.$
Therefore,
By substituting the value of $(\Delta {n_g})$, $R$ and $T$
$\Delta H - \Delta U = \left( { - 3} \right) \times 8.314 \times 2.98.15{\text{ J}}{\text{.}}$
By solving the above values.
$\Delta H - \Delta U = - 7.43{\text{ KJ}}$
So, the correct answer is Option A.
Note: At constant volume, the heat of reaction is equal to the change in the internal energy of the system. At constant pressure, the heat of reaction is equal to the enthalpy change of the system.
If $\Delta H > 0 \Rightarrow $ Therefore, the change in the enthalpy is positive which means that energy is being absorbed, this is called an endothermic reaction.
If $\Delta H < 0 \Rightarrow $The change in enthalpy is negative which means that energy is being expelled, this is called an exothermic reaction.
Complete step by step answer:
Basically, the heat of reaction at constant pressure is $\Delta H$ and at constant volume is $\Delta U$.
Since, for any reaction the relation between this heat is given by the following equation.
$\Delta H = \Delta U + 9(\Delta {n_g})RT$
As we have to calculate the difference,
Therefore,
$\Delta H - \Delta U = 9(\Delta {n_g})RT$
Where $(\Delta {n_g})$= change in number of gaseous moles in any reaction.
While as the given reaction is:
$2{C_6}{H_6}(l) + 15{O_2}(g) \to 12C{O_2}(g) + 6{H_2}O(l)$
Since in the above reaction, the only gaseous species are $C{O_2}$ and ${O_2}$. So
$(\Delta {n_g}) = 12 - 15$ which is equal to $ - 3.$
Hence,
By using $\Delta H - \Delta U = \left( {\Delta {n_g}} \right)RT.$
Therefore,
By substituting the value of $(\Delta {n_g})$, $R$ and $T$
$\Delta H - \Delta U = \left( { - 3} \right) \times 8.314 \times 2.98.15{\text{ J}}{\text{.}}$
By solving the above values.
$\Delta H - \Delta U = - 7.43{\text{ KJ}}$
So, the correct answer is Option A.
Note: At constant volume, the heat of reaction is equal to the change in the internal energy of the system. At constant pressure, the heat of reaction is equal to the enthalpy change of the system.
If $\Delta H > 0 \Rightarrow $ Therefore, the change in the enthalpy is positive which means that energy is being absorbed, this is called an endothermic reaction.
If $\Delta H < 0 \Rightarrow $The change in enthalpy is negative which means that energy is being expelled, this is called an exothermic reaction.
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