Question

If $\Delta E$ is the heat of reaction for ${{C}_{2}}{{H}_{3}}O{{H}_{(l)}}+3{{O}_{2(g)}}\to 2C{{O}_{2(s)}}+3{{H}_{2}}{{O}_{(l)}}$ at constant volume, the $\Delta H$ (heat of reaction at constant pressure) at constant temperature is:
(a) $\Delta H=\Delta E+2RT$
(b) $\Delta H=\Delta E-2RT$
(c) $\Delta H=\Delta E+RT$
(d) $\Delta H=\Delta E-2RT$

Answer 1

Hint: Heat of reaction or the enthalpy of reaction is the heat released or absorbed during the chemical reaction at constant pressure and we can easily find the heat of reaction at constant temperature by using the formula as: $\Delta H = \Delta U + nRT$, we have given internal energy of reaction and we can find the value of n by using the given equation. Now solve it.

Complete step by step answer:
First of all, let’s discuss what is the heat of reaction. Heat of reaction may be defined as the heat evolved or released during the chemical reaction or heat absorbed during the chemical reaction at the constant temperature. It is also known as the enthalpy of reaction and is denoted as $\Delta H$. Now, if $\Delta U$ is the internal energy i.e. heat of reaction at constant volume and $p\Delta v$ is constant, then the enthalpy of the reaction is given by the formula as;
$\Delta H=\Delta E+p\Delta v$ ----------(1)
If we know the heat of reaction and values of pressure and temperature, we can easily calculate the heat of enthalpy at constant pressure.
- Now considering the statement;
For the reaction given in statement;
${{C}_{2}}{{H}_{3}}O{{H}_{(l)}}+3{{O}_{2(g)}}\to 2C{{O}_{2(s)}}+3{{H}_{2}}{{O}_{(l)}}$
We know that;
$\Delta U=\Delta E$(given)
Since, $p\Delta v$ is constant , then; ------(2)
From ideal gas, we know that;
$pv=nRT$
Put this value in equation(2), we get;
$p\Delta v=nRT$
Here, n is the no. of moles of gaseous products minus the number of moles gaseous reactants. R is the gas and T is the temperature.
For the given reaction;
No. of moles of gaseous products$=2$
No. of moles of gaseous products$=3$
So, then;
 $\begin{align}
 & n=2-3 \\
 & \text{ =-1} \\
\end{align}$
Now, put these both values in equation (1), we get;
$\begin{align}
 & \Delta H=\Delta E+nRT \\
 & \text{ =}\Delta E-RT \\
\end{align}$
So, thus if $\Delta E$ is the heat of reaction for ${{C}_{2}}{{H}_{3}}O{{H}_{(l)}}+3{{O}_{2(g)}}\to 2C{{O}_{2(s)}}+3{{H}_{2}}{{O}_{(l)}}$ at constant volume, the $\Delta H$ (heat of reaction at constant pressure) at constant temperature is $\Delta H=\Delta E-RT$.
The correct answer is option “C” .

Note: For exothermic reactions, $\Delta H=-ve$ which indicates that the heat is evolved during the chemical reaction and for endothermic reactions, $\Delta H=+ve$ which indicates that the heat is absorbed from the surroundings during the chemical reaction.