Question

The enthalpy of formation of methane at constant pressure and $300K$ is$ - 75.83KJ$ .
What will be the heat of formation at constant volume?
$\left( {R = 8.3{\text{ }}Jmo{l^{ - 1}}{K^{ - 1}}} \right)$
A.\[53.34{\text{ kJ}}\]
B.$ - 53.34{\text{ kJ}}$
C.$73.34{\text{ kJ}}$
D.$ - 73.34{\text{ kJ}}$

Answer 1

Hint: First use the formula of ideal gas equation for one mole-
$ \Rightarrow \Delta PV = RT$
Then put this value in the formula of enthalpy of formation which is given as-
$ \Rightarrow \Delta H = \Delta U - \Delta PV$Where $\Delta H$ is change in enthalpy, $\Delta U$ is change in internal energy
Then solve for$\Delta U$. Use the formula of heat of formation-
$ \Rightarrow Q = \Delta U + P\Delta V$ Where notations have usual meaning
Now we know that at constant volume, the heat of formation is equal to the change in internal energy of the system$\Delta V = 0$. On applying this we get, the heat of formation at constant volume.

Complete step by step answer:
Given, enthalpy of formation of methane at constant pressure=$ - 75.83KJ$
And temperature=$300K$
We have to find the heat formation at constant volume.
First we will convert the kilojoule into joule.
We know that $1{\text{kJ = 1000J}}$
So on applying this we get,
The enthalpy of formation of methane at constant pressure=$ - 75.83 \times 1000J$
On multiplication we get,
Enthalpy of formation of methane at constant pressure=$ - 75830{\text{ J}}$
Now we know from ideal gas equation that-
$ \Rightarrow \Delta PV = RT$ For one mole --- (i)
Now we also know the formula of enthalpy of formation at constant pressure is given as-
$ \Rightarrow \Delta H = \Delta U - \Delta PV$ -- (ii) Where $\Delta H$ is change in enthalpy, $\Delta U$ is change in internal energy
On substituting the value of eq. (i) in eq. (ii) we get,
$ \Rightarrow \Delta H = \Delta U - RT$
Now putting the given values in the equation we get,
$ \Rightarrow - 75830 = \Delta U - \left( {8.3 \times 300} \right)$
On adjusting we get,
$ \Rightarrow - \Delta U = + 75830 - \left( {8.3 \times 300} \right)$
On multiplying by negative sign both side we get,
$ \Rightarrow \Delta U = - 75830 + \left( {8.3 \times 300} \right)$
Now on solving we get,
$ \Rightarrow \Delta U = - 75830 + 2490$
On subtraction we get,
$ \Rightarrow \Delta U = - 73340$ J
Now we know that at constant volume $\Delta V = 0$ then heat of formation is given as-,
$ \Rightarrow Q = \Delta U + P\Delta V = \Delta U$
It means at constant volume the heat of formation is equal to the change in internal energy of the system. And we already know the value of $\Delta U$
So we can write-
$ \Rightarrow Q = - 73340$ J
Now we know that $1{\text{kJ = 1000J}}$
Then $1{\text{ J = }}\frac{1}{{1000}}{\text{kJ}}$
Then $Q = \frac{{ - 73340}}{{1000}}{\text{kJ}}$
On solving we get,
$ \Rightarrow Q = - 73.340{\text{ kJ}}$

Hence the correct answer is D.

Note:
Here the student may get confused between the heat of formation and enthalpy of formation. So remember that enthalpy of formation is represented by $\Delta H$ while$Q$ represents the heat required or released. They are not the same.
The value of heat of formation comes from first law of thermodynamics which is given as-
$ \Rightarrow \Delta U = Q - P\Delta V$ where the notations have usual meaning.