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Question: The enthalpy of formation of methane at constant pressure and \(300K\) is\( - 75.83KJ\) . What wil...

The enthalpy of formation of methane at constant pressure and 300K300K is−75.83KJ - 75.83KJ .
What will be the heat of formation at constant volume?
(R=8.3 Jmol−1K−1)\left( {R = 8.3{\text{ }}Jmo{l^{ - 1}}{K^{ - 1}}} \right)
A.53.34 kJ53.34{\text{ kJ}}
B.−53.34 kJ - 53.34{\text{ kJ}}
C.73.34 kJ73.34{\text{ kJ}}
D.−73.34 kJ - 73.34{\text{ kJ}}

Explanation

Solution

First use the formula of ideal gas equation for one mole-
⇒ΔPV=RT\Rightarrow \Delta PV = RT
Then put this value in the formula of enthalpy of formation which is given as-
⇒ΔH=ΔU−ΔPV\Rightarrow \Delta H = \Delta U - \Delta PVWhere ΔH\Delta H is change in enthalpy, ΔU\Delta U is change in internal energy
Then solve forΔU\Delta U. Use the formula of heat of formation-
⇒Q=ΔU+PΔV\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 systemΔV=0\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 - 75.83KJ
And temperature=300K300K
We have to find the heat formation at constant volume.
First we will convert the kilojoule into joule.
We know that 1kJ = 1000J1{\text{kJ = 1000J}}
So on applying this we get,
The enthalpy of formation of methane at constant pressure=−75.83×1000J - 75.83 \times 1000J
On multiplication we get,
Enthalpy of formation of methane at constant pressure=−75830 J - 75830{\text{ J}}
Now we know from ideal gas equation that-
⇒ΔPV=RT\Rightarrow \Delta PV = RT For one mole --- (i)
Now we also know the formula of enthalpy of formation at constant pressure is given as-
⇒ΔH=ΔU−ΔPV\Rightarrow \Delta H = \Delta U - \Delta PV -- (ii) Where ΔH\Delta H is change in enthalpy, ΔU\Delta U is change in internal energy
On substituting the value of eq. (i) in eq. (ii) we get,
⇒ΔH=ΔU−RT\Rightarrow \Delta H = \Delta U - RT
Now putting the given values in the equation we get,
⇒−75830=ΔU−(8.3×300)\Rightarrow - 75830 = \Delta U - \left( {8.3 \times 300} \right)
On adjusting we get,
⇒−ΔU=+75830−(8.3×300)\Rightarrow - \Delta U = + 75830 - \left( {8.3 \times 300} \right)
On multiplying by negative sign both side we get,
⇒ΔU=−75830+(8.3×300)\Rightarrow \Delta U = - 75830 + \left( {8.3 \times 300} \right)
Now on solving we get,
⇒ΔU=−75830+2490\Rightarrow \Delta U = - 75830 + 2490
On subtraction we get,
⇒ΔU=−73340\Rightarrow \Delta U = - 73340 J
Now we know that at constant volume ΔV=0\Delta V = 0 then heat of formation is given as-,
⇒Q=ΔU+PΔV=ΔU\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 ΔU\Delta U
So we can write-
⇒Q=−73340\Rightarrow Q = - 73340 J
Now we know that 1kJ = 1000J1{\text{kJ = 1000J}}
Then 1 J = 11000kJ1{\text{ J = }}\frac{1}{{1000}}{\text{kJ}}
Then Q=−733401000kJQ = \frac{{ - 73340}}{{1000}}{\text{kJ}}
On solving we get,
⇒Q=−73.340 kJ\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 ΔH\Delta H whileQQ 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-
⇒ΔU=Q−PΔV\Rightarrow \Delta U = Q - P\Delta V where the notations have usual meaning.