JEE Main 2026 — Chemical Equilibrium Question with Solution

Let the initial moles of $N_2{O}_5$ be $2$.

The given reaction is:
$2N_2{O}_5(g)\rightleftharpoons 2N_2{O}_4(g)+O_2(g)$

Since $N_2{O}_5$ is $50\%$ dissociated (degree of dissociation $\alpha =0.5$), the moles at equilibrium are:
Moles of $N_2{O}_5=2(1-0.5)=1$
Moles of $N_2{O}_4=2(0.5)=1$
Moles of $O_2=0.5$

Total moles at equilibrium $=1+1+0.5=2.5$

Given the total pressure $P=2$ atm, the partial pressures of the gases are:
$P_{N_2{O}_5}=\frac{1}{2.5}\times 2=0.8$ atm
$P_{N_2{O}_4}=\frac{1}{2.5}\times 2=0.8$ atm
$P_{O_2}=\frac{0.5}{2.5}\times 2=0.4$ atm

The equilibrium constant $K_p$ is:
$K_p=\frac{(P_{N_2{O}_4}{)}^2\cdot P_{O_2}}{(P_{N_2{O}_5}{)}^2}=\frac{(0.8{)}^2\times 0.4}{(0.8{)}^2}=0.4$

The standard free energy change $\Delta G^{\circ }$ is given by:
$\Delta G^{\circ }=-RT\ln K_p$

Using the given values:
$\ln (0.4)=2.303\log (0.4)=2.303(\log 4-\log 10)$
$\ln (0.4)=2.303(2\log 2-1)=2.303(2\times 0.30-1)=2.303(-0.4)=-0.9212$

Now, substituting the values into the $\Delta G^{\circ }$ equation ($T=50^{\circ }C+273=323$ K):
$\Delta G^{\circ }=-8.314\times 323\times (-0.9212)$
$\Delta G^{\circ }=2473.81$ J mol${}^{-1}$

The magnitude of the standard free energy change to the nearest integer is $2474$ J mol${}^{-1}$.

Answer: $2474$