JEE Main 2026 — Chemistry Physical Chemistry

Let the initial moles of $N_2O_5$N2​O5​ be $2$2.

The given reaction is:

$$2N_2O_5(g) \rightleftharpoons 2N_2O_4(g) + O_2(g)$$2N2​O5​(g)⇌2N2​O4​(g)+O2​(g)

Since $N_2O_5$N2​O5​ is $50\%$50% dissociated (degree of dissociation $\alpha = 0.5$α=0.5), the moles at equilibrium are:

Moles of $N_2O_5 = 2(1 - 0.5) = 1$N2​O5​=2(1−0.5)=1

Moles of $N_2O_4 = 2(0.5) = 1$N2​O4​=2(0.5)=1

Moles of $O_2 = 0.5$O2​=0.5

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

Given the total pressure $P = 2$P=2 atm, the partial pressures of the gases are:

$P_{N_2O_5} = \dfrac{1}{2.5} \times 2 = 0.8$PN2​O5​​=2.51​×2=0.8 atm

$P_{N_2O_4} = \dfrac{1}{2.5} \times 2 = 0.8$PN2​O4​​=2.51​×2=0.8 atm

$P_{O_2} = \dfrac{0.5}{2.5} \times 2 = 0.4$PO2​​=2.50.5​×2=0.4 atm

The equilibrium constant $K_p$Kp​ is:

$$K_p = \dfrac{(P_{N_2O_4})^2 \cdot P_{O_2}}{(P_{N_2O_5})^2} = \dfrac{(0.8)^2 \times 0.4}{(0.8)^2} = 0.4$$Kp​=(PN2​O5​​)2(PN2​O4​​)2⋅PO2​​​=(0.8)2(0.8)2×0.4​=0.4

The standard free energy change $\Delta G^\circ$ΔG∘ is given by:

$$\Delta G^\circ = -RT \ln K_p$$ΔG∘=−RTlnKp​

Using the given values:

$$\ln(0.4) = 2.303 \log(0.4) = 2.303 (\log 4 - \log 10)$$ln(0.4)=2.303log(0.4)=2.303(log4−log10)

$$\ln(0.4) = 2.303 (2 \log 2 - 1) = 2.303 (2 \times 0.30 - 1) = 2.303 (-0.4) = -0.9212$$ln(0.4)=2.303(2log2−1)=2.303(2×0.30−1)=2.303(−0.4)=−0.9212

Now, substituting the values into the $\Delta G^\circ$ΔG∘ equation ($T = 50^\circ C + 273 = 323$T=50∘C+273=323 K):

$$\Delta G^\circ = -8.314 \times 323 \times (-0.9212)$$ΔG∘=−8.314×323×(−0.9212)

$\Delta G^\circ = 2473.81$ΔG∘=2473.81 J mol$^{-1}$−1

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

Answer: $2474$2474