Given are the following informations at 298 K. N2O4(g)⇌2NO2(g)ΔfH∘/kJmol−1 9.16 33.18ΔfG∘/kJmol−1 97.54 51.31The reaction is carried out in a closed vessel starting from N2O4.The standard equilibrium constant Kp∘ at 298 K is

Kp∘=exp⁡−ΔfG∘/RT =exp⁡−5.08×103Jmol−1/8.314JK−1mol−1(298K) =exp⁡(−2.05)=0.129

Given are the following informations at 298 K. N2O4(g)⇌2NO2(g)ΔfH∘/kJmol−1 9.16 33.18ΔfG∘/kJmol−1 97.54 51.31The reaction is carried out in a closed vessel starting from N2O4.The standard equilibrium constant Kp∘ at 298 K is

Kp∘=exp⁡−ΔfG∘/RT =exp⁡−5.08×103Jmol−1/8.314JK−1mol−1(298K) =exp⁡(−2.05)=0.129

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Given are the following informations at 298 K.
$\begin{array}{llr} N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g) \\ Δ_{f} H^{\circ} / kJ {mol}^{- 1} 9.16 33.18 \\ Δ_{f} G^{\circ} / kJ {mol}^{- 1} 97.54 51.31 \end{array}$
The reaction is carried out in a closed vessel starting from $N_{2} O_{4}$ .
The standard equilibrium constant $K_{p}^{\circ}$ at 298 K is

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0.29

0.129

2.43

5.67

answer is B.

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$K_{p}^{\circ} = \exp (- Δ_{f} G^{\circ} / R T) = \exp (- 5.08 \times 10^{3} J {mol}^{- 1} / (8.314 {JK}^{- 1} {mol}^{- 1}) (298 K)) = \exp (- 2.05) = 0.129$

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Given are the following informations at 298 K. N2O4(g)⇌2NO2(g)ΔfH∘/kJmol−1 9.16 33.18ΔfG∘/kJmol−1 97.54 51.31The reaction is carried out in a closed vessel starting from N2O4.The standard equilibrium constant Kp∘ at 298 K is

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