Consider the following reaction at 298 K. $\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{O}_{3(\mathrm{~g})} . \mathrm{K}_{\mathrm{p}}=2.47 \times 10^{-29}$. $\Delta_{\mathrm{r}} \mathrm{G}^{\Theta}$ for the reaction is ______ kJ. (Given R $=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$)

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Chemical Thermodynamics

Consider the following reaction at 298 K.

\frac{3}{2} O_{2 (g)} ⇌ O_{3 (g)} . K_{p} = 2.47 \times 1 0^{- 29}

Δ_{r} G^{Θ}

for the reaction is ______ kJ. (Given R

= 8.314 JK^{- 1} mol^{- 1}

)

Answer: 163

The relationship between standard Gibbs free energy change and equilibrium constant is given by:

Δ_{r} G^{⊖} = - RT ln K_{P}

Given data:

Calculation:

Δ_{r} G^{⊖} = - 8.314 \times 1 0^{- 3} \times 298 \times ln (2.47 \times 1 0^{- 29})

ln (2.47 \times 1 0^{- 29}) = ln (2.47) + ln (1 0^{- 29})

= 0.904 - 29 \times 2.303

= 0.904 - 66.787

= - 65.883

Δ_{r} G^{⊖} = - 8.314 \times 1 0^{- 3} \times 298 \times (- 65.883)

= 8.314 \times 1 0^{- 3} \times 298 \times 65.883

= 163.19 kJ

\approx 163 kJ

Answer: 163 kJ

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Integer answer