For the reaction $\mathrm{A}_{(\mathrm{g})} \rightarrow \mathrm{B}_{(\mathrm{g})}$, the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of $\Delta_{\mathrm{r}}\mathrm{G}$ for the reaction at 300 K and 1 atm in $\mathrm{J}\mathrm{~mol}^{-1}$ is -xR, where x is ____. (Rounded off to the nearest integer) $\left[\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\text{ and }\ln 10=2.3\right]$

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Chemical Thermodynamics

For the reaction

A_{(g)} \to B_{(g)}

, the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of

Δ_{r} G

for the reaction at 300 K and 1 atm in

J mol^{- 1}

is -xR, where x is ____.
(Rounded off to the nearest integer)

[R = 8.31 J mol^{- 1} K^{- 1} and ln 10 = 2.3]

Answer: 1380

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

Δ_{r} G^{\circ} = - R T ln K_{e q}

Given:

Calculation:

Δ_{r} G^{\circ} = - R T ln (1 0^{2})

= - R T \times 2 ln (10)

= - 8.31 \times 300 \times 2 \times 2.3

= - 300 \times 2 \times 2.3 \times R

= - 1380 R J mol^{- 1}

Therefore,

x = 1380

Answer: 1380

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Integer answer