Solve the following


For the reaction $\mathrm{A}_{(g)} \rightarrow \mathrm{B}_{(\mathrm{g})}$, the value of the equilibrium constant at $300 \mathrm{~K}$ and $1 \mathrm{~atm}$ is equal to $100.0$. The value of $\Delta_{r} G$ for the reaction at $300 \mathrm{~K}$ and 1 atm in $\mathrm{Jmol}^{-1}$ is $-\mathrm{xR}$, where $\mathrm{x}$ is__________________. (Rounded off to the nearest integer)

$\left[\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right.$ and $\left.\ln 10=2.3\right]$



$\Delta G^{\circ}=-R T$ In Keq.

$=-R \times 300 \times \ln \left(10^{2}\right)$

$=300 \times 2 \times 2.3 \times(-R)$

$=-1380 \mathrm{R}$

$x=1380$ ans.

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