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Chemistry · 2021

JEE Main · 18 March 2021, Shift 2 · Q53

The gas phase reaction 2 A ( g ) ⇌ A_2( g ) at 400 K has Δ G^°=+25.2 kJ mol^-1. The equilibrium constant K_C for this reaction is \ \ \ \ × 10^-2.…

The gas phase reaction $$2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons \mathrm{A}_2(\mathrm{~g}) $$ at $\displaystyle 400 \mathrm{~K}$ has $\displaystyle \Delta \mathrm{G}^{\circ}=+25.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The equilibrium constant $\displaystyle \mathrm{K}_{\mathrm{C}}$ for this reaction is \_\_\_\_$\displaystyle \times 10^{-2}$. (Round off to the Nearest Integer). $$\begin{gathered} \text { [Use : } \mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \ln 10=2.3 \\ \log _{10} 2=0.30,1 \mathrm{~atm}=1 \mathrm{bar} \text { ] } \end{gathered} $$ $\displaystyle [$ antilog $\displaystyle (-0.3)=0.501]$
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JEE Main 2021 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.