PYQ NEET- Chemical Equilibrium-1 L-8

Question: Using the Gibb’s energy change, $\Delta \mathrm{G}^0=+63.3 \mathrm{~kJ}$, for the following reaction,#

$\mathrm{Ag}2 \mathrm{CO}{3(\mathrm{~s})} \rightleftharpoons 2 \mathrm{Ag}^{+}{ }{(\mathrm{aq})}+\mathrm{CO}3{ }^{2-}{ }{(\mathrm{aq})}$ the $\mathrm{K}{\mathrm{sp}}$ of $\mathrm{Ag}2 \mathrm{CO}{3(\mathrm{~s})}$ in water at $25^{\circ} \mathrm{C}$ is $\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$

A) $3.2 \times 10^{-26}$

B) $8.0 \times 10^{-12}$

C) $2.9 \times 10^{-3}$

D) $7.9 \times 10^{-2}$

Answer: $8.0 \times 10^{-12}$#

Solution:#

We know, $\Delta \mathrm{G}^0=-2.303 \mathrm{RT} \log \mathrm{K}{\mathrm{sp}}$ $$ \begin{aligned} & \therefore 63300=-2.303 \times 8.314 \times 298 \log \mathrm{K}{\mathrm{sp}} \ & \Rightarrow \log \mathrm{K}{\mathrm{sp}}=-11.09 \ & \Rightarrow \mathrm{K}{\mathrm{sp}}=10^{-11.09} \ & =8.0 \times 10^{-12} \end{aligned} $$