NEET Solved Paper 2014 Question 10
Question: Using the Gibbs energy change $ \Delta {G^{{}^\circ }}=+63.3kJ $ for the following reaction, $ Ag _2CO _3(s)r,2A{g^{+}}(aq)+CO_3^{2-}(aq) $ the $ K _{sp} $ of $ Ag _2CO _3(s) $ in water at $ 25^{o}C $ is $ (R=8.314J{K^{-1}}mo{l^{-1}}) $ [AIPMT 2014]
Options:
A) $ 3.2\times {10^{-26}} $
B) $ 8.0\times {10^{-12}} $
C) $ 2.9\times {10^{-3}} $
D) $ 7.9\times {10^{-2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \Delta {G^{{}^\circ }} $ is related to $ K _{sp} $ by the equation, $ \Delta {G^{{}^\circ }}=-2.303RT,\log K _{sp} $ Given, $ \Delta {G^{{}^\circ }}=+,63.3\text{kJ=63}\text{.3}\times 1{0^{3}}J $
Thus, substitute $ \Delta {G^{{}^\circ }}=63.3\times 10^{3}\text{J,} $ $ R=8.314J{K^{-1}}mo{l^{-1}} $ and $ T=298K,[25+273K] $ into the .
above equation to get, $ 63.3\times 10^{3}=-2.303\times 8.314\times 298,log,K _{sp} $
$ \therefore $ $ \log ,K _{sp}=-11.09 $
$ \Rightarrow ,K _{sp}=\text{antilog(-11}\text{.09)} $ $ K _{sp}=8.0\times {10^{-12}} $
BETA
