A) \[3.2\times {{10}^{-26}}\]
B) \[8.0\times {{10}^{-12}}\]
C) \[2.9\times {{10}^{-3}}\]
D) \[7.9\times {{10}^{-2}}\]
Correct Answer: B
Solution :
\[\Delta {{G}^{{}^\circ }}\] is related to \[{{K}_{sp}}\] by the equation, \[\Delta {{G}^{{}^\circ }}=-2.303RT\,\log {{K}_{sp}}\] Given, \[\Delta {{G}^{{}^\circ }}=+\,63.3\text{kJ=63}\text{.3}\times \text{1}{{\text{0}}^{3}}\text{J}\] 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, \[63.3\times {{10}^{3}}=-2.303\times 8.314\times 298\,log\,{{K}_{sp}}\] \[\therefore \] \[\log \,{{K}_{sp}}=-11.09\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,{{K}_{sp}}=\text{antilog(-11}\text{.09)}\] \[{{K}_{sp}}=8.0\times {{10}^{-12}}\]
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