| Using the Gibbs energy change [AIPMT 2014] |
| \[\Delta {{G}^{{}^\circ }}=+63.3\,\,kJ\] for the following reaction,\[A{{g}_{2}}C{{O}_{3}}(s)r\,2A{{g}^{+}}(aq)+CO_{3}^{2-}(aq)\] |
| the \[{{K}_{sp}}\] of \[A{{g}_{2}}C{{O}_{3}}(s)\] in water at \[{{25}^{o}}C\] is \[(R=8.314\,J{{K}^{-1}}\,mo{{l}^{-1}})\] |
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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