| The rate constant for forward and backward reactions of hydrolysis of ester are \[1.1\times {{10}^{-2}}\] and \[1.5\times {{10}^{-3}}\] per minute respectively. Equilibrium constant for the reaction is |
| \[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O\]\[\rightleftharpoons \]\[C{{H}_{3}}COOH\]\[+{{C}_{2}}{{H}_{5}}OH\] |
A) 4.33
B) 5.33
C) 6.33
D) 7.33
Correct Answer: D
Solution :
| [d] \[{{K}_{f}}=1.1\times {{10}^{-2}}\]; \[{{K}_{b}}=1.5\times {{10}^{-3}}\] |
| \[{{K}_{c}}=\frac{{{K}_{f}}}{{{K}_{b}}}=\frac{1.1\times {{10}^{-2}}}{1.5\times {{10}^{-3}}}=7.33\]. |
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