InterviewSolution
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InterviewSolution
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The hydrolyiss of an ester was carried out with `0.1 M H_(2)SO_(4)` and `0.1 M HCl` separately. Which of the following expresison between the rate constants is expected ? The rate expresison being rate `= k[H^(o+)]["ester"]`A. `k_(HCl) = k_(H_(2)SO_(4))`B. `k_(HCl) gt k_(H_(2)SO_(4))`C. `k_(HCl) lt k_(H_(2)SO_(4))`D. `k_(H_(2)SO_(4)) = 2k_(HCl)` |
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Answer» Correct Answer - B `[H_(2)SO_(4)] = 0.1 M = 0.1 xx 2 = 0.2 N` `[HCl] = 0.1 N` In case of `H_(2)SO_(4)` `r_(1) = k[H^(o+)]["Ester"]` `k_(H_(2)SO_(4)] = (r_(1))/(2 N xx ["Ester"])` In case of `HCl`, `r_(2) = k[H^(o+)]["Ester"]` `k_(HCl) = (r_(2))/(1 N xx ["Ester"])` Hence `K_(HCl) gt K_(H_(2)SO_(4)`Correct Answer - B `[H_(2)SO_(4)] = 0.1 M = 0.1 xx 2 = 0.2 N` `[HCl] = 0.1 N` In case of `H_(2)SO_(4)` `r_(1) = k[H^(o+)]["Ester"]` `k_(H_(2)SO_(4)] = (r_(1))/(2 N xx ["Ester"])` In case of `HCl`, `r_(2) = k[H^(o+)]["Ester"]` `k_(HCl) = (r_(2))/(1 N xx ["Ester"])` Hence `K_(HCl) gt K_(H_(2)SO_(4)` |
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