A) Rate \[=k[C{{H}_{3}}COC{{H}_{3}}][{{H}^{+}}]\]
B) Rate\[=k[CH=COC{{H}_{3}}][B{{r}_{2}}]\]
C) Rate \[=k[C{{H}_{3}}COC{{H}_{3}}][B{{r}_{2}}]{{[{{H}^{+}}]}^{2}}\]
D) Rate \[=k[C{{H}_{3}}COC{{H}_{3}}][B{{r}_{2}}]{{[{{H}^{+}}]}^{2}}\]
Correct Answer: A
Solution :
Key Idea: By comparing the rate and concentration, the order of the reaction can be calculated. Let the rate of the reaction wrt \[[C{{H}_{3}}COC{{H}_{3}}],\]\[[B{{r}_{2}}]\] and \[[{{H}^{+}}]\] are x, y and z respectively. Thus, Rate \[\propto {{[C{{H}_{3}}COC{{H}_{3}}]}^{x}}{{[B{{r}_{2}}]}^{y}}{{[{{H}^{+}}]}^{z}}\] \[5.7\times {{10}^{-5}}={{[0.30]}^{x}}{{[0.05]}^{y}}{{[0.05]}^{z}}\] ?(i) \[5.7\times {{10}^{-5}}={{[0.30]}^{x}}{{(0.10)}^{y}}{{(0.05)}^{z}}\] ?(ii) \[1.2\times {{10}^{-4}}={{[0.30]}^{x}}{{(0.10)}^{y}}{{(0.10)}^{z}}\] ?(iii) \[3.1\times {{10}^{-4}}={{[0.40]}^{x}}{{(0.05)}^{y}}{{(0.20)}^{z}}\] ?(iv) From Eqs (i) and (ii) y = 0 From Eqs (ii) and (iii) x = 1 From, Eqs (i) and (iv) x = 1 Thus, rate law \[\propto [C{{H}_{3}}COC{{H}_{3}}][{{H}^{+}}]\] \[=k[C{{H}_{3}}COC{{H}_{3}}][{{H}^{+}}]\]You need to login to perform this action.
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