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03 AL chem pp 問題
發問:
03II 3a iv) (請自行找這條的詳細題目, 謝謝) suggest how u would show experimentally that the rate of the hydrolysis is also first order with respect to hydrochloric acid. 我不明白 why the ans.: k' valuses obtained vary as [H+(aq)] 就代表 the reaction is also first order w.r.t. H+(aq). 更新: Re 老爺子 thanks so much, now i know how the k' comes from, but i still dont understand: 你寫: when the concentration of H+ varies, the k' values obtained varies as [H+(aq)] and thus a = 1 can a equal to 2 or other value? isn't it because the word" varies" means a must equal to 1?
The rate equation is written as: Rate = k[H+(aq)]a[H2O(l)]b[C12H22O11(aq)]c ...... (1) a, b and c are the orders of reaction with respect to [H+(aq)], [H2O(l)] and [C12H22O11(aq)] respectively. Put k' = [H+(aq)]a[H2O(l)]b ...... (2) Then, (1) becomes: Rate = k'[C12H22O11(aq)]c ...... (3) The concentration of H2O(l) is nearly constant because H2O(l) is the solvent of the dilute solution. When a constant concentration of H+(aq) is used, k' is a constant. Thus, the rate equation can be written as (3). In this question, c is found to be 1 in (3). Therefore, the reaction is first order with respect to C12H22O11(aq). k' = [H+(aq)]a[H2O(l)]b ...... (2) When the concentration of H+ varies, the k' values obtained varies as [H+(aq)] and thus a = 1. Therefore, (2) becomes: k' = [H+(aq)] [H2O(l)]b ...... (2') Put (2') into (1), the rate equation becomes: Rate = k[H+(aq)][H2O(l)]b[C12H22O11(aq)]c The reaction is first order with respect to H+(aq). All H+(aq) ions are formed from HCl. In other words, the reaction is first order with respect to HCl. =
其他解答:
03 AL chem pp 問題
發問:
03II 3a iv) (請自行找這條的詳細題目, 謝謝) suggest how u would show experimentally that the rate of the hydrolysis is also first order with respect to hydrochloric acid. 我不明白 why the ans.: k' valuses obtained vary as [H+(aq)] 就代表 the reaction is also first order w.r.t. H+(aq). 更新: Re 老爺子 thanks so much, now i know how the k' comes from, but i still dont understand: 你寫: when the concentration of H+ varies, the k' values obtained varies as [H+(aq)] and thus a = 1 can a equal to 2 or other value? isn't it because the word" varies" means a must equal to 1?
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最佳解答:The rate equation is written as: Rate = k[H+(aq)]a[H2O(l)]b[C12H22O11(aq)]c ...... (1) a, b and c are the orders of reaction with respect to [H+(aq)], [H2O(l)] and [C12H22O11(aq)] respectively. Put k' = [H+(aq)]a[H2O(l)]b ...... (2) Then, (1) becomes: Rate = k'[C12H22O11(aq)]c ...... (3) The concentration of H2O(l) is nearly constant because H2O(l) is the solvent of the dilute solution. When a constant concentration of H+(aq) is used, k' is a constant. Thus, the rate equation can be written as (3). In this question, c is found to be 1 in (3). Therefore, the reaction is first order with respect to C12H22O11(aq). k' = [H+(aq)]a[H2O(l)]b ...... (2) When the concentration of H+ varies, the k' values obtained varies as [H+(aq)] and thus a = 1. Therefore, (2) becomes: k' = [H+(aq)] [H2O(l)]b ...... (2') Put (2') into (1), the rate equation becomes: Rate = k[H+(aq)][H2O(l)]b[C12H22O11(aq)]c The reaction is first order with respect to H+(aq). All H+(aq) ions are formed from HCl. In other words, the reaction is first order with respect to HCl. =
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