標題:
chemistry integrated
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發問:
1.)The following reaction:A + 2B → Cwas studied by measuring the [A] as a function of time. Reactant B was always present in excess and it was assumed that the [B] remained essentially constant during the reaction. Shown below is a table of the [A] as a function of time, in seconds. In this experiment,... 顯示更多 1.)The following reaction: A + 2B → C was studied by measuring the [A] as a function of time. Reactant B was always present in excess and it was assumed that the [B] remained essentially constant during the reaction. Shown below is a table of the [A] as a function of time, in seconds. In this experiment, [B] = 1.0 M. Also shown is the plot of these data which gave a linear relationship, ln[A] versus time. t (s) [A] (M) 40 0.220 80 0.162 120 0.119 160 0.088 200 0.065 Click the following link to see the graph: https://access1.lon-capa.uiuc.edu/res/uiuc/cyerkes/post_lab_4/pl4ver1-2.gif When the [B] was doubled, the rate of the reaction doubled, making it first order in B. Use these data to solve for kapp and k. Remember that you are measuring the apparent rate constant (kapp) which is related to the true rate constant (k) by the equation kapp = k [B] The apparent rate constant (kapp) and true rate constant (k) are: a.)kapp = 9.69 x 10-4 s-1, k = 1.94 x 10-3 M-1 s-1 b.)kapp = 9.69 x 10-4 s-1, k = 4.84 x 10-4 M-1 s-1 c.)kapp = 7.62 x 10-3 s-1, k = 3.81 x 10-3 M-1 s-1 d.)kapp = 7.62 x 10-3 s-1, k = 1.52 x 10-3 M-1 s-1 e.)kapp = 7.62 x 10-3 s-1, k = 7.62 x 10-3 M-1 s-1 更新: I rearrange it , hope it will make you easier understand t (s) [A] (M) 40 0.220 80 0.162 120 0.119 160 0.088 200 0.065 更新 2: t (s)--- [A] (M) 40--- 0.220 80--- 0.162 120 ---0.119 160 ---0.088 200 ---0.065
最佳解答:
The answer is e). When keeping [B] constant and plotting ln[A] against t, a st. line is obtained, this means that the reaction is first order with respect to A. i.e. Rate = kapp [A] where kapp is the slope of the above graph. Slope of the graph, kapp = (-2.73 + 1.51)/(200 - 40) = 7.63 x 10^-3 s^-1 It is first order with respect to B. Rate = k [B] [A] = kapp [A] Hence, k [B] = kapp Since [B] = 1.0 M, thus k(1.0) = 7.63 x 10^-3 s^-1 k = 7.63 x 10^-3 M^-1 s^-1 (7.63 x 10^-3 is very close to 7.62 x 10^-3.)
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