標題:
chemistry integrated 2
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發問:
2.)The following reaction:X + 2Y → Zwas studied by measuring the [X] as a function of time. Reactant Y was always present in excess and it was assumed that the [Y] remained essentially constant during the reaction. Shown below is a table of the [X] as a function of time, in seconds. In the first... 顯示更多 2.)The following reaction: X + 2Y → Z was studied by measuring the [X] as a function of time. Reactant Y was always present in excess and it was assumed that the [Y] remained essentially constant during the reaction. Shown below is a table of the [X] as a function of time, in seconds. In the first experiment, [Y] = 1.0 M and in the second experiment, [Y] = 2.0 M. Also shown are the plots of these data which gave linear relationships, 1/[X]versus time. ---------[Y] = 1.0 M ------[Y] = 2.0 M t (s) -----[X] (M) -------[X] (M) 2 -----0.448 ----------0.392 4 -----0.427 ---------0.336 6 -----0.409---------0.294 8 -----0.392 ---------0.261 10 -----0.376 ---------0.235 Click the link and see the graph: https://access1.lon-capa.uiuc.edu/res/uiuc/cyerkes/post_lab_4/pl4ver1-5.gif https://access1.lon-capa.uiuc.edu/res/uiuc/cyerkes/post_lab_4/pl4ver2-5a.gif Use these data to determine the form of the rate equation for this reaction. The experimental rate equation has the form: a.)rate = k [X][Y] b.)rate = k [X]2[Y] c.)rate = k [X]2[Y]2 d.)rate = k [X]2 e.)rate = k [X][Y]2 Hint:Look at the plot first. If [] vs t is linear, it is zero order in what is plotted. If ln[] vs time is linear, it is first order. If 1/[] is linear it is second order. Then find the slope of the two experiments when the concentration of the other reactant is varied. If the ratio of the slopes doesn't change, it is zero order in this second component. If the slope doubles when you double the concentration, it is first order. If the slope goes up by 4 when you double the concentration it is second order.
最佳解答:
The plot giving linear relationships between 1/[X] and time implies that the raction is of second order of [X]. Make the plot of 1/[X] vs. time for the first and second experiment to get below plot. Equations of linear regression are also shown in it. 圖片參考：http://imgcld.yimg.com/8/n/AF03901095/o/701003290019613873370730.jpg From the 1st linear equation, 1/[X] = 0.0532t + 2.1267 From the 2nd linear equation, 1/[X] = 0.2132t + 2.1239 The difference between exp1 and exp2 is [Y] concentration. From exp1 to exp2, [Y] increases by 2 times. However, coefficients of time in above two equations increases by 4 times. Therefore, the reaction rate should be of second order for [Y], i.e. rate = k[X]2[Y]2.
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