Question

2. The results of the alkaline hydrolysis of ethyl nitrbenzoate at various times are reported below. Time / sec (CH3NC]/M 0.0500 100 0.0355 200 0.0275 300 0.0225 400 0.0185 500 0.0168 600 0.0148 700 0.0140 800 0.0138 a) Use a spreadsheet to plot concentration vs. time. b) Use a spreadsheet to plot the data as i) zeroth order ii) first order and iii) second order. Which graph appears to be the best line? What is the order of reaction with respect to [benzoate)? c) Use your graph to determine the rate constant. Attach your graphs.

Answer 1

(a) The plot of Concentration vs time is in the following image.

9(b)For zeroth order [CH3NC] vs t should be straight line, for first order ln[CH3NC] vs t should be straight line and for second order 1/[CH3NC] vs t should be straight line.

t(s) 20 In[CH3NC] vs t Concentration (M) vs time(s) o 200 500 1000 0.06 + [CH3NC](M) In[CH3NC] 1/[CH3NC] 0 0.05 -2.995732 100 0.0355 -3.338223 28.16901 0.0275 -3.593569 36.36364 300 0.0225 -3.79424 44.44444 400 0.0185 -3.989985 54.05405 500 0.0168 -4.086376 59.52381 600 0.0148 -4.213128 67.56757 700 0.014 -4.268698 71.42857 800 0.0138 -4.283087 72.46377 y = -4E-05x + 0.0397 R2 = 0.8075 z 0.047 y = -0.0016x - 3.2089 R2 = 0.9174 Concentration(M) S In[CH3NC] do 4 500 1000 an t(s) t(s) 1/(CH3NC] vs t y = 0.0695x + 22.638 R2 = 0.976 1/[CH3NC] + 200 400 600 800 1000 t(s)

The three plots are as drawn above in part (a). We can see that plot of 1/[CH3NC] vs t is closest to straight line whereas rest two are not. Thus the reaction is of second order.

(c) Now integrated rate law of second order reaction is 1/[CH3NC] = 1/[CH3NC]₀ +kt, so slope of plot for second order reaction would give the value of k(rate constant).

From the trendline equation obtained in the plot we can see slope is 0.0695. Thus k= 0.0695 M^-1 s^-1.

Comment in case of any doubt.