Question

Data for the following reaction was collected at 25.0 degrees celsius.

Br_2(aq) -> products

Time (s)	[Br2] (M)
0	0.0120
50	0.0101
100	0.00846
150	0.00710
200	0.00596
250	0.00500
300	0.00420
350	0.00353
400	0.00296

(a) Make three plots of this data and use the appropriate plot to determine the rate law expression. (please tell me how to graph this on excel?)

(b) What is the value of the rate constant?

Answer 1

(a)

For a zero order reaction, the integrated rate law expression is written as

Where

k is the rate constant

Hence, if we plot [Br₂] in y axis vs time t in x axis, we should get a straight line with slope -k if the reaction is zero order.

Similarly, for a first order reaction, the integrated rate law expression is

Hence, if we plot ln[Br₂] in y axis and t in x-axis, we should get a straight line with slope -k if the reaction follows first order kinetics.

Now, for a second order reaction, the integrated rate law expression is

Hence, if we plot 1/[Br₂] in y axis vs t in x-axis, we will get a straight line, if the reaction is indeed second order.

Hence, we can tabulate the required data points to plot the three graphs as follows:

Now, using excel, we select the first column and second column and insert a scatter plot as follows to see if the reaction is zero order. We get the following result.

Note that we have also inserted a linear trendline to the data set to see how the plotted data deviates from straight line.

Clearly, the data points do not conform to a straight line. Hence, the reaction is not a zero order reaction.

Now, we take the first column and third column to check if the reaction is first order.

The result is

Note that the data points for the third column clearly conforms to the straight line of the trendline with a R² - value of 1.

Hence, the reaction must be first order with respect to [Br₂].

Now, we can also plot column 1 with column 4 to confirm that the reaction is not a second order reaction.

Hence, the result is

Hence, clearly the data points do not conform to a straight line predicted by the trendline.

Hence, the reaction is not second order.

Hence,since the reaction order is 1 with respect to [Br₂], the rate law expression can be written as

(b)

Note that the slope of the plot of ln[Br2] vs time equals the negative of rate constant k.

Hence, we can write

Now, the trendline equation for the first order graph is

Where y is ln[Br₂] and x is time in seconds.

Hence, the slope of the plot is -0.0035 with a unit of s^-1.

Hence,

Hence, the rate constant of the reaction is