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5. Using the decomposition reaction and the table below, (1) calculate the rate constant of the...

Question

5. Using the decomposition reaction and the table below, (1) calculate the rate constant of the...

5. Using the decomposition reaction and the table below, (1) calculate the rate constant of the reaction, (2) calculate the r
5. Using the decomposition reaction and the table below, (1) calculate the rate constant of the reaction, (2) calculate the rate of reaction when the molar concentration of H202 is 0.500 M. (3) plot a graph of the decomposition of H202 and then calculate the approximate concentration of H2O2 at 200 seconds. Decomposition reaction: H202 (1) ► H2O(l) + O2(g) Time, s 0 60 120 180 240 300 360 420 480 540 600 [H2O2], M 0.882 0.697 0.566 0.458 0.372 0.298 0.236 0.188 0.152 0.120 0.094

Answers

To predict the rate constant of a reaction graphically using concentration and time data, we can use the following rate equations for different order reactions.

Zero \ Order : [A]_t = [A]_0 - kt

First \ Order : \ln[A]_t =\ln [A]_0 - kt

Second \ Order : \frac{1}{[A]_t} =\frac{1}{\ln [A]_0} + kt

Where

[A]_t = concentration \ at \ time \ t

[A]_0=initial \ concentration

k = rate \ constant

Hence, plotting [A]_t \ vs \ t yields a straight line with slope -k for zero order reaction.

Plotting \ln[A]_t \ vs \ t yields a straight line with slope -k for first order reaction.

Plotting \frac{1}{[A]_t} \ vs \ t yields a straight line with slope k for second order reaction.

Now, for our given data set, we can plot the above three graphs and determine the order of the reaction and rate constant by whichever plots happen to be a straight line indicating the order.

Hence, the data points we need for the plot are

Now, plotting [H2O2] vs t, we get the following graph.

[H202],M 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 100 200 300 400 500 600 700 time, s H202], M

Note that the graph is not a straight line. Hence, clearly the reaction is not zero order.

Now, we will test for first order kinetics by plotting ln[H2O2] vs. t. Hence, the plot we get is

php8EPC4a.png

(1).

Note that the plot is a straight line represented by the fit equation, y = -0.0037 x -0.1195 .

Hence, the reaction obeys first order kinetics.

Note: we need not check for the second order kinetics as we found out that the reaction is indeed first order.

We can calculate the rate constant of the reaction from the slope of the graph.

Note that the slope equals negative of the rate constant. Also note that kt is unitless. Hence, unit of the rate constant must be sec-1.

Since the slope is -0.0037, the rate constant of the reaction is k = 0.0037 \ sec^{-1} .

(2).

Now that we know the reaction is first order with respect to H2O2, we can write the following rate law.

rate = k [H_2O_2]^1

Hence, we can calculate the rate of the reaction at [H_2O_2] = 0.500 \ M as follows

rate = k [H_2O_2] = 0.0037 \ sec^{-1} \times 0.500 \ M = {\color{Red} 1.85 \times 10^{-3} \ M \cdot s^{-1}}

(3).

The graph of decomposition of H2O2 showing the approximate concentration of H2O2 at t =200 s can be shown as follows:

Decomposition of H2O2 1 0.9 0.8 0.7 0.42 M, 200 s 0.6 0.5 0.4 0.3 0.2 0.1 0 100 200 300 400 500 600 700 time, s H202], M

Hence, the approximate concentration of H2O2 at t=200 s is about 0.42 M.


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