Question

Make appropriate plots or perform linear regression using these data to test them for fitting zero-, first-, and second-order rate laws. Test all three even if you happen to guess the correct rate law on the lint trial. Determine the rate constant for the reaction. Using the rate law that you have determined, calculate the half-life for the reaction. At what time will the concentration of A be 0.380?

Answer 1

Question a

part 1: zero order rate law

The integrated zero order rate law is as follows.

$[A_t]=-kt+[A_0]$

The graph between [At] and t should be linear.

1.4 1.2 0.8 0.6 0.4 0.2 8 10 12 0 4 Time

Part 2: First order rate law

The integrated first order rate law is as follows.

$ln[A_t]=-kt+ln[A_0]$

Therefore, a graph between ln [At] vs t must be linear if the reaction is first order.

0.4 0.2 12 0.2 0.4 -0.6 0.8 -1 1.2 1A

Part 3: Second Order rate law

The integrated second order rate law is as follows.

$\frac{1}{[A_t]}=kt+\frac{1}{[A_0]}$

Thus, the graph between 1/[At] vs t should be linear for a second order reaction.

3.5 0.242k+0 8032 2 = 0.9999 1.5 0.5 0 4 6 8 10 12

Thus, the highest linear regression was obtained for the second order graph (r²=0.9999). Thus, this is a second order equation.

QUESTION B

From the integrated rate law for the second order reaction, the rate constant k is given by the gradient of the graph.

$\frac{1}{[A_t]}=kt+\frac{1}{[A_0]}$

From the graph, the rate constant is,

$k=0.242/(Ms)$

QUESTION C

The half-life for the second order reactions is as follows.

$t=\frac{1}{k[A_0]}$

where [A0] is the initial concentration. Substituting,

$t=\frac{1}{0.242/(Ms)*1.24M]}$

$t=0.300s$

QUESTION D

Using the second order equation,

$\frac{1}{[A_t]}=kt+\frac{1}{[A_0]}$

Substituting,

$\frac{1}{0.380M}=0.242/Ms*t+\frac{1}{1.24M}$

$t=7.54s$