Question

Does the following data fit a second-order rate law?

Trial	Time(s)	(A) (M)
1	5	0.952
2	10	0.625
3	15	0.465
4	20	0.370
5	25	0.308
6	35	0.230

Short answer

Yes. The graph between 1 / (A) vs. t is linear.

Step-by-step solution

Rate of a Reaction

The rate of reaction may be defined as the speed of the reactant reacting to obtain a product in a particular reaction at a particular time. The concentration of the reactant and product are represented into mole/L.

\({\bf{Rate = k}}{\left( {\bf{A}} \right)^{\bf{m}}}{\left( {\bf{B}} \right)^{\bf{n}}}^{}\)

Second-Order Reaction

In this reaction, the second-order reaction depend upon only one concentration of the reactant:

\({\bf{Rate = k}}{\left( {\bf{A}} \right)^{\bf{2}}}\)

Second-order reactions having the integrated rate law:

\({\bf{1 / }}\left( {\bf{A}} \right){\bf{ = kt + 1 / }}{\left( {\bf{A}} \right)_{\bf{0}}}\)

Graph

Trial	Time(s)	(A) (M)	1 / (A)
1	5	0.952	1.050
2	10	0.625	1.60
3	15	0.465	2.15
4	20	0.370	2.70
5	25	0.308	3.25
6	35	0.230	4.34