Question

A certain reaction has the following general form. aA → bB At a particular temperature and [A]0 = 2.80 ✕ 10−3 M, concentration versus time data were collected for this reaction, and a plot of 1/[A] versus time resulted in a straight line with a slope value of +3.40 ✕ 10−2 L mol−1 s−1. (a) Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. (Rate expressions take the general form: rate = k . [A]a . [B]b.) rate law:

Answer 1

Answer:

Rate law: $r=k[A]^2$

Integrated rate law: $\frac{1}{[A]}=kt+ \frac{1}{[A]_0}$

Rate constant: $k=3.40x10^{-2}\frac{L}{mol*s}$

Explanation:

Hello,

In this case, since the slope is obtained by plotting 1/[A] and it has the units L/(mol*s) or 1/(M*s), we can infer the reaction is second-order, therefore, its rate law is:

$r=k[A]^2$

The integrated rate law:

$\frac{1}{[A]}=kt+ \frac{1}{[A]_0}$

That is obtained from the integration of:

$\frac{d[A]}{dt}=-k[A]^2$

And of course, since the slope equals the rate constant, its value is:

$k=3.40x10^{-2}\frac{L}{mol*s}$

Regards.