Question

A certain reaction has the following general form:

$\mathbf{aA}\mathbf{\to }\mathbf{bB}$

At a particular temperature and ${\left[A\right]}_{\mathbf{0}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{M}$, concentration v/s time data were collected for this reaction and a plot of $\mathbf{ln}\left[A\right]$ versus time resulted in a straight line with a slope value.

a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction.

b. Calculate the half-life for this reaction.

c. How much time is required for the concentration of A to decrease to $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{M}$?

Short answer

a.

Rate law will be $\mathrm{Rate}=K\left[A\right]$$Rate=K\left[A\right]$

Integrated rate law $\ln \left[A\right]=-Kt+\ln \left[A_0\right]$

Rate constantK =$-2.97\times {10}^{-2}{\min }^{-1}$.

b. The half-life for this reaction is 23.3 min

c. The time required for the concentration of A to decrease to $2.50\times {10}^{-3}M$is 70.0 min

Step-by-step solution

Determination of the rate law 

Rate law: Rate of reaction is directly proportional to the reactant of concentration

Order of reaction: The sum of the power of the reactants that takes part in a chemical reaction

$\mathrm{aA}\to \mathrm{bB}$

The reaction is first order with respect to A, as the slope is obtained by plotting $\ln \left[\mathrm{A}\right] \mathrm{vs} \mathrm{time}$. The slope is equal to K.

Therefore, the rate law will be

$\mathrm{Rate}=\mathrm{K}\left[\mathrm{A}\right]$

Determination of the integrated rate law and the value of the rate constant for the reaction

First order integrated rate law derivation

We have

$\begin{array}{rcl}\mathrm{Rate}& =& \mathrm{K}\left[\mathrm{A}\right]\\ \frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}& =& -\mathrm{K}\left[\mathrm{A}\right]\\ \underset{\left[{\mathrm{A}}_0\right]}{\overset{\left[\mathrm{A}\right]}{\int }}\frac{\mathrm{d}\left[\mathrm{A}\right]}{\left[\mathrm{A}\right]}& =& -\mathrm{K}\underset{0}{\overset{\mathrm{t}}{\int }}\mathrm{dt}\end{array}$

After integrating, we have

$\begin{array}{rcl}\ln \left[A\right]-\ln \left[A_0\right]& =& -Kt\\ \ln \left[A\right]& =& -Kt+\ln \left[A_0\right]\end{array}$

Therefore, the graph or plot between $\ln \left[\mathrm{A}\right] \mathrm{vs} \mathrm{time}$is found to be a straight line.

Therefore, the value of slope (K) =$-2.97\times {10}^{-2}{\min }^{-1}$.

Determine the half-life for the reaction

Half-life: The time required for a reactant to react to half its original concentration is called the half-life.

The formula of half-life for first order reaction:

$\begin{array}{rcl}{t}_{\frac{1}{2}}& =& \frac{0.693}{K}\\ & =& \frac{0.693}{\left(2.97\times {10}^{-2}\right)}\\ & =& 23.3 \min \end{array}$

Time required for the concentration of A to decrease to  2.50×10-3M

For a first-order reaction $\mathrm{aA}\to \mathrm{bB}$

$\ln \left[A\right]=-Kt+\ln \left[A_0\right]$(Integrated rate law)

We can calculate time for the reactant concentration decreases to$2.50\times {10}^{-3}\mathrm{M}$.

Given data

[A₀] =initial concentration=$2.00\times {10}^{-2}\mathrm{M}$

[A]=final concentration=$2.50\times {10}^{-3}\mathrm{M}$

K =$-2.97\times {10}^{-2}{\min }^{-1}$.

Plugging respective values in the integrated rate law equation, we get

$\begin{array}{rcl}\ln \left[2.50\times {10}^{-3}\right]& =& -\left(2.97\times {10}^{-2}\right)t+\ln \left[2.00\times {10}^{-2}\right]\\ t& =& 70.0 \min \end{array}$