Question

The initial rate for a reaction is equal to the slope of the tangent line at t< 0 in a plot of [A] versus time. From calculus,

$\mathbf{Initial}\mathbf{}\mathbf{rate}\mathbf{=}\mathbf{-}\mathbf{ }\frac{\mathbf{d}\mathbf{\left[}\mathbf{A}\mathbf{\right]}}{\mathbf{dt}}$

Therefore, the differential rate law for a reaction is

$\mathbf{Rate}\mathbf{=}\mathbf{-}\mathbf{ }\frac{\mathbf{d}\mathbf{\left[}\mathbf{A}\mathbf{\right]}}{\mathbf{dt}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{k}\mathbf{[}\mathbf{A}{\mathbf{]}}^{\mathbf{n}}$

Assuming you have some calculus in your background, derive the zero, first, and second-order integrated rate laws using the differential rate law.

Short answer

The integrated rate law for zero, first, and second order reactions are derived in the following steps using the differential rate law using the calculus.

Step-by-step solution

Deriving the zero-order integrated rate law

$\mathrm{Initial}\mathrm{rate}=- \frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}$

$\mathrm{Differential}\mathrm{rate}\mathrm{law}=- \frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}} = \mathrm{k}[\mathrm{A}{]}^{\mathrm{n}}$

For zero order, n = 0

$\mathrm{Rate}=- \frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}} = \mathrm{k}[\mathrm{A}{]}^0 = \mathrm{k}$

$- \frac{d\left[A\right]}{dt} = k$

Integrating both sides with initial concentration $\left[A\right]={\left[A\right]}_0$at $t=0$and final concentration $\left[A\right]={\left[A\right]}_t\mathrm{at}t=t$

$\begin{array}{rcl}- d\left[A\right] & =& kdt\\ \underset{{\left[A\right]}_0}{\overset{{\left[A\right]}_t}{\int }}d\left[A\right] & =& - k\underset{0}{\overset{t}{\int }}dt\\ \left[A\right] - [A{]}_0 & =& - k (t-0)\\ [A{]}_0 - \left[A\right] & =& kt\end{array}$

This is the integrated rate law for the zero-order reaction.

Deriving the first-order integrated rate law

$\mathrm{Initial}\mathrm{rate}=- \frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}$

$\mathrm{Differential}\mathrm{rate}\mathrm{law}=- \frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}} = \mathrm{k}[\mathrm{A}{]}^{\mathrm{n}}$

For the first order, n = 1

$\mathrm{Rate}=- \frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}} = \mathrm{k}[\mathrm{A}{]}^1$

$- \frac{\mathrm{d}\left[\mathrm{A}\right]}{\left[\mathrm{A}\right]} = \mathrm{kdt}$

Integrating both sides with initial concentration$\left[A\right]={\left[A\right]}_0$ at$t=0$ and final concentration $\left[A\right]={\left[A\right]}_t\mathrm{at}t=t$.

$\begin{array}{rcl} \underset{{\left[A\right]}_0}{\overset{\left[A\right]}{\int }}\frac{d\left[A\right]}{\left[A\right]} & =& - k\underset{0}{\overset{t}{\int }}t\\ \left[ In\left[A\right] \right]{ }_{{\left[A\right]}_0}^{\left[A\right]}& =& - k{\left[t\right]}_0^t\\ In\left[A\right] - In[A{]}_0 & =& - k(t - 0)\end{array}$

$In[A{]}_0 - In\left[A\right] = kt$

This is the integrated rate law for the first-order reaction.

Deriving the second-order integrated rate law

$\mathrm{Initial}\mathrm{rate}=- \frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}$

$\mathrm{Differential}\mathrm{rate}\mathrm{law}=- \frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}} = \mathrm{k}[\mathrm{A}{]}^{\mathrm{n}}$

For the first order, n = 2

$\mathrm{Rate}=- \frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}} = \mathrm{k}[\mathrm{A}{]}^2$

$- \frac{\mathrm{d}\left[\mathrm{A}\right]}{{\left[\mathrm{A}\right]}^2} = \mathrm{kdt}$

Integrating both sides with initial concentration localid="1663759549607" $\left[A\right]={\left[A\right]}_0$at localid="1663760185402" $t=0$and final concentration $\left[\mathrm{A}\right]={\left[\mathrm{A}\right]}_{\mathrm{t}}\mathrm{at}\mathrm{t}=\mathrm{t}$

$\begin{array}{rcl}\underset{{\left[A\right]}_0}{\overset{\left[A\right]}{\int }}\frac{d\left[A\right]}{{\left[A\right]}^2} & =& k(t-0)\\ {\left[\frac{1}{\left[A\right]}\right]}_{\left[A\right]{ }_0}^{\left[A\right]} & =& - k[t{]}_0^t\\ \frac{1}{\left[A\right]} - \frac{1}{{\left[A\right]}_0}& =& k(t-0)\\ \frac{1}{\left[A\right]} - \frac{1}{{\left[A\right]}_0}& =& kt\\ \therefore \frac{1}{\left[A\right]} & =& kt + \frac{1}{{\left[A\right]}_0}\end{array}$

This is the integrated rate law for the second-order reaction.