Question

Which is the correct expression for integrated rate law of $n^{th}$ order reaction?
$A\to \text{Product(s)}$

Here,
$a$ is initial concentration of the reactant
$x$ is concentration of reactant consumed at time ‘t’
$k$ is the rate constant

• A. $\frac{1}{(a-x{)}^{n-1}}-\frac{1}{a^{n-1}}=(n-1)kt$
• B. $\frac{1}{(a+x{)}^{n-1}}-\frac{1}{a^{n-1}}=(n+1)kt$
• C. $(a-x{)}^{n-1}=nkt$
• D. $\frac{1}{(a-x{)}^{n+1}}+\frac{1}{a^{n-1}}=(n-1)kt$

Answer 1

The correct option is A $\frac{1}{(a-x{)}^{n-1}}-\frac{1}{a^{n-1}}=(n-1)kt$

For general reaction,
$A\to \text{Product(s)}$

According to rate law
$Rate\left(R\right)=k[A{]}^n$
where,
n : Order
k : Rate constant
[A] : Concentration of reactant A

$A\to \text{Product(s)}$
$time=0atime=\left(t\right)a-x$

a : Initial concentration of A
x : Concentration of reactant consumed at time (t)

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

At any time t,
$-\frac{d(a-x)}{dt}=k(a-x{)}^n....eqn(1)$

As 'a' is constant, equation (1) becomes,
$\frac{dx}{dt}=k(a-x{)}^n$
$\frac{dx}{(a-x{)}^n}=kdt$

Integrating from time, $t=0tot$
$-\underset{0}{\overset{x}{\int }}\frac{dx}{(a-x{)}^n}=k\underset{0}{\overset{t}{\int }}dt$

$-\frac{1}{-n+1}\left[\right(a-x{)}^{-n+1}{]}_0^x=k[t{]}_0^t$

$\frac{1}{n-1}\left[\right(a-x{)}^{-n+1}-a^{-n+1}]=kt$

$\frac{1}{n-1}\left[\frac{1}{(a-x{)}^{n-1}}\right]-\left[\frac{1}{a^{n-1}}\right]=kt$

$\left[\frac{1}{(a-x{)}^{n-1}}\right]-\left[\frac{1}{a^{n-1}}\right]=(n-1)kt.....eqn\left(2\right)$

Equation (2) gives the expression for integrated rate law of $n^{th}$ order reaction.

Another form of integrated rate law for $n^{th}$ order reaction is
$kt=\frac{1}{n-1}[\frac{1}{[A{]}^{n-1}}-\frac{1}{[A{]}_0^{n-1}}]$

The equation is valid when $n\ne 1$