Question

The rate of a certain reaction is given by, rate $=k[H^{+}{]}^n$.
The rate increases $100$ times when the $pH$ changes from $3$ to $1$. The order $(n)$ of the reaction is

• A. 2
• B. 0
• C. 1
• D. 1.5

Answer 1

Answer: (C)

Given, rate $=k{\left[H^{+}\right]}^n$
Initial $pH=3$ so $\left[H^{+}\right]=1\times 10^{-3}$, initial rate $=r_1$
Final $pH=1$ so $\left[H^{+}\right]=1\times 10^{-1}$
Final rate $=r_2=100r_1$
On substituting values, we get
$r_1=k{\left[1\times 10^{-3}\right]}^n...(i)$
$r_2=100r_1=k{\left[10^{-1}\right]}^n...(ii)$
Dividing eqs (i) by (ii),
$\frac{r_1}{100r_1}={\left[\frac{1\times 10^{-3}}{1\times 10^{-1}}\right]}^n$
$\frac{1}{100}={\left[\frac{10^{-2}}{1}\right]}^n$
$\Rightarrow {\left[\frac{1}{100}\right]}^1={\left[\frac{1}{100}\right]}^n$
$\therefore n=1$
Thus, the reaction is of first order.