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 $=r_{2}=100 r_{1}$
On substituting values, we get
$r_{1}=k\left[1 \times 10^{-3}\right]^{n}\,\,\,...(i)$
$r_{2}=100 r_{1}=k\left[10^{-1}\right]^{n}\,\,\,...(ii)$
Dividing eqs (i) by (ii),
$ \frac{r_{1}}{100 r_{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.