Question

Ammonium ion $\left(N{H}_4^{+}\right)$ reacts with nitrite ion $\left(N{O}_2^{-}\right)$ in aqueous solution according to the equation
$N{H}_4^{+}\left(aq\right)+N{O}_2^{-}\left(aq\right)\to N_2\left(g\right)+2H_{2}O\left(l\right)$
The following initial rates of reaction have been measured for the given reactant concentrations.

Expt No.	$\left[N{H}_4^{+}\right],\left(M\right)$	$\left[N{O}_2^{-}\right],\left(M\right)$	Rate $\left(M/hr\right)$
$1$	$0.010$	$0.020$	$0.020$
$2$	$0.015$	$0.020$	$0.030$
$3$	$0.010$	$0.010$	$0.005$

Which of the following is the rate law for this reaction?

• A. $Rate=k\left[N{H}_4^{+}\right][N{O}_2^{-}{]}^4$
• B. $Rate=k\left[N{H}_4^{+}\right]\left[N{O}_2^{-}\right]$
• C. $Rate=k\left[N{H}_4^{+}\right][N{O}_2^{-}{]}^2$
• D. $Rate=k\left[N{H}_4^{+}{]}^2\right[N{O}_2^{-}]$
• E. $Rate=k\left[N{H}_4^{+}{]}^{\frac{1}{2}}\right[N{O}_2^{-}{]}^{\frac{1}{4}}$

Answer 1

The correct option is C $Rate=k\left[N{H}_4^{+}\right][N{O}_2^{-}{]}^2$

Let the order of reaction $wrt\left[N{H}_4^{+}\right],\left[N{O}_2^{-}\right]$ be $x$ and $y$ respectively.
Then, rate law, $r=k\left[N{H}_4^{-}{]}^x\right[N{O}_2^{-}{]}^y$

On substituting the values of different experiments,
(i) $0.020=k\left[0.010{]}^x\right[0.020{]}^y$
(ii) $0.030=k\left[0.015{]}^x\right[0.020{]}^y$
(iii) $0.005=k\left[0.010{]}^x\right[0.010{]}^y$

Dividing Eq. (i) by Eq. (ii) gives
$\left(\frac{0.020}{0.030}\right)={\left(\frac{0.010}{0.015}\right)}^x$
${\left(\frac{2}{3}\right)}^1={\left(\frac{2}{3}\right)}^x$ or $x=1$

Dividing Eq. (i) by (iii) gives
$\frac{0.020}{0.005}={\left(\frac{0.020}{0.010}\right)}^y$
$=4=2^y$
$(2{)}^2=(2{)}^y$
or $y=2$
On substituting the values of $x$ and $y$ in the rate law, we get
$r=k\left[N{H}_4^{+}\right][N{O}_2^{-}{]}^2$.