Question

Ammonium ion $\left( NH _{4}^{+}\right)$ reacts with nitrite ion $\left( NO _{2}^{-}\right)$ in aqueous solution according to the equation $NH _{4}^{+}(a q)+ NO _{2}^{-}(a q) \longrightarrow N _{2}(g)+2 H _{2} O (l)$ The following initial rates of reaction have been measured for the given reactant concentrations. Which of the following is the rate law for this reaction?

• A. Rate $= k [NH_4^{+}][NO_2^{-}]^4$
• B. Rate $= k [NH_4^{+}][NO_2^{-}]$
• C. Rate $= k [NH_4^{+}][NO_2^{-}]^2$
• D. Rate $= k [NH_4^{+}]^2 [NO_2^{-}]$

Answer 1

Answer: (C)

Rate $= k [NH_4^{+}][NO_2^{-}]^2$

Answer 2

Let, the order of reaction wrt $\left[ NH _{4}^{+}\right]_{\text {v }}\left[ NO _{2}^{-}\right]$ be $x$

and $y$ respectively.

Then, rate law, $r=k\left[ NH _{4}^{+}\right]^{X}\left[ NO _{2}^{-}\right]^{Y}$

On substituting the values of different experiments,

(i) $0.020=k[0.010]^{X}[0.020]^{Y}$

(ii) $0.020=k[0.015]^{x}[0.020]^{Y}$

(iii) $0.005=k[0.010]^{x}[0.010]^{Y}$

Dividing E (i) by E (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 E (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

$r=k\left[ NH _{4}^{+}\right]\left[ NO _{2}^{-}\right]^{2}$