Question

The rate expression for the reaction:
$N{H}_{4}CNO\rightleftharpoons N{H}_{2}CON{H}_2$ can be derived from the mechanism:
i. $N{H}_{4}CNO\underset{k_2}{\overset{k_1}{\rightleftharpoons }}N{H}_{4}NCO$ (Fast)
ii. $N{H}_{4}NCO\stackrel{k_3}{\to }N{H}_3+HNCO$ (Fast)
iii. $N{H}_3+HNCO\stackrel{k_4}{\to }N{H}_{2}CON{H}_2$ (Slow)
Which of the following statement(s) is/are correct about the rate expression?

• A. $\frac{d_{[\text{urea }]}}{dt}=\frac{k_1{k}_3}{k_2}\left[N{H}_{4}NCO\right]$
• B. $\frac{d_{[\text{urea] }}}{dt}=\frac{k_1{k}_3}{k_2{k}_4}\left[N{H}_{4}NCO\right]$
• C. $\frac{d_{[\text{urea }]}}{dt}=k\left[N{H}_{4}NCO\right]$
• D. $\frac{d_{[\text{urea] }}}{dt}=\frac{k_1\times k_2}{k_3\times k_4}\left[N{H}_{4}NCO\right]$

Answer 1

Answer: (A), (C)

$\frac{d}{dt}\left[N{H}_{2}CON{H}_2\right]=k_4\left[N{H}_3\right][HNCO]$ from (iii)
Applying steady state approximately to HNCO or
$N{H}_3$ $\frac{d[HNCO]}{dt}=0=k_3\left[N{H}_{4}NCO\right]-k_4\left[N{H}_3\right][HNCO]$
$\therefore \frac{k_3}{k_4}=\frac{\left[N{H}_3\right][HNCO]}{\left[N{H}_{4}CNO\right]}$
$\frac{d_{[\text{urea }]}}{dt}=k_4\times \left[N{H}_3\right][HNCO]$
$=k_4\times \frac{k_3}{k_4}\left[N{H}_{4}NCO\right]$
Also, $\left[N{H}_{4}NCO\right]=\frac{k_1}{k_2}\times \left[N{H}_{4}CNO\right]$
$\therefore \frac{d_{\text{[urea] }}}{dt}=k_3\times \frac{k_1}{k_2}\times \left[N{H}_{4}CNO\right]$
$=k\left[N{H}_{4}CNO\right]$