Question

The rate expression for the reaction:
$NH _4 CNO \rightleftharpoons NH _2 CONH _2$ can be derived from the mechanism:
i. $NH _4 CNO \xrightleftharpoons[k_2]{k_1} NH _4 NCO$ (Fast)
ii. $NH _4 NCO \xrightarrow{k_3} NH _3+ HNCO \quad$ (Fast)
iii. $NH _3+ HNCO \xrightarrow{ k _4} NH _2 CONH _2$ (Slow)
Which of the following statement(s) is/are correct about the rate expression?

• A. $\frac{d_{[\text {urea }]}}{d t}=\frac{k_1 k_3}{k_2}\left[ NH _4 NCO \right]$
• B. $\frac{d_{[\text {urea] }}}{d t}=\frac{k_1 k_3}{k_2 k_4}\left[ NH _4 NCO \right]$
• C. $\frac{d_{[\text {urea }]}}{d t}=k\left[ NH _4 NCO \right]$
• D. $\frac{d_{[\text {urea] }}}{d t}=\frac{k_1 \times k_2}{k_3 \times k_4}\left[ NH _4 NCO \right]$

Answer 1

Answer: (A), (C)

$\frac{d}{d t}\left[ NH _2 CONH _2\right]=k_4\left[ NH _3\right][ HNCO ]$ from (iii)
Applying steady state approximately to HNCO or
$NH _3$ $\frac{d[ HNCO ]}{d t}=0=k_3\left[ NH _4 NCO \right]-k_4\left[ NH _3\right][ HNCO ]$
$\therefore \frac{k_3}{k_4}=\frac{\left[ NH _3\right][ HNCO ]}{\left[ NH _4 CNO \right]}$
$\frac{d_{[\text {urea }]}}{d t}=k_4 \times\left[ NH _3\right][ HNCO ]$
$=k_4 \times \frac{k_3}{k_4}\left[ NH _4 NCO \right]$
Also, $\left[ NH _4 NCO \right]=\frac{k_1}{k_2} \times\left[ NH _4 CNO \right]$
$\therefore \frac{d_{\text {[urea] }}}{d t}=k_3 \times \frac{k_1}{k_2} \times\left[ NH _4 CNO \right]$
$=k\left[ NH _4 CNO \right]$