The rate expresison for the reaction:
`NH_(4)CNO hArr NH_(2)CONH_(2)` can be derived form the mechanism:
(i) `NH_(4)CNO underset(k_(2))overset(k_(1))hArr NH_(4)NCO ("Fast")`
(ii) `NH_(4)CNO overset(k_(3))rarrNH_(3) + HNCO` ("fast")
(iii) `NH_(3)+HNCo overset(k_(4))rarr NH_(2)CONH_(2)` (Slow)
Which of the following statement `(s)` is/are correct about the rate expresison?
a.`(d_(["urea]))/(dt) = (k_(1)k_(3))/(k_(2))[NH_(4)NCO]`
b. `(d_(["urea]))/(dt) = (k_(1)k_(3))/(k_(2)k_(4))[NH_(4)NCO]`
c.`(d_(["urea]))/(dt) = k[NH_(4)NCO]`
d.`(d_(["urea]))/(dt) = (k_(1) xx k_(2))/(k_(3) xx k_(4))[NH_(4)NCO]`

Options

A. `(d_(["urea]))/(dt) = (k_(1)k_(3))/(k_(2))[NH_(4)NCO]`
B. `(d_(["urea]))/(dt) = (k_(1)k_(3))/(k_(2)k_(4))[NH_(4)NCO]`
C. `(d_(["urea]))/(dt) = k[NH_(4)NCO]`
D. `(d_(["urea]))/(dt) = (k_(1) xx k_(2))/(k_(3) xx k_(4))[NH_(4)NCO]`

Explanation

To derive the rate expression for the reaction \( NH_4CNO \rightleftharpoons NH_2CONH_2 \) based on the given mechanism, we will follow these steps:

### Step 1: Identify the Rate-Determining Step
The slow step of the mechanism is the third step:
\[ NH_3 + HNCO \overset{k_4}{\rightarrow} NH_2CONH_2 \]
Since this is the slow step, the rate of formation of urea (\( NH_2CONH_2 \)) can be expressed as:
\[ \frac{d[\text{urea}]}{dt} = k_4 [NH_3][HNCO] \]

### Step 2: Apply the Steady-State Approximation
To express the concentrations of \( NH_3 \) and \( HNCO \) in terms of \( NH_4NCO \), we will apply the steady-state approximation to \( HNCO \):
\[ \frac{d[HNCO]}{dt} = 0 \]
From the second step of the mechanism:
\[ NH_4CNO \overset{k_3}{\rightarrow} NH_3 + HNCO \]
and the third step:
\[ NH_3 + HNCO \overset{k_4}{\rightarrow} NH_2CONH_2 \]
We can write:
\[ k_3 [NH_4CNO] - k_4 [NH_3][HNCO] = 0 \]
Rearranging gives:
\[ k_3 [NH_4CNO] = k_4 [NH_3][HNCO] \]
Thus, we can express \( [HNCO] \) in terms of \( [NH_4CNO] \):
\[ [HNCO] = \frac{k_3}{k_4} [NH_4CNO] \cdot \frac{1}{[NH_3]} \]

### Step 3: Substitute Back into the Rate Expression
Now, substituting \( [HNCO] \) back into the rate equation:
\[ \frac{d[\text{urea}]}{dt} = k_4 [NH_3] \left(\frac{k_3}{k_4} [NH_4CNO] \cdot \frac{1}{[NH_3]}\right) \]
This simplifies to:
\[ \frac{d[\text{urea}]}{dt} = k_3 [NH_4CNO] \]

### Step 4: Relate \( [NH_4CNO] \) to the Initial Reactant
From the first step of the mechanism, we can express \( [NH_4NCO] \) in terms of \( [NH_4CNO] \):
\[ [NH_4NCO] = \frac{k_1}{k_2} [NH_4CNO] \]
Substituting this into our rate expression gives:
\[ \frac{d[\text{urea}]}{dt} = k_3 \left(\frac{k_1}{k_2} [NH_4CNO]\right) \]
This can be simplified to:
\[ \frac{d[\text{urea}]}{dt} = \frac{k_1 k_3}{k_2} [NH_4CNO] \]

### Conclusion
The final rate expression for the formation of urea is:
\[ \frac{d[\text{urea}]}{dt} = \frac{k_1 k_3}{k_2} [NH_4CNO] \]