It is known that the heat is needed to dissociate ammonia into `N_(2)` and `H_(2)`. For the reaction `N_(2)+3H_(3) hArr 2NH_(3), K_(f)` is the velocity constant for forward reaction and `K_(b)` is velocity constant for backward reaction, `K_(c)` is equilibrium constant for the reaction shown. Then `(dk_(f))/(dT)` (where T is symbol for absolute temp.):

Options

A. Is greater than `dk_(b)//dT`
B. Is less than `dk_(b)//dT`
C. Is equal to `dk_(b)//dT`
D. Cannot be compared with `dk_(b)//dT`

Answer

To solve the problem, we need to analyze the relationship between the equilibrium constant and the temperature for the given reaction. The reaction is:

\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

### Step 1: Understand the nature of the reaction
The dissociation of ammonia (NH₃) into nitrogen (N₂) and hydrogen (H₂) requires heat, indicating that this reaction is endothermic. Therefore, the enthalpy change (\( \Delta H \)) for the dissociation reaction is positive.

### Step 2: Relate the forward and backward reactions
For the forward reaction:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]
This reaction is exothermic since it is the reverse of the endothermic dissociation of NH₃. Thus, the enthalpy change (\( \Delta H \)) for this reaction is negative.

### Step 3: Write the relationship between the equilibrium constant and the rate constants
The equilibrium constant (\( K_c \)) is related to the forward rate constant (\( K_f \)) and the backward rate constant (\( K_b \)) as follows:
\[ K_c = \frac{K_f}{K_b} \]

### Step 4: Apply the van 't Hoff equation
The van 't Hoff equation relates the change in the equilibrium constant with temperature:
\[ \frac{d \ln K_c}{dT} = \frac{\Delta H}{RT^2} \]

Since \( \Delta H \) for the forward reaction is negative, this means that as the temperature increases, \( K_c \) decreases.

### Step 5: Differentiate the rate constants with respect to temperature
From the relationship \( K_c = \frac{K_f}{K_b} \), we can differentiate both sides with respect to temperature:
\[ \frac{dK_c}{dT} = \frac{dK_f}{dT} \cdot \frac{1}{K_b} - \frac{K_f}{K_b^2} \cdot \frac{dK_b}{dT} \]

### Step 6: Analyze the signs of the derivatives
Since we established that \( K_c \) decreases with increasing temperature, we can conclude that:
\[ \frac{dK_f}{dT} < \frac{dK_b}{dT} \]

This implies that the rate constant for the forward reaction (\( K_f \)) decreases with temperature at a slower rate than the rate constant for the backward reaction (\( K_b \)) increases with temperature.

### Conclusion
Thus, the relationship we derive is:
\[ \frac{dK_f}{dT} < \frac{dK_b}{dT} \]