In the synthesis of ammonia by Harber's process. If 60 moles of ammonia is obtained in one hour, then the rate of disappearence of nitrogen is:

Options

A. 30 mol/min
B. 6 mol/min
C. 0.5 mol/min
D. 60 mol/min

Answer

To find the rate of disappearance of nitrogen in the synthesis of ammonia by the Haber process, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the Haber process is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 2: Determine the relationship between ammonia and nitrogen
From the balanced equation, we can see that:
- 1 mole of nitrogen (N₂) produces 2 moles of ammonia (NH₃).

### Step 3: Calculate the moles of nitrogen consumed
Given that 60 moles of ammonia are produced in one hour, we can calculate the moles of nitrogen consumed:
- Since 2 moles of NH₃ are produced from 1 mole of N₂, we can set up the ratio:
\[ \text{Moles of } N_2 = \frac{\text{Moles of } NH_3}{2} \]
\[ \text{Moles of } N_2 = \frac{60 \text{ moles of } NH_3}{2} = 30 \text{ moles of } N_2 \]

### Step 4: Calculate the rate of disappearance of nitrogen
The rate of disappearance of nitrogen can be calculated using the formula:
\[ \text{Rate of disappearance of } N_2 = -\frac{d[N_2]}{dt} \]
Where:
- \( d[N_2] \) is the change in moles of nitrogen,
- \( dt \) is the change in time.

In this case:
- \( d[N_2] = 30 \text{ moles} \)
- \( dt = 1 \text{ hour} = 60 \text{ minutes} \)

Thus, the rate is:
\[ \text{Rate} = -\frac{30 \text{ moles}}{60 \text{ minutes}} = -0.5 \text{ moles per minute} \]

### Conclusion
The rate of disappearance of nitrogen is:
\[ 0.5 \text{ moles per minute} \]