The decomposition of ammonia on tungsten surface at 500 K follows zero order kinetics. The half-life period of this reaction is 45 minutes when the initial pressure is 4 bar. Find the half-life period (in minutes) of the reaction when the initial pressure is 16 bar at the same temperature.

Explanation

To solve the problem, we need to determine the half-life period of the decomposition of ammonia on a tungsten surface at 500 K, given that it follows zero-order kinetics. We know the half-life period when the initial pressure is 4 bar, and we need to find it when the initial pressure is 16 bar.

### Step-by-Step Solution:

1. **Understanding Zero-Order Kinetics**:
- For a zero-order reaction, the half-life (T_half) is directly proportional to the initial concentration (or pressure in this case).
- The relationship can be expressed as:
\[ T_{1/2} \propto \frac{1}{[A]_0^{n-1}} \]
- For zero-order reactions, n = 0, hence:
\[ T_{1/2} \propto [A]_0 \]

2. **Given Data**:
- Half-life period (T1) when initial pressure (P1) = 4 bar is 45 minutes.
- We need to find the half-life period (T2) when the initial pressure (P2) = 16 bar.

3. **Setting Up the Proportionality**:
- From the relationship, we can write:
\[ \frac{T_1}{T_2} = \frac{P_1}{P_2} \]
- Substituting the known values:
\[ \frac{45 \text{ min}}{T_2} = \frac{4 \text{ bar}}{16 \text{ bar}} \]

4. **Simplifying the Ratio**:
- The ratio simplifies to:
\[ \frac{4}{16} = \frac{1}{4} \]
- Therefore:
\[ \frac{45}{T_2} = \frac{1}{4} \]

5. **Cross-Multiplying to Solve for T2**:
- Cross-multiplying gives:
\[ 45 = \frac{T_2}{4} \]
- Multiplying both sides by 4:
\[ T_2 = 45 \times 4 = 180 \text{ minutes} \]

6. **Final Answer**:
- The half-life period of the reaction when the initial pressure is 16 bar is **180 minutes**.