The half-period T for the decomposition of ammonia on tungsten wire, was measured for different initial pressure P of ammonia at `25^@C`.Then
`{:("P(mm Hg)",11,21,48,73,120),("T(sec)",48,92,210,320,525):}`

Options

A. Zero order reaction
B. First order reaction
C. Rate constant for reaction is 0.114 mol `"lit."^(-1) sec^(-1)`
D. Rate constant for reaction is 1.14 seconds.

Explanation

To solve the problem, we need to determine the order of the reaction and the rate constant for the decomposition of ammonia on tungsten wire based on the given half-period data at different pressures.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The half-period \( T_{1/2} \) is inversely proportional to the concentration of the reactant raised to the power \( n-1 \), where \( n \) is the order of the reaction. When dealing with gases, pressure can be used in place of concentration. Thus, we can write:
\[ T_{1/2} \propto \frac{1}{P^{(n-1)}} \]
This implies:
\[ T_{1/2} = k \cdot \frac{1}{P^{(n-1)}} \]
where \( k \) is a proportionality constant.

2. **Setting Up Ratios**:
We will use two sets of data to find \( n \):
- For \( P_1 = 11 \, \text{mm Hg} \), \( T_{1/2,1} = 48 \, \text{sec} \)
- For \( P_2 = 21 \, \text{mm Hg} \), \( T_{1/2,2} = 92 \, \text{sec} \)

We set up the ratio:
\[ \frac{T_{1/2,1}}{T_{1/2,2}} = \frac{P_2^{(n-1)}}{P_1^{(n-1)}} \]
Substituting the values:
\[ \frac{48}{92} = \left(\frac{21}{11}\right)^{(n-1)} \]

3. **Calculating the Left Side**:
Calculate \( \frac{48}{92} \):
\[ \frac{48}{92} = 0.5217 \]

4. **Taking Logarithms**:
Taking logarithms on both sides:
\[ \log(0.5217) = (n-1) \log\left(\frac{21}{11}\right) \]

5. **Calculating the Right Side**:
Calculate \( \frac{21}{11} \):
\[ \frac{21}{11} = 1.9091 \]
Now, calculate \( \log(1.9091) \):
\[ \log(1.9091) \approx 0.2804 \]

6. **Substituting Values**:
Now substitute back into the equation:
\[ \log(0.5217) \approx -0.2804 = (n-1) \cdot 0.2804 \]

7. **Solving for \( n \)**:
Rearranging gives:
\[ n - 1 = \frac{-0.2804}{0.2804} \implies n - 1 = -1 \implies n = 0 \]
This means the reaction is of **zero order**.

8. **Finding the Rate Constant \( k \)**:
For a zero-order reaction, the half-life is given by:
\[ T_{1/2} = \frac{P}{2k} \]
Rearranging gives:
\[ k = \frac{P}{2T_{1/2}} \]
Using \( P = 11 \, \text{mm Hg} \) and \( T_{1/2} = 48 \, \text{sec} \):
\[ k = \frac{11}{2 \times 48} = \frac{11}{96} \approx 0.1146 \, \text{mm Hg sec}^{-1} \]

### Final Answers:
- Order of the reaction \( n = 0 \) (zero order)
- Rate constant \( k \approx 0.1146 \, \text{mm Hg sec}^{-1} \)