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 determine the order of the reaction and calculate the rate constant for the decomposition of ammonia on tungsten wire, we can follow these steps:

### Step 1: Understand the relationship between half-life and pressure
The half-life \( T_{1/2} \) for a reaction can be expressed as:
\[ T_{1/2} \propto P_0^{1-n} \]
where \( P_0 \) is the initial pressure and \( n \) is the order of the reaction.

### Step 2: Take logarithms of both sides
Taking the logarithm of the half-life expression gives us:
\[ \log T_{1/2} \propto (1-n) \log P_0 \]
This implies that we can express it as:
\[ \log T_{1/2} = k \log P_0 + C \]
where \( k = 1 - n \) and \( C \) is a constant.

### Step 3: Use the given data to find the order of the reaction
We have the following data for \( P_0 \) (in mm Hg) and \( T_{1/2} \) (in seconds):
- \( P_0 \): 11, 21, 48, 73, 120
- \( T_{1/2} \): 48, 92, 210, 320, 525

We can calculate \( \log T_{1/2} \) and \( \log P_0 \) for each pair:
- For \( P_0 = 11 \), \( T_{1/2} = 48 \): \( \log(11) \) and \( \log(48) \)
- For \( P_0 = 21 \), \( T_{1/2} = 92 \): \( \log(21) \) and \( \log(92) \)

### Step 4: Calculate the values
Using a calculator, we find:
- \( \log(11) \approx 1.041 \), \( \log(48) \approx 1.681 \)
- \( \log(21) \approx 1.322 \), \( \log(92) \approx 1.964 \)

Now we can use the formula:
\[ 1 - n = \frac{\log T_{1/2,1} - \log T_{1/2,2}}{\log P_{0,1} - \log P_{0,2}} \]
Using the first two data points:
\[ 1 - n = \frac{\log(48) - \log(92)}{\log(11) - \log(21)} = \frac{1.681 - 1.964}{1.041 - 1.322} \]

### Step 5: Solve for \( n \)
Calculating the above:
\[ 1 - n = \frac{-0.283}{-0.281} \approx 1.007 \]
Thus, \( n \approx 0 \).

### Step 6: Conclusion on the order of the reaction
Since \( n \approx 0 \), this indicates that the reaction is a **zero-order reaction**.

### Step 7: Calculate the rate constant \( k \)
For a zero-order reaction, the half-life is given by:
\[ T_{1/2} = \frac{a}{2k} \]
where \( a \) is the initial concentration (or pressure in this case). Rearranging gives:
\[ k = \frac{a}{2T_{1/2}} \]
Using the first data point where \( P_0 = 11 \) mm Hg and \( T_{1/2} = 48 \) sec:
\[ k = \frac{11}{2 \times 48} = \frac{11}{96} \approx 0.114 \, \text{mm Hg sec}^{-1} \]

### Final Results
- The order of the reaction is **zero-order**.
- The rate constant \( k \) is approximately **0.114 mm Hg sec\(^{-1}\)**.