For `2NH_(3)(g) overset(Pt)underset(Delta)to` Products order kinetics. If `t_(1//2)` at p = 4 atm is 25 sec, `t_(1//2)` at p = 16 atm will be (in sec)

Options

A. 6.25
B. 625
C. 100
D. `(25)^(1//4)`

Explanation

To solve the problem, we need to find the half-life (\(t_{1/2}\)) of the reaction at a pressure of 16 atm, given that the half-life at 4 atm is 25 seconds. The reaction follows zero-order kinetics.

### Step-by-Step Solution:

1. **Understand the relationship for zero-order kinetics:**
The half-life (\(t_{1/2}\)) for a zero-order reaction is given by the formula:
\[ t_{1/2} = \frac{[A]_0}{2k} \]
where \([A]_0\) is the initial concentration (or pressure in this case) and \(k\) is the rate constant.

2. **Identify the relationship between half-life and pressure:**
Since the half-life is directly proportional to the initial concentration (or pressure), we can express this relationship as:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \frac{P_1}{P_2} \]
where:
- \(t_{1/2,1}\) is the half-life at pressure \(P_1\) (4 atm),
- \(t_{1/2,2}\) is the half-life at pressure \(P_2\) (16 atm).

3. **Substitute the known values:**
We know:
- \(t_{1/2,1} = 25 \, \text{seconds}\)
- \(P_1 = 4 \, \text{atm}\)
- \(P_2 = 16 \, \text{atm}\)

Plugging these values into the equation gives:
\[ \frac{25}{t_{1/2,2}} = \frac{4}{16} \]

4. **Simplify the equation:**
The ratio \(\frac{4}{16}\) simplifies to \(\frac{1}{4}\). Thus, we can rewrite the equation as:
\[ \frac{25}{t_{1/2,2}} = \frac{1}{4} \]

5. **Cross-multiply to solve for \(t_{1/2,2}\):**
Cross-multiplying gives:
\[ 25 \cdot 4 = t_{1/2,2} \]
Therefore:
\[ t_{1/2,2} = 100 \, \text{seconds} \]

### Final Answer:
The half-life at a pressure of 16 atm will be **100 seconds**.