At certain temperature, the half life period in the thermal decomposition of a gaseous substance as follows :
`"P (mm Hg) "500" "250`
`t_(1//2)" (in min) "235" "950`
Find the order of reaction [Given log (23.5)=1.37, log (95)=1.97]

Options

A. 1
B. 2
C. `2.5`
D. 3

Answer

To determine the order of the reaction based on the given half-life periods and pressures, we can follow these steps:

### Step 1: Write the relationship for half-life in nth order reactions
For a reaction of order \( n \), the relationship between half-life (\( t_{1/2} \)) and pressure (\( P \)) is given by:

\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \left( \frac{P_2}{P_1} \right)^{n-1} \]

### Step 2: Substitute the given values
From the problem, we have:
- \( P_1 = 500 \, \text{mm Hg} \), \( t_{1/2,1} = 235 \, \text{min} \)
- \( P_2 = 250 \, \text{mm Hg} \), \( t_{1/2,2} = 950 \, \text{min} \)

Substituting these values into the equation:

\[ \frac{235}{950} = \left( \frac{250}{500} \right)^{n-1} \]

### Step 3: Simplify the equation
Calculating the left side:

\[ \frac{235}{950} = 0.2474 \]

Calculating the right side:

\[ \frac{250}{500} = 0.5 \]

Thus, the equation becomes:

\[ 0.2474 = (0.5)^{n-1} \]

### Step 4: Take logarithm on both sides
Taking logarithm on both sides:

\[ \log(0.2474) = (n-1) \log(0.5) \]

### Step 5: Calculate the logarithms
Using the logarithm values provided:
- \( \log(23.5) = 1.37 \) implies \( \log(0.2474) \approx \log(23.5) - 1 = 1.37 - 1 = 0.37 \)
- \( \log(0.5) = -0.301 \)

Substituting these values:

\[ 0.37 = (n-1)(-0.301) \]

### Step 6: Solve for \( n \)
Rearranging gives:

\[ n - 1 = \frac{0.37}{-0.301} \approx -1.23 \]

Thus,

\[ n = -1.23 + 1 = -0.23 \]

This value seems incorrect based on the earlier calculations. Let's recalculate correctly:

### Correct Calculation:
Using the correct logarithmic values:

\[ \log(0.2474) \approx -0.606 \]
\[ \log(0.5) \approx -0.301 \]

Now substituting back:

\[ -0.606 = (n-1)(-0.301) \]

Solving for \( n-1 \):

\[ n - 1 = \frac{-0.606}{-0.301} \approx 2.01 \]

Thus,

\[ n \approx 3.01 \]

### Conclusion
The order of the reaction \( n \) is approximately **3**.

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