For a `1^(st)` order reaction (gaseous) (constant V, T) :
`aAto(b-1)B+1 C` (with `bgta`) the pressure of the system rose by `50(b/a-1)` % in a time of 10 min. The half life of the reaction is therefore.

Options

A. 10 min
B. 20 min
C. 30 min
D. 40 min

Explanation

To solve the problem, let's break it down step by step.

### Step 1: Understand the Reaction
We have a first-order reaction:
\[ aA \rightarrow (b-1)B + C \]
where \( b > a \).

### Step 2: Initial and Final Pressures
At time \( t = 0 \), the pressure of the system is \( P_0 \). As the reaction proceeds, the pressure changes due to the consumption of \( A \) and the formation of \( B \) and \( C \).

### Step 3: Change in Pressure
Let \( x \) be the change in pressure due to the reaction. The pressure of \( A \) at time \( t \) will be:
\[ P_A = P_0 - x \]
The pressure of \( B \) will be:
\[ P_B = \frac{(b-1)}{a} x \]
And the pressure of \( C \) will be:
\[ P_C = \frac{x}{a} \]

### Step 4: Total Pressure at Time \( t \)
The total pressure at time \( t \) can be expressed as:
\[ P_t = P_A + P_B + P_C \]
Substituting the expressions we derived:
\[ P_t = (P_0 - x) + \frac{(b-1)}{a} x + \frac{x}{a} \]
This simplifies to:
\[ P_t = P_0 + x \left( \frac{(b-1) + 1}{a} \right) - x \]
\[ P_t = P_0 + x \left( \frac{b}{a} - 1 \right) \]

### Step 5: Given Change in Pressure
According to the problem, the pressure rose by \( 50 \left( \frac{b}{a} - 1 \right) \% \) in 10 minutes. This means:
\[ P_t = P_0 + \frac{50}{100} \left( P_0 \left( \frac{b}{a} - 1 \right) \right) \]
This can be rewritten as:
\[ P_t = P_0 + \frac{1}{2} P_0 \left( \frac{b}{a} - 1 \right) \]
This simplifies to:
\[ P_t = P_0 \left( 1 + \frac{1}{2} \left( \frac{b}{a} - 1 \right) \right) \]

### Step 6: Equating Pressures
Now, we equate the two expressions for \( P_t \):
\[ P_0 + x \left( \frac{b}{a} - 1 \right) = P_0 \left( 1 + \frac{1}{2} \left( \frac{b}{a} - 1 \right) \right) \]

### Step 7: Solve for \( x \)
Rearranging gives us:
\[ x \left( \frac{b}{a} - 1 \right) = P_0 \frac{1}{2} \left( \frac{b}{a} - 1 \right) \]
Dividing both sides by \( \left( \frac{b}{a} - 1 \right) \) (assuming \( \frac{b}{a} - 1 \neq 0 \)):
\[ x = \frac{P_0}{2} \]

### Step 8: Relate \( x \) to Half-Life
Since \( x = P_0 - P_A \), we have:
\[ P_A = P_0 - x = P_0 - \frac{P_0}{2} = \frac{P_0}{2} \]
This indicates that the pressure of \( A \) is half of the initial pressure after 10 minutes.

### Step 9: Conclusion on Half-Life
In a first-order reaction, the half-life is the time required for the concentration (or pressure in this case) to decrease to half its initial value. Since we found that this occurs in 10 minutes, the half-life of the reaction is:
\[ \text{Half-life} = 10 \text{ minutes} \]

### Final Answer
The half-life of the reaction is **10 minutes**.

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