For a `1^(st)` order reaction (gaseous) (cont. V, T) `aA rarr (b-1) B+C` (with `b gt a`) the pressure of the system increased by `50((b)/(a)-1)%` in a time of 10 min. The halflife of the reaction is therefore (in min).

Answer

To solve the problem, we need to determine the half-life of a first-order reaction given the increase in pressure over a specified time. Let's break this down step by step.

### Step 1: Understand the Reaction
The reaction is given as:
\[ aA \rightarrow (b-1)B + C \]
where \( b > a \).

### Step 2: Initial and Final Pressure
At time \( t = 0 \), the initial pressure of the system is \( P_0 \). After some time \( t \), the pressure increases by \( 50\left(\frac{b}{a} - 1\right)\% \).

### Step 3: Calculate the Increase in Pressure
The increase in pressure can be expressed as:
\[ \Delta P = P_0 \cdot 0.5\left(\frac{b}{a} - 1\right) \]
Thus, the final pressure \( P \) after time \( t \) can be written as:
\[ P = P_0 + \Delta P = P_0 + 0.5\left(\frac{b}{a} - 1\right) P_0 = P_0 \left(1 + 0.5\left(\frac{b}{a} - 1\right)\right) \]

### Step 4: Relate the Change in Pressure to the Reaction Progress
Let \( x \) be the change in moles of \( A \) that have reacted. The change in pressure due to the reaction can be expressed as:
\[ P = P_0 - \frac{x}{a} + \frac{(b-1)x}{a} + \frac{x}{a} = P_0 + \frac{(b-1)x}{a} \]
Setting the two expressions for \( P \) equal gives:
\[ P_0 + \frac{(b-1)x}{a} = P_0 \left(1 + 0.5\left(\frac{b}{a} - 1\right)\right) \]

### Step 5: Solve for \( x \)
From the above equation, we can isolate \( x \):
\[ \frac{(b-1)x}{a} = 0.5\left(\frac{b}{a} - 1\right) P_0 \]
This simplifies to:
\[ x = \frac{0.5\left(\frac{b}{a} - 1\right) a P_0}{b-1} \]

### Step 6: Determine the Half-Life
For a first-order reaction, the half-life \( t_{1/2} \) is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Where \( k \) is the rate constant.

Since we know that the pressure increased by \( 50\left(\frac{b}{a} - 1\right)\% \) in 10 minutes, and this corresponds to a decrease in concentration of \( A \) to half its initial value, we can conclude:
\[ t_{1/2} = 10 \text{ minutes} \]

### Final Answer
The half-life of the reaction is:
\[ \boxed{10 \text{ minutes}} \]