The reaction A `to`B follows first order reaction. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. What is the time taken for conversion of 0.9 mole of A to produce 0.675 moles of B?

Options

A. 1 hour
B. 0.5 hour
C. 0.25 hour
D. 2 hours

Answer

To solve the problem, we will use the first-order reaction kinetics formula:

\[ k t = 2.303 \log \left( \frac{[A]_0}{[A]} \right) \]

Where:
- \( k \) is the rate constant
- \( t \) is the time taken
- \( [A]_0 \) is the initial concentration of A
- \( [A] \) is the concentration of A at time \( t \)

### Step 1: Determine the initial and final moles for the first reaction.

For the first reaction:
- Initial moles of A, \( [A]_0 = 0.8 \)
- Final moles of A after 1 hour, \( [A] = 0.8 - 0.6 = 0.2 \)

### Step 2: Write the equation for the first reaction.

Using the first-order kinetics formula:

\[ k \cdot 1 = 2.303 \log \left( \frac{0.8}{0.2} \right) \]

### Step 3: Calculate the ratio for the first reaction.

Calculate the logarithm:

\[ \frac{0.8}{0.2} = 4 \]

Thus, the equation becomes:

\[ k = 2.303 \log(4) \]

### Step 4: Determine the initial and final moles for the second reaction.

For the second reaction:
- Initial moles of A, \( [A]_0 = 0.9 \)
- Final moles of A after time \( t \), \( [A] = 0.9 - 0.675 = 0.225 \)

### Step 5: Write the equation for the second reaction.

Using the first-order kinetics formula again:

\[ k \cdot t = 2.303 \log \left( \frac{0.9}{0.225} \right) \]

### Step 6: Calculate the ratio for the second reaction.

Calculate the logarithm:

\[ \frac{0.9}{0.225} = 4 \]

Thus, the equation becomes:

\[ k \cdot t = 2.303 \log(4) \]

### Step 7: Relate the two equations.

From the first reaction, we have:

\[ k = 2.303 \log(4) \]

From the second reaction, we have:

\[ k \cdot t = 2.303 \log(4) \]

### Step 8: Solve for time \( t \).

Since both equations have \( 2.303 \log(4) \) on the right side, we can equate them:

\[ k = k \cdot t \]

This simplifies to:

\[ t = 1 \text{ hour} \]

### Conclusion

The time taken for the conversion of 0.9 moles of A to produce 0.675 moles of B is **1 hour**.

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