Question

The reaction, $A \rightarrow B$ follows first order kinetics. The time taken for $0.8$ mole of A to produce $0.6$ mole of B is 1h. What is the time taken for the conversion of $0.9$, mole of A to $0.675$ mole of $B$?

• A. 0.25 h
• B. 2 h
• C. 1 h
• D. 0.5 h

Answer 1

The Correct Option is C

To solve this problem, we need to apply the formula used for first-order reactions. For a first-order reaction, the rate law is given by:

k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]}\right)

where:

• k is the rate constant,
• [A]_0 is the initial concentration of reactant A,
• [A] is the concentration of A at time t,
• t is the time.

For the first situation, where 0.8 mole of A produces 0.6 mole of B, the remaining concentration of A would be 0.8 - 0.6 = 0.2 mole. Hence:

k = \frac{1}{1} \ln\left(\frac{0.8}{0.2}\right) = \ln(4)

For the second situation, where 0.9 mole of A converts to 0.675 mole of B, the remaining concentration of A would be 0.9 - 0.675 = 0.225 mole.

We need to find the time t for this conversion using the same rate constant:

\ln(4) = \frac{1}{t} \ln\left(\frac{0.9}{0.225}\right)

\ln(4) = \frac{1}{t} \ln(4)

Simplifying the equation:

t = 1 hour

Thus, the time taken for the conversion of 0.9 mole of A to 0.675 mole of B is indeed 1 hour, confirming the correct answer as 1 h.