Question

The reaction \(\text {A→B}\) follows first order kinetics. 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 mole of B.

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

Answer 1

The Correct Option is A

The reaction \(\text {A → B}\) follows first-order kinetics. This means that the rate of the reaction is directly proportional to the concentration of A. The first-order rate equation can be expressed as:
\(\frac {-d[A]}{dt} = kA\)
\(ln \frac {[A]_0}{[A]} = kt\)
Where:
[A]₀ = is the initial concentration of A.
[A] is the concentration of A at time t.
k is the rate constant.
t is the time.
Given that the time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour, we can find the rate constant:
ln (\(\frac {0.8}{0.8-0.6}\)) = kx₁
ln (\(\frac {o.8}{0.2}\)) = k
k = ln (4)                ........(1)
Now, we can use this rate constant to find the time (t) it takes to convert 0.9 mole of A to produce 0.675 mole of B:
ln(\(\frac {0.9}{0.9-0.675}\)) = kt
ln (\(\frac {0.9}{0.225}\)) = kt
ln (4) = kt                ......... (2)
From eq (1) and (2)
k = kt
t = \(\frac kk\)
t = 1 hour

So, the time taken for the conversion of 0.9 mole of A to produce 0.675 mole of B is 1 hour, which corresponds to option (A).