The reaction 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
The answer is A.
Solution
For a first order reaction, $t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}$. The time depends only on the fraction converted, not on the initial amount. Case 1: 0.8 mol A gives 0.6 mol B, so A left $= 0.2$ mol. Fraction converted $= \dfrac{0.6}{0.8} = 0.75$, and $\dfrac{[A]_0}{[A]} = \dfrac{0.8}{0.2} = 4$. Case 2: 0.9 mol A gives 0.675 mol B, so A left $= 0.225$ mol. Fraction converted $= \dfrac{0.675}{0.9} = 0.75$, and $\dfrac{[A]_0}{[A]} = \dfrac{0.9}{0.225} = 4$. Both cases correspond to 75% conversion (two half lives), so the time taken is the same. So the time taken is 1 hour.
ConceptTime for same fractional conversion
Chemical Kinetics in past papers
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