Chemical Reaction Engineering MCQ Questions & Answers | Chemical Engineering

For a first-order reaction, the relationship between time and conversion can be derived using the half-life formula.

The half-life (t_1/2) of a first-order reaction is constant and is given by:

t_1/2 = 0.693 / k

Let the initial concentration be C₀ and concentration at time t be C.

Using the first-order equation: C = C₀ * e^-kt

If t = 3 * t_1/2, then:

C / C₀ = e^-k(3t_1/2) = e^{-3 * 0.693} = e^-2.079 ≈ 0.125

Conversion (X) = 1 - (C / C₀) = 1 - 0.125 = 0.875

Therefore, the conversion at time equal to three times the half-life is 0.875.

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Alternate Solution

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First-order reactions have a special property: their half-life (t_1/2) is constant.
This means it always takes the same amount of time for half of the reactant to be consumed, regardless of the initial amount.

The question states the time is three times the half-life (3 * t_1/2).

Let's break it down step-by-step:
After 1 half-life (t_1/2), the conversion is 50% (0.5).
After 2 half-lives (2 * t_1/2), half of the remaining reactant is converted.
So, half of 0.5 (which is 0.25) is converted, resulting in a total conversion of 0.5 + 0.25 = 0.75.
After 3 half-lives (3 * t_1/2), half of the remaining reactant is converted again.
Half of what's left after 2 half lives = Half of 0.25 = 0.125
The total conversion after 3 half lives is 0.5 + 0.25 + 0.125 = 0.875

Therefore, the conversion after three half-lives is 0.875.

So the answer is: Option A: 0.875