Question

The time taken for 80% of a first order reaction to be complete is approximately ____________.

Options

• 3.3 × t_1/2 × log₁₀ 5

• 3.3 × t_1/2 × log₁₀ 10

• 3.3 × t_1/2 × log₁₀ 20

• 3.3 × t_1/2 × log₁₀ 80

Answer 1

The time taken for 80% of a first order reaction to be complete is approximately 3.3 × t_1/2 × log₁₀ 5.

Explanation:

For a first order reaction, the integrated rate law is,

k = `2.303/"t" log_10  (["A"]_0)/(["A"]_"t")`

or t = `2.303/"k" log_10  (["A"]_0)/(["A"]_"t")`

When reaction is 80% complete, if a = 100 mol dm⁻³, then x = 80 mol dm⁻³.

Substituting values in the integrated rate law,

t = `2.303/"k" log_10  100/((100 - 80))`

= `2.303/"k" log_10  100/20`

Also substituting k = `0.693/"t"_(1//2)`,

∴ t = `(2.303 xx "t"_(1//2))/(0.693) xx log  100/20`

= 3.3 × t_1/2 × log₁₀ 5