Question

A first order reaction is 50% completed in 16 minutes. The percentage of reactant that will react in 32 minutes is ____________.

Options

• 75%

• 12.5%

• 25%

• 100%

Answer 1

A first order reaction is 50% completed in 16 minutes. The percentage of reactant that will react in 32 minutes is 75%.

Explanation:

For the first order reaction

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

`1/16 log_10  100/50 = 1/32 log_10  100/(["A"]_"t")`

0.0188 × 32 = log₁₀ 100 − log₁₀ [A]_t

0.0188 × 32 = 2 − log₁₀ [A]_t

0.602 = 2 − log₁₀ [A]_t

∴ log₁₀ [A]_t = 2 − 0.602 = 1.398

By taking antilog

[A]_t = 25.00

∴ % of reactant that will react after 32 minutes = 100 − 25 = 75%