Question

The decomposition reaction of a to b has a rate constant of 0.00132 s-1: a → 2b if the initial concentration of a is 0.156 m, how long will it take for the concentration of a to decrease by 78.1%

Answer 1

Answer: -

92.4 s

Explanation: -

The decomposition reaction of a → b has a rate constant k = 0.00132 s⁻¹

From the rate constant we see that the reaction is of zero order.

The rate equation for a zero order reaction is

A₀ - A = kt

where A₀ = initial concentration.

T = time passed since start of reaction,

A is the amount present after t time passed.

A₀ = 0.156 M

A = A₀ - 78.1% of A₀

  = 0.156 - $\frac{78.1}{100}$ x 0.156

 = 0.156 - 0.122

 = 0.034 M

Plugging into the formula

A₀ - A = kt

0.156 - 0.034 = 0.00132 x t

t = $\frac{0.122}{0.00132}$

= 92.4 s

t =