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makvit [3.9K]
3 years ago
5

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, ho

w long will it take for the concentration of a to decrease by 78.1%
Chemistry
1 answer:
Misha Larkins [42]3 years ago
8 0

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 =

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