Question

The decomposition reaction of NOBr is second order in NOBr, with a rate constant at 20°C...

The decomposition reaction of NOBr is second order in NOBr, with a rate constant at 20°C of 25 M-1 min-1. If the initial concentration of NOBr is 0.025 M, find

(a) the time at which the concentration will be 0.010 M.


(b) the concentration after 145 min of reaction.

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Homework Answers

Answer #1

a)

we have:

[NOBr]o = 0.025 M

[NOBr] = 0.01 M

k = 25 M-1.min-1

use integrated rate law for 2nd order reaction

1/[NOBr] = 1/[NOBr]o + k*t

1/(0.01) = 1/(0.025) + 25*t

100 = 40 +25*t

25*t = 60

t = 2.40 min

Answer: 2.40 min

b)

we have:

[NOBr]o = 0.025 M

t = 145.0 min

k = 25 M-1.min-1

Given:

[NOBr]o = 0.025 M

use integrated rate law for 2nd order reaction

1/[NOBr] = 1/[NOBr]o + k*t

1/[NOBr] = 1/(0.025) + 25*145

1/[NOBr] = 40 + 25*145

1/[NOBr] = 3665

[NOBr] = 2.73*10^-4 M

Answer: 2.73*10^-4 M

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