Question

The decomposition of P2O5 is given by: rate = k[P2O5]. If the half-life for this reaction...

The decomposition of P2O5 is given by: rate = k[P2O5]. If the half-life for this reaction is 1400 s, about how long does it take for the P2O5 concentration to drop from 3.0 M to 0.3 M?

(1) 200 s

(2) 3000 s

(3) 4000s

(4) 4210 s

(5) 4650 s

The answer is (5) 4650s but I'm not sure how they got that. Please help, thanks!

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Homework Answers

Answer #1

By seeing the rate law, we can say that it is 1st order reaction

Given:

Half life = 1400 s

use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(1400)

= 4.95*10^-4 s-1

we have:

[P2O5]o = 3.0 M

[P2O5] = 0.3 M

k = 4.95*10^-4 s-1

use integrated rate law for 1st order reaction

ln[P2O5] = ln[P2O5]o - k*t

ln(0.3) = ln(3) - 4.95*10^-4*t

-1.204 = 1.0986 - 4.95*10^-4*t

4.95*10^-4*t = 2.3026

t = 4650 s

Answer: option 5

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