2N2O5 --> 2 N2O4 + O2 (k=2.85x10^-4 s^-1 at 25 degrees Celsius) what would the [N2O5]...

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2N2O5 --> 2 N2O4 + O2 (k=2.85x10^-4 s^-1 at 25 degrees Celsius) what would the [N2O5]...

2N2O5 --> 2 N2O4 + O2 (k=2.85x10^-4 s^-1 at 25 degrees Celsius)

what would the [N2O5] be after 150 s given that the initial [N2O5] was 0.35? what would be the [N2O4] after 150 s?

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Answer 1

Answer #1

1)

we have:

[N2O5]o = 0.35 M

t = 150 s

k = 2.85*10^-4 s-1

Unit of k suggests it is 1st order reaction.

use integrated rate law for 1st order reaction

ln[N2O5] = ln[N2O5]o - k*t

ln[N2O5] = ln(0.35) - 2.85*10^-4*1.5*10^2

ln[N2O5] = -1.05 - 2.85*10^-4*1.5*10^2

ln[N2O5] = -1.093

[N2O5] = e^(-1.093)

[N2O5] = 0.3354 M

Answer: 0.34 M

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Answer 2

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2N2O5 --> 2 N2O4 + O2 (k=2.85x10^-4 s^-1 at 25 degrees Celsius) what would the [N2O5]...

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2N2O5 --> 2 N2O4 + O2 (k=2.85x10^-4 s^-1 at 25 degrees Celsius) what would the [N2O5]...