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16.41 For the first-order decomposition of gaseous dinitrogen
pentoxide,
2Nâ‚‚Os(g)
4NO2(g) + O2(g)
the rate constant is k = 2.8x10 s at 60°C. The initial concen-
tration of N2O is 1.58 mol/L. (a) What is [N2O5] after 5.00 min?
(b) How long will it take for 35.0% of the N2O5 to decompose?

Respuesta :

JohnPositivity
(a) To find the concentration of N2O5 after 5.00 minutes:

We'll use the integrated rate law equation:

1/[N2O5]_t = 1/[N2O5]_0 + kt

Given:
- Initial concentration [N2O5]_0 = 1.58 mol/L
- Rate constant k = 2.8 × 10^-2 s^-1
- Time t = 5.00 minutes = 5.00 × 60 seconds

Plugging in the values:

1/[N2O5]_t = 1/1.58 + (2.8 × 10^-2) × (5.00 × 60)

[N2O5]_t ≈ 1/9.4

[N2O5]_t ≈ 0.106 mol/L

So, the concentration of N2O5 after 5.00 minutes is approximately 0.106 mol/L.

(b) To find the time it takes for 35.0% of N2O5 to decompose:

We'll use the integrated rate law again:

1/[N2O5]_final = 1/[N2O5]_0 + kt_final

Given:
- Final fraction remaining = 0.35
- Initial concentration [N2O5]_0 = 1.58 mol/L
- Rate constant k = 2.8 × 10^-2 s^-1

Plugging in the values:

t_final = 1/(0.35 × (2.8 × 10^-2))

t_final ≈ 1027.67 seconds

So, it will take approximately 1027.67 seconds for 35.0% of the N2O5 to decompose.

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