The second-order decomposition of NO2 has a rate constant of 0.255 M-1s-1. How much NO2 decomposes in 4.00 s if the initial concentration of NO2 (1.00 L volume) is 1.33 M?

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Chemistry

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29 August, 21:42

The second-order decomposition of NO2 has a rate constant of 0.255 M-1s-1. How much NO2 decomposes in 4.00 s if the initial concentration of NO2 (1.00 L volume) is 1.33 M?

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Hebert · Answer 1

C NO2 = 0.5644 M

Explanation:

The second order decomposition of NO2 (A):

- rA = K (CA) ² = - δCA/δt

⇒ Kδt = - δCA/CA²

⇒ K∫δt = - ∫δCA/CA²

⇒ K*t = 1/CA - 1/CAo

∴ K = 0.255/M. s

∴ t = 4.00 s

∴ CAo = 1.33 M

⇒ 1/CA = K*t + 1/CAo

⇒ 1/CA = ((0.255/M. s) (4.00 s)) + (1/1.33 M)

⇒ 1/CA = 1.02/M + 0.752/M

⇒ 1/CA = 1.772/M

⇒ CA = 0.5644 M