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10. At 573K, NO2(g) decomposes forming NO and O2. The decomposition reaction is second order in NO2 with a rate constant of 1.1 M-1s-1. If the initial concentration of NO2 is 0.056 M, how long will it take for 75% of the NO2 to decompose

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Answer:

48.67 seconds

Explanation:

From;

1/[A] = kt + 1/[A]o

[A] = concentration at time t

t= time taken

k= rate constant

[A]o = initial concentration

Since [A] =[A]o - 0.75[A]o

[A] = 0.056 M - 0.042 M

[A] = 0.014 M

1/0.014 = (1.1t) + 1/0.056

71.4 - 17.86 = 1.1t

53.54 = 1.1t

t= 53.54/1.1

t= 48.67 seconds

Hence,it takes 48.67 seconds to decompose.

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