The decomposition of N_2O_4 is represented below. Its half-life is independent o

Question

The decomposition of N_2O_4 is represented below. Its half-life is independent of concentration and the rate constant is 5.21 Times 10^-5 s^-1 at a temperature of 28.0 degree C. The activation energy is 46.2 kJ/mol. 2N_2O_5(g) rightarrow 4NO_2(g) + O_2(g) Calculate the concentration of NO_2(g) produced after 35 minutes if the reaction starts with 0.110 M N_2O_5(g) at 28.0 degree C. Calculate the rate constant for the reaction at 55.0 degree C. Fully explain why the rate constant at 55.0 degree C differs from the rate constant at 28.0 degree C.

Explanation / Answer

a)

rate constant k = 5.21 x 10^-5 s-1

k = 1 / t ln (Ao / At)

5.21 x 10^-5 = 1 / 2100 ln (0.110 / At)

At = 0.0986 M

remaing N2O5 = 0.0986 M

2 N2O5 --------------> 4 NO2 + O2

   0.110                               0        0

0.110 - 2x                         4x        x

here remaining N2O5 = 0.110 - 2x = 0.0986 M

x = 0.0057

[NO2] = 4x = 0.0228 M

b)

T1 = 28 + 273 = 301 K

T2 = 55 + 273 = 328 K

ln (k2 / k1) = Ea / R [1/T1 - 1/T2]

ln (k2 / 5.1 x 10^-5) = 46.2 / (8.314 x 10^-3) [1/301 - 1/328]

rate constant k2 = 2.33 x 10^-4 s-1

c)

on increasing temperature rate of the reaction increases. so rate constant increases

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