The reaction 2NO2 2NO + O2 obeys the rate law: rate = 1.4 x 10-2[NO2]2 at 500 K

Question

The reaction 2NO2 2NO + O2 obeys the rate law: rate = 1.4 x 10-2[NO2]2 at 500 K What would be the rate constant at 283 K if the activation energy is 80. kJ/mol? This is a second order reaction, giving k the units of M1s1 This will not change with the change in temperature. Do not include units in your answer. Exponential numbers need to be entered like this: 2 E-1 means 2 x 10-1 The rate constant, k, at 283 K equals 7.98e12 Use the Arrhenius Equation to solve for the value of k2 at T2 In (k2/k1) = (Ea/R)(1/T1-1/T2) Pay attention to units Submit AnswerIncorrect. Tries 6/45 Previous Tries

Explanation / Answer

we have:

T1 = 500 K

T2 = 283 K

K1 = 1.4*10^-2 M-1.s-1

Ea = 80 KJ/mol

= 80000 J/mol

we have below equation to be used:

ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

ln(K2/1.4*10^-2) = (80000.0/8.314)*(1/500.0 - 1/283.0)

ln(K2/1.4*10^-2) = 9622*(-1.534*10^-3)

K2 = 5.46*10^-9 M-1.s-1

Answer: 5.46E-9

Feel free to comment below if you have any doubts or if this answer do not work

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