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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 333 K if the activation energy is 80. kJ/mol?
This is a second order reaction, giving k the units of M-1S-1 This will not change with the change in temperature
Jan 2nd, 2015

According to the description,

when T1=500 K, k1=1.4x10^-2 M-1 S-1.

Now since k=Ae^(-Ea/RT)

k2/k1=e^(Ea/RT1-Ea/RT2)

when T2=333 K, Ea=80. kJ/mol=80x10^3 J/mol

k2=k1xe^(Ea/RT1-Ea/RT2)

    =1.4x10^-2xe^(80x10^3/ 8.314x500 - 80x10^3/ 8.314x333)

    =1.4x10^-2xe^(-9.65)

    =1.4x10^-2x6.44x10^-5

    =9.02x10^-7 M-1 S-1

Jan 2nd, 2015

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Jan 2nd, 2015
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