The reaction 2NO 2 2NO + O 2 obeys the rate law: rate = 1.4 x 10 -2 [NO 2 ] 2 at
ID: 900885 • Letter: T
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 277 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. Do not include units in your answer. Exponential numbers need to be entered like this: 2 E-1 means 2 x 10-1.
Use the Arrhenius Equation to solve for the value of k2 at T2
ln (k2/k1) = (Ea/R)(1/T1 - 1/T2)
Pay attention to units.
The rate constant, k, at 277 K equals: ??
Explanation / Answer
ln (k2/k1) = (Ea/R)(1/T1 - 1/T2)
ln (K2/1.4*10^-2) = 80*1000/8.314 (1/500-1/277)
= ln (K2/1.4*10^-2)= -15.492
= (K2/1.4*10^-2) = 1.87*10^-7
= K2 = 2.618 * 10^-9 or
K2 = 2.6E-9
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