NO2 ? 2NO + O2 obeys the rate law: rate = 1.4 x 10-2[NO2]2 at 500 K . What would

Question

NO2 ? 2NO + O2 obeys the rate law: rate = 1.4 x 10-2[NO2]2 at 500 K . What would be the rate constant at 186 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. The rate constant, k, at 186 K equals:

Explanation / Answer

No2---->2NO+O2 rate=kC^2= 1.4 x 10-2[NO2]2 ?k1=1.4*10^-2 @ 500 K we know from Arrehenius equation k=Aexp(-E/RT) A= pre-exponential factor, E= activation energy taking ln in both side we get ln(K2/K1)= E/R (1/T1 - 1/t2) ln(K2/K1) = 80E3/8.314 ( (1/500) - (1/186) ) = -32.488 K2/k1= exp (-32.488)=7.77E-15 K2 = K1* 1.28E14 =1.4e-2*7.77e-15 = 1.08e-16

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