Question

For the reaction 2NO₂ (g) -> 2NO (g) + O₂ (g), data for rate constants was obtained at the following temperatures.

L1	L2	L3	L4
Temperature, K	k, M-1 s-1	1/T, K-1	Ln k
330	0.77
354	1.8
378	4.1
383	4.7

What happens to the rate constant as temperature increases?

Find the A and E_a graphically using linear regression.

Find E_a using the two-data point approach with the data at 330 K and 383 K. Why is there a slight difference?

Answer 1

Arhenius equation is K=Ko*e^(-Ea/RT), Ko= Frequency factor and Ea= activation energy and R= gas constant =8.314 J/mole.K

takin ln, lnK= lnKo-Ea/RT, so a plot of lnK vs 1/T gives straight line whose slope is -Ea/R and intercept is lnK

the plot of lnK vs 1/T is generated and shown below.

from the plot of best fit, -Ea/R= slope =-4344.2, Ea= 4344.2*8.314 J/mole =36118 J/mole= 36.12 Kj/mole

for two different temperatrues, T1 and T2, Arhenius equation is

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

given T1= 330K and T2= 383K, K2= 4.7 and K1= 0.77

ln(4.7/0.77)= (Ea/8.314)*(1/330-1/383)

Ea= 35864.76 J/mole= 35.864 KJ/mole

the activation energy calculated accounting for all the points is based on the best fit of the plot of lnK vs 1/T.