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