In: Chemistry
The data below shows the production of CO2 from carbonic aanhydrase. The zero-order rate constant was measured at different temperatures. Calculate EA , Delta H+ , the Arrhenius Constant, and Delta S+.
T(K) | 289 | 293.5 | 298.1 | 303.2 | 308 | 113.5 |
k (10-6) | 1.04 | 1.34 | 1.53 | 1.89 | 2.29 | 2.84 |
Solution:
Zero order reaction is said in the problem so the rate expression for zero order reaction is given as,
-rA = K
Arhenius equation is,
K = Ae-EA/RT
take log on both side
lnK = lnA + lne-EA/RT
lnK = -EA/RT + lnA
we have Temperature and K data so draw graph of lnK vs 1/T then the slope of the line will be, -EA/R
from the above table,
T (K) | 1/T | K*10-6 | lnK | |
289 | 0.00346 | 1.04 | 13.85 | |
293.5 | 0.003407 | 1.34 | 14.10 | |
298.1 | 0.003354 | 1.53 | 14.24 | |
303.2 | 0.003298 | 1.89 | 14.45 | |
308 | 0.003246 | 2.29 | 14.64 | |
113.5 | 0.00881 | 2.84 | 14.85 |
Draw a graph if you want and slope of that line will be,
Slope = -EA/R = Y/X
= (14.24 - 14.10)/(0.003354 - 0.003407)
= 0.14/-0.000053
-EA/R = -2641.5
EA/R = 2641.5
EA = 2641.* 8.314
EA = 21961 KJ/Kmol
2.Hcan be calculated by formula we have,
EA = H + RT
H = EA - RT
and Arhenius constant A can be calculated by
K = e-EA/RT
lnK = lnA - EA/RT
lnA = lnK + EA/RT
T | K*10-6 | EA | H | lnA | A |
289 | 1.04 | 21961 | 19558 | 9.17 | 9693 |
293.5 | 1.34 | 21961 | 19520 | 9.29 | 10856 |
298.1 | 1.53 | 21961 | 19482 | 9.28 | 10788 |
303.2 | 1.89 | 21961 | 19440 | 9.34 | 11481 |
308 | 2.29 | 21961 | 19400 | 9.40 | 12145 |
113.5 | 2.84 | 21961 | 21017 | 24.31 | 3.64*1010 |