Rate law and Rate Constant

Question

1. The ra‎te of the rea‎ction between hemo‎globin (Hb) and ca‎rbon monox‎ide (CO) was stud‎ied at 20^oC. The fo‎llowing data‎ were collectp‎ed with all concentra‎tion units in micromol/L.

[Hb]0 (micromol/L)

[CO]0 (micromol/L)

Initial Rate (micromol/L -s)

2.21                    1.00                0.619

4.42                    1.00                 1.24

4.42                    3.00                 3.71

a. Determ‎ine the rate law an‎d the value of the‎ rate cons‎tant. 

b. Wha‎t is the Activation‎ energy of above‎ reaction i‎f rate constant was calcul‎ated to be at 1.0‎0 x 10³  L/ (micromol^-s) at 30^o C .

Answer 1

First, determine the order with respect to both reactants.

For Hb: compare first 2 experiments where [Hb] doubles from 2.21 uM to 4.42 uM and [CO] remains constant. The rate increases from 0.619 to 1.24, which is also a doubling. So, this tells us the reaction is FIRST ORDER with respect to Hb

For CO: compare experiments 2 and 3 where [CO] triples from 1.00 uM to 3.00 uM and [Hb] remains constant. The rate increases from 1.24 to 3.71, which is also a tripling. So, this tells us the reaction is FIRST ORDER with respect to CO.

(a). The rate law is..

Rate = k[Hb][CO]

To find the value of the rate constant (k), simply choose an experiment and plug in the values. Using the first experiment, we have

0.619 uM/s = k(2.21 uM)(1.00 uM)

k = 0.619 uM/s / (2.21 uM)(1.00 uM)

k = 0.280 uM^-1s^-1

(b). To solve this part of the question, we will use the Arrhenius equation.

Arrhenius equation:

ln (k2/k1) = -Ea/R(1/T2 - 1/T1)

k1 = 0.280 uM^-1s^-1

k2 = 1.00x10³ uM^-1s^-1

Ea = ?

R = 0.008314 kJ/Kmol (note units are in kJ, not J)

T1 = 20ºC + 273 = 293K

T2 = 30ºC + 273 = 303K

ln (1.00x10³ / 0.280) = -Ea/0.008314 (1/303 - 1/293)

8.18 = -Ea/0.008314 (0.0033 - 0.0034)

8.18 = -Ea/0.008314 (-0.0001)

8.18 =0.0001Ea / 0.008314

Ea = 680 kJ/molK (better check the math as this seem rather high, but not impossible)