Question

Please show work and answer all parts! (finding Ea and A)

6.39 Make an appropriate Arrhenius plot of the following data for the binding of an inhibitor to the enzyme carbonic anhydrase and calculate the activation energy for the reaction. 308.0 313.5 T/K 289.0 293.5 298.1 303.2 2.29 k,/(106 dm3 mol"' s) 1.04 1.34 1.53 1.89 2.84

Answer 1

The Arrhenius rate equation is given by $k = Ae^{^{\frac{-E_{a}}{RT}}}$

Where k is the rate of the reaction, A is the pre-exponential factor, E_{​​​​​​a} is the activation energy, R is the universal gas constant and T is the temperature.

To find the value of A and E_{​​​​​​a}, we would have to plot a ln (k) versus 1/T graph.

Applying logarithm, the equation changes to

ln(k) = ln (A) - E_{​​​​​​a}/RT.

This a straight line with slope as -Ea/R and y intercept as ln (A)

The new points are: (1/T, ln(k)) = (0.00346, 13.84), (0.003407, 14.1081), (0.00335, 14.241), (0.003298, 14.452), (0.00325, 14.644), (0.003189, 14.859)

Using Excel or any linear regression calculator, we get the equation of the line as y = -3678.36762x + 26.59037, where x is 1/T and y is ln(A).

ln(A) = 26.59037 => A = e^{​​​​​​26.59037} = 3.53 * 10¹¹

-E_{​​​​​a}​/R = -3678.36762 => E_{​​​​​​a} = R * 3678.36762 = 8.314*3678.36762 = 30581.95

The values are A = 3.53*10¹¹ dm^{​​​​​3}​​/mol.s

Ea = 30581.95 J/mol

The plot is attached below:

6 -4 6 5 2- 3 4 2 0 200 400 LO

The point numbers corresponds to the order of the points