Question

A. Using a first order plot, determine the rate constant, k_1, for the oxidation of benzyl alcohol with potassium permanganate at the temperature, T_1 = 20.8 degrees celcius, from the information in the table below. ( Molar absorptivity = 2350M^-1cm^-1 ; path length, 1=1cm.

Time (Minutes)	Absorbance
15	.785
30	.500
45	.0.333

B.) Using a second rate constant, K_2 = .0609min^-1, at the temperature, T_2 = 37.9 degrees celcius, determine the activation energy for the oxidation of benzyl alcohol with potassium permanganate. (Gas constant, R = 8.314 J/(K*mol))

Answer 1

Solution :-

A) To calculate the rate constant K1 lets plot the data of ln[A] vs time

Then we get the following graph

Where the K= - slope

The equation of the straight line is y=-0.0286x+0.1793

So the

Slope = -0.0286

K1 = - (-0.0286)

K1= 0.0286 min-1

B) Now lets calculate the activation energy Ea

K1 = 0.0286 min -1

K2 = 0.0609 min-1

T1 = 20.8 C +273 = 293.8 K

T2 = 37.9 C +273 = 310.9 K

Ea= ?

Ln(K2/K1) = Ea /R [(1/T1)-(1/T2)]

Ln[0.0609/0.0286] = Ea / 8.314 J per mol K [(1/293.8)-(1/310.9)]

0.7558 = (Ea / 8.314 J per mol K ) * 0.0001872

0.7558 * 8.314 J per mol K / 0.0001872 K = Ea

33570 J /mol = Ea

Lets convert it to the kJ

33570 J/mol * 1 kJ / 1000 J = 33.6 kJ/mol

So the activation energy Ea = 33.6 kJ/mol