Question

A student collected conducted 6 kinetics experiments for one reaction by holding the concentrations of ALL the reactants and catalyst constant and varying the temperature. After determining the rate constant for this reaction at the 6 different temperatures, an Arrhenius Plot was made. The graph resulted in a straight line. The equation for the straight line was: y= -3,920.54 x + 14.326 What is the Activation Energy for this reaction at 30 oC?

Answer 1

From Ahrrenius equation;

K1 = A*exp(-Ea/(RT1))

K2 = A*exp(-Ea/(RT2))

Note that A and Ea are the same, they do not depend on Temperature ( in the range fo temperature given)

Then

Divide 2 and 1

K2/K1 = A/A*exp(-Ea/(RT2)) / exp(-Ea/(RT1))

Linearize:

ln(K2/K1) = -Ea/R*(1/T2-1/T1)

get rid of negative sign

ln(K2/K1) = Ea/R*(1/T1-1/T2)

for a generic value:

ln(K) = -Ea/R*(1/T) + ln(A)

where:

x-axis = 1/T ; inverse value of absolute temperature

y-axis = ln(K) , natural log of the rate constant

slope = -E/R or Acitvation Energy divided by Ideal Gas Constant

y-intercept = ln(A) ; natural log of the Frequency Factor

now...

ln(k) = -Ea/R*(1/T1) + ln(A)

has the shape of

y = m*x + b

slope = -Ea/R

slope = -3920.54

Ea/R = 3920.54

Ea = 8.314 J/mol * 3920.54

Ea = 32595.37 J