C2. (5) The reaction: (CH3)3C-Br + NaOH (CH3)3C-OH + NaBr in a certain solvent i

Question

C2. (5) The reaction: (CH3)3C-Br + NaOH (CH3)3C-OH + NaBr in a certain solvent is first order with respect to (CH3)&C-Br; and zero order with respect to NaOH. In several experiments, the rates and rate constants k were determined at several temperatures and a plot of In(k) versus 1/T was constructed resulting in a straight line with a slope value of -1.10 x 10 K and a y-intercept of 33.5. Assume k has units of s (a) Determine the activation energy (in Joules/mol) for this reaction. (b) Determine the value of the Arrhenius frequency factor A. (c) Calculate the value of the rate constant k at 25°C.

Explanation / Answer

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,

a)

-Ea/R = slope

-Ea/R = -1.1*10^4

Ea = ( -1.1*10^4)(-8.314) = 91454 J/mol

Ea = 91454 J/mol

b)

for A

ln(A) = y.itnercept

ln(A) = 33.5

A = exp(33.5) = 3.53*10^14

c)

get k at T = 25°C

y = m*x + y

y = (-1.1*10^4)(1/(295)) + 33.5

y = -3.78813

ln(K) = -3.78813

k = exp(-3.78813)

k = 0.02263 1/s

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