In a certain solvent is first order with respect to (CH3)3CBr and zero order with respect to OH2. In several experiments the rate constant k was determined at different temperatures. A plot of ln(k) versus 1/T was constructed that resulted in a straight line with a slope of 21.10 3 104 K and a y intercept of 33.5. Assume that k has units of s21. a. Determine the activation energy for this reaction. b. Determine the value of the frequency factor A. c. Calculate the value of k at 258C.

Ea= -175.45J

A= 3.5×10^14

k=3.64 ×10^14 s^2.

Explanation:

From

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

This is similar to the equation of a straight line:

y= mx + c

Where m= -(Ea/R)

c= ln A

y= ln k

a)

Therefore

21.10 3 104= -(Ea/8.314)

Ea=-( 21.10 3 104×8.314)

Ea= -175.45J

b) ln A= 33.5

A= e^33.5

A= 3.5×10^14

c)

k= Ae^-Ea/RT

k= 3.5×10^14 × e^ -(-175.45/8.314×531)

k = 3.64 ×10^14 s^2.

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