How do I solve this Chemistry problem on activation energy and calculating the value of k?

Question

 The reaction

(CH3)3CBr+ OH==>(CH3)3COH+Be

 

In a certain solvent is first order with respect to (CH3)3CBr and zero order with respect to  OH-. In several experiments, the rate constant k was determined at different temperatures. A plot of ln (k) vs. 1/T was constructed resulting in a straight line with a slope of -1.10×10^4K and y intercept of 33.5. Assume k has unit of s^-1.

 

A. Determine the activation energy for this reaction

 

B. Determine the value of the frequency factor A.

 

C. Calculate the value of k at 25°C

Answer 1

A.  The activation energy (Ea) is the slope of the line = -Ea/R so calculate Ea from that.

 

B.  The y-intercept is equal to ln A, so find A from that.

 

C.  At 25ºC, you convert that to 298ºK and take the reciprocal.  1/298 = 0.003356 and then find this value on the graph and read the corresponding ln k and from that find k.  If not required to use the graph, you can find k from ln (k2/k1) = -Ea/R(1/T2 - 1/T1)