Question

The table below gives the values for the rate constant, k, of the reaction between potassium hydroxide and bromoethane in ethanol at a series of temperatures. Use these data to determine the activation energy of the reaction.

Explain your answer in brief.

T/K 305.0 313.0 323.1 332.7 343.6 353.0

k/M-1 s-1 0.182 0.466 1.35 3.31 10.2 22.6

A 80J

B 90J

C 80 J mol-1

D 80 kJ mol-1

E 90 kJ mol-1

Answer 1

Calculate the values of ln K and 1/T

K	ln K	T	1/T
0.182	-1.7037	305	0.00327
0.466	-0.7636	313	0.00319
1.35	0.3	323.1	0.00309
3.31	1.197	332.7	0.00301
10.2	2.322	343.6	0.00291
22.6	3.118	353	0.00283

From the Arrhenius equation

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

?Y = m*X + c

Slope = -(y2 - y1) / (x1 - x2)

= -(2.322 - 1.197)/(0.00301 - 0.00291)

= -11250 K-1

?Slope = - E/R

-11250 K-1 = - E/(8.314 J/mol·K)

?E = 93532.5 J/mol x 1kJ/1000J

?E = 93.5 kJ

?Option E is the correct answer (nearest)