Question

# The table below gives the values for the rate constant, k, of the reaction between potassium...

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.

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

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)

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