The activation energy for the isomerization ol cyclopropane to propene is 274 kJ/mol. By what factor does the rate of this reaction increase as the temperature rises from 231 to 293 oC?

By a factor of 2.25.

Explanation:

Using the Arrhenius equations for the given conditions:

k1 = A * (exp^ (-Ea / (RT1))

k2 = A * (exp^ (-Ea / (RT2))

T1 = 231°C

= 231 + 273.15 K

= 574.15 K

T2 = 293°C

= 293 + 273.15 K

= 566.15 K

Ea = 274 kJ mol^-1

R = 0.008314 kJ/mol. K

Now divide the second by the first:

k2/k1 = exp^ (-Ea/R * (1/T2 - (1/T1))

= 0.444

2.25k2 = k1

The rate of the reaction increased by a factor of 1012.32

Explanation:

Applying Arrhenius equation

ln (k₂/k₁) = Ea/R (1/T₁ - 1/T₂)

where;

k₂/k₁ is the ratio of the rates which is the factor

Ea is the activation energy = 274 kJ/mol.

T₁ is the initial temperature = 231⁰C = 504 k

T₂ is the final temperature = 293⁰C = 566 k

R is gas constant = 8.314 J/Kmol

Substituting this values into the equation above;

ln (k₂/k₁) = 274000/8.314 (1/504 - 1/566)

ln (k₂/k₁) = 32956.4589 (0.00198-0.00177)

ln (k₂/k₁) = 6.92

k₂/k₁ = exp (6.92)

k₂/k₁ = 1012.32

The rate of the reaction increased by 1012.32