Question

Part A

The activation energy of a certain reaction is 48.4 kJ/mol . At 24  ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast

T2 =

35 ∘C

Part B

Given that the initial rate constant is 0.0190s−1 at an initial temperature of 24  ∘C , what would the rate constant be at a temperature of 190.  ∘C for the same reaction described in Part A?

k2 =?

Answer 1

Apply Ahrrenius

ln(k2/k1) = E/R*(1/T1-1/T2)

ln(2k1/k1) = (48400)/(8.314)*(1/(24+273) - 1/T2)

Solve for T2

1/T = (ln(2k1/k1)*8.314/(48400) - 1/(24+273)) = -0.00324

T = -(-0.00324^-1) = 308.6419 K

T = 308.6419-373 = 35.6419 °C

B)

E = 48400 and R = 8.314

if rate K = 0.019 and T = 24 °C = 299 K

then

find k2 at T = 190 ° C = 463

ln(k2/k1) = E/R*(1/T1-1/T2)

ln(k2/(0.019)) = 48400/8.314*(1/(299)-1/463)

solve for k2

ln(k2/(0.019)) = 6.896472

k2 = 0.019*exp(6.896472) = 18.7868 s^-1