Question

The Arrhenius Equation is typically written as

k=Ae−Ea/RT

However, the following more practical form of this equation also exists:

lnk2k1=EaR(1T1−1T2)

where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1and T2).

Part A

The activation energy of a certain reaction is 32.1 kJ/mol . At 20 ∘C, the rate constant is 0.0130 s−1. At what temperature would this reaction go twice as fast?

Express your answer numerically in degrees Celsius

Part B

Given that the initial rate constant is 0.0130 s−1 at an initial temperature of 20 ∘C, what would the rate constant be at a temperature of 100 ∘C?

Express your answer numerically in inverse seconds.

Answer 1

A)

we have:

T1 = 20 oC

=(20+273)K

= 293 K

K2/K1 = 2/1

Ea = 32.1 KJ/mol

= 32100 J/mol

we have below equation to be used:

ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

ln(2/1) = (32100.0/8.314)*(1/293.0 - 1/T2)

0.6931 = 3860.9574*(1/293.0 - 1/T2)

T2 = 309 K

= (309-273) oC

= 36 oC

Answer: 36 oC

B)

we have:

T1 = 20 oC

=(20+273)K

= 293 K

T2 = 100 oC

=(100+273)K

= 373 K

K1 = 1.3*10^-2 s-1

Ea = 32.1 KJ/mol

= 32100 J/mol

we have below equation to be used:

ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

ln(K2/1.3*10^-2) = (32100.0/8.314)*(1/293.0 - 1/373.0)

ln(K2/1.3*10^-2) = 3861*(7.32*10^-4)

K2 = 0.2195 s-1

Answer: 0.220 s-1