Question

A reaction has a rate constant of 1.22×10−4 s−1 at 26  ∘C and 0.234 s−1 at 75  ∘C...

A reaction has a rate constant of 1.22×10−4 s−1 at 26  ∘C and 0.234 s−1 at 75  ∘C .

A) Determine the activation barrier for the reaction.

B)What is the value of the rate constant at 15  ∘C ?

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Homework Answers

Answer #1

A)

we have:

T1 = 26 oC

=(26+273)K

= 299 K

T2 = 75 oC

=(75+273)K

= 348 K

K1 = 1.22*10^-4 s-1

K2 = 0.234 s-1

we have below equation to be used:

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

ln(0.234/1.22*10^-4) = ( Ea/8.314)*(1/299.0 - 1/348.0)

7.5591 = (Ea/8.314)*(4.709*10^-4)

Ea = 133454 J/mol

Ea = 133 KJ/mol

Answer: 133 KJ/mol

B)

we have:

T1 = 26 oC

=(26+273)K

= 299 K

T2 = 15 oC

=(15+273)K

= 288 K

K1 = 1.22*10^-4 s-1

we have below equation to be used:

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

ln(K2/1.22*10^-4) = (133454.0/8.314)*(1/299.0 - 1/288.0)

ln(K2/1.22*10^-4) = 16052*(-1.277*10^-4)

K2 = 1.57*10^-5 s-1

Answer: 1.57*10^-5 s-1

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