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A reaction has a rate constant of 1.26×10−4 s−1 at 28 ∘C and 0.233 s−1 at...

A reaction has a rate constant of 1.26×10−4 s−1 at 28 ∘C and 0.233 s−1 at 79 ∘C .

1.Determine the activation barrier for the reaction. Express your answer in units of kilojoules per mole.

2. What is the value of the rate constant at 17 ∘C ? Express your answer in units of inverse seconds.

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Answer #1

1. From Arhenius Equation K= koe-Ea/RT

where Ko =Frequency factor and Ea = activation energy and R= gas constant =8.314 J/mole.KT= temperature in K.

at two different temperatures T1= 79 deg. = 79+273.15=352.15 K and T1= 28 deg.c =28+273.15=301.15K

K2= 0.233 sec-1 and K1= 1.26*10-4 sec-1

the equation can be written as ln (K2/K1)= (Ea/R)*(1/T1-1/T2)

ln (0.233/1.26*10-4)= (Ea/R)*(1/301.15-1/352.15)

7.52 = (Ea/R)*0.000481

Ea/R =7.52/0.000481=15637.17

Ea= 15637.17*8.314= 130007.5 Joules/Mole=130.0075 Kj/mol

2. Let the temperature at which rate contant need to be evaluated be T3 =17 deg.c =17+273.15= 290.15 and the rate constant be K3

Writing the equation for two diferent temperatures T1 and T3 gives

ln (K1/K3)= (Ea/R)*(1/T3-1/T1) =15637.17*(1/290.15-1/301.15)= 15637.17*0.000126=1.968

K1/K3= 7.16

K3= K1/7.16= 1.26*10-4/7.16= 1.759*10-5 sec-1

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