Question

In: Chemistry

A reaction has a rate constant of 1.22*10^-4/s at 27 0C and 0.230/s at 78 0C....

A reaction has a rate constant of 1.22*10^-4/s at 27 ⁰C and 0.230/s at 78 ⁰C. Determine the activation barrier for the reaction. What is the value of the rate constant at 18 ⁰C?

Expert Solution

Solution :-

Given data

K1 = 1.22*10^-4 s-1

K2 = 0.230 s-1

Temperature T1 = 27 C +273 = 300 K

Temperature T2 = 78 C +273 = 351 K

Activation energy Ea = ?

Using the Arrhenius equation we can calculate the activation energy as follows

ln[k2/k1] = Ea/R [(1/T1)-(1/T2)]

R= constant = 8.314 J permol K

Lets put the values in the formula

ln[0.230/1.22*10^-4] = Ea/8.314 J per mol K [(1/300)-(1/351)]

7.5418=Ea/8.314 J permol K *0.000484 K

Ea = 7.5418*8.314 J per mol K / 0.000484 K

Ea = 129551 J permol

Lets convert it to okJ

129551 J per mol * 1 kJ / 1000 J = 129.6 kJ per mol

Therefore the activation barrier is 129.6 kJ per mol

Second part

Calculating the rate constant at 18 C

18 C +273 = 291 K

Lets use the initial temperature and rate constant at 27 C that is 300 K and K1 = 1.22*10^-4 s-1

Using the same equation we can calculate the rate constant K2 at 18 C that is 291 K

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

lets put the values in the formula

ln[K2/1.22*10^-4]= 129551 J per mol / 8.314 J per mol K *[(1/300)-(1/291)]

ln[K2/1.22*10^-4]= -1.61

[K2/1.22*10^-4]= anti ln -1.61

K2/1.22*10^-4= 0.1999

K2 = 0.1999 * 1.22*10^-4

K2=2.44*10^-5 s-1

Therefore the rate constant at 18 C = 2.44*10^-5 /s

queen_honey_blossom answered 1 year ago

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