Question

An Arrhenius plot (ln k vs. 1/T) for a second order reaction (2A -> C) produced...

An Arrhenius plot (ln k vs. 1/T) for a second order reaction (2A -> C) produced a straight line with a slope of -8.23e+03 K. What is the value of the Arrhenius pre-exponential if k = 4500 M-1 s-1 at 690 K?

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Answer #1

The relation between K and T is:

K = A*e^(-Ea/RT)

taking ln on both sides, above expression becomes,

ln K = ln A - Ea/RT

So, slope between ln K and 1/T will have a value of -Ea/R

Here slope = -8.23*10^3

So,

-Ea/R = -8.23*10^3

-Ea/8.314 = -8.23*10^3

Ea = 6.84*10^4 J/mol

we have:

T= 690.0 K

K = 4.5*10^3 M-1.s-1

Ea = 6.84*10^4 J/mol

R = 8.314 J/mol.K

we have below equation to be used:

K = A*e^(-Ea/RT)

4.5*10^3 = A*e^(-68400.0/(8.314*690.0))

4.5*10^3 = A*e^(-11.9233)

4.5*10^3 = A*6.634*10^-6

A = 6.78*10^8 M-1.s-1

Answer: 6.78*10^8 M-1.s-1

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