Question

calculate at room temperature: a) the ratio of the rate constants for two reactions that have the same pre-exponential value but have activation energies that differ by 40.0 kJ/mole b) the difference in the activation energies (Ea1-Ea2) when the first reaction has a pre-exponential value that is 5 times the pre-exponential value of the second reaction and where the rate constant of the first reaction is 100 times larger than the rate constant of the second reaction.

Answer #1

a) k = A e^{-Ea/RT}

k1= A e^{-Ea1/RT}

k2= A e^{-Ea2/RT}

k1/k2 = A e^{-Ea1/RT} / A
e^{-Ea2/RT}

= e^{-(Ea1-Ea2)/RT}

= e^-40000 J/(8.314 J/K/mol) (298)

= 9.73 x10^{-8}

k1/k2 = 9.73 x10^{-8}

b) Given that

A1= 5A2

k1 = 100 k2

k1= A1 e^{-Ea1/RT}

k2= A2 e^{-Ea2/RT}

k1/k2 = A1 e^{-Ea1/RT} / A2
e^{-Ea2/RT}

100k2/k2 = 5A2 e^{-(Ea1-Ea2)/RT} /A2

100 = 5 e^{-(Ea1-Ea2)/RT}

20 = e^{-(Ea1-Ea2)/(8.314) (298)}

On solving,

Ea1-Ea2 = -7422.1 J/mol

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equation shows the relationship between the rate constant
k and the temperature T in kelvins
and is typically written as
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where R is the gas constant (8.314 J/mol⋅K), A
is a constant called the frequency factor, and Ea
is the activation energy for the reaction.
However, a more practical form of this equation is
lnk2k1=EaR(1T1−1T2)
which is mathematically equivalent to
lnk1k2=EaR(1T2−1T1)
where k1 and k2 are the rate constants for a
single reaction at two different...

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k=Ae−Ea/RT
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lnk2k1=EaR(1T1−1T2)
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