The Arrhenius equation shows the relationship between the rate constant k and th

Question

The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as

k=AeEa/RT

where R is the gas constant (8.314 J/molK), 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(1T11T2)

which is mathmatically equivalent to

lnk1k2=EaR(1T21T1)

where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2).

Part A

The activation energy of a certain reaction is 31.2 kJ/mol . At 25  C , the rate constant is 0.0190s1. At what temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

Part B

Given that the initial rate constant is 0.0190s1 at an initial temperature of 25  C , what would the rate constant be at a temperature of 200.  C for the same reaction described in Part A?

Express your answer with the appropriate units.

Explanation / Answer

As the arhenius equation,

k= Ae- E/RT

0.0190 s-1 = A e- (31200 J mol-1 ÷ (8.314 J mol-1 k-1 ))

A = 5594.82 s

When the rate become 2 times faster,

0.0190× 2 s-1 = 5594.82 s × e- (31200 Jmol-1 ÷(8.314 Jmol-1 K-1 × T)

TTherefore T = 315.45 K

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