Find the rate constant, given temperature and activation energy.

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Homework Statement



A reaction is found to have an activation energy of 38.0 kJ/mol. If the rate constant for this reaction is 1.60 × 102 M-1s-1 at 249 K, what is the rate constant at 436 K?

Homework Equations



[tex]ln\frac{K_{2}}{K_{1}}=\frac{E_{a}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)[/tex]

The Attempt at a Solution



Given:

R=8.314
T1=249K
T2=436K
Ea=160

[tex]ln(K_{2})=\frac{38}{8.314}\left(\frac{1}{249}-\frac{1}{436}\right)+ln(160)[/tex]

Which equals 161.257 which is incorrect. Any clues where I went wrong would be greatly appreciated. Thanks in advance.

Joe
 
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Ah ha! I should have noticed that being that R has units of J not Kj. Thank you very much Borek.

Joe