Find the rate constant, given temperature and activation energy.

[Agent M27]

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

$$ln\frac{K_{2}}{K_{1}}=\frac{E_{a}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)$$

The Attempt at a Solution

Given:

R=8.314
T₁=249K
T₂=436K
E_a=160

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

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

Joe

 

[Borek]

38000

 

[Agent M27]

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

Joe

 

[Borek]

kJ, not Kj...

 

[DeeNos]

I would like to point out that your attempt at a solution is close, but there are a few errors in your calculations.

Firstly, the activation energy (Ea) should be in units of J/mol, not kJ/mol. So it should be 38,000 J/mol.

Secondly, the natural logarithm function (ln) should be applied to the ratio of the rate constants, not just the second rate constant. So your equation should be:

ln\frac{K_{2}}{K_{1}}=\frac{38,000}{8.314}\left(\frac{1}{249}-\frac{1}{436}\right)

This will give you a value of 3.009 for the natural logarithm of the ratio of rate constants.

To find the actual ratio of rate constants, you will need to take the inverse natural logarithm (e^x) of this value. So your final equation should be:

\frac{K_{2}}{K_{1}}=e^{3.009}

This will give you a value of approximately 20.26 for the ratio of rate constants.

To find the rate constant at 436 K, you can then multiply the rate constant at 249 K (1.60 x 10^2 M^-1s^-1) by this ratio. This will give you a value of approximately 3.24 x 10^3 M^-1s^-1 for the rate constant at 436 K.

I hope this helps and good luck with your calculations!