Posted by **Amy** on Thursday, January 31, 2013 at 12:36pm.

Q: A reaction rate has a constant of 1.23 x 10-4/s at 28 degrees C and 0.235 at 79 degrees C. Determine the activation barrier for the reaction.

A: My work thus far:

ln(1.23 x 10-4/0.235)=(Ea/8.314)(1/79-1/28)

-7.551=(Ea/8.314)(0.01265-0.03571)

-7.551=(Ea/8.314)(-0.02305)

327.53=(Ea/8.314)

Ea=2,723.08 J/mol

2,723 J/mol = 0.002723 kJ/mol

My answer is incorrect, and I would like to know where I went wrong and what the correct answer is.

## Answer this Question

## Related Questions

- Kinetics Problem I - A reaction has a rate constant of 1.23×10-4 s at 28 degrees...
- How to Plug these values into the Arrhenius Formul - A reaction has a rate ...
- Temperature Problem - Q: A reaction rate has a constant of 1.23 x 10-4/s at 28 ...
- Temperature - Q: A reaction rate has a constant of 1.23 x 10-4/s at 28 degrees C...
- science - If a temperature increase from 19.0 degrees C to 36.0 degrees C ...
- Chemistry - Can someone please check the following answers for me...THANK YOU! 1...
- rate constant - A reaction has a rate constant of 1.25×10−4 at 28 and 0....
- CH102 - If a temperature increase from 18.0 to 34.0 triples the rate constant ...
- chemistry - I have two questions that I really don't understand. The first one ...
- ch102 - The rate constant of a reaction at 30 was measured to be 5.7×10^-2. If ...

More Related Questions