Homework Help: Temperature Problem

Posted by Amy on Saturday, February 2, 2013 at 10:16am.

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)(1/301.15-1/352.15)
-7.551=(Ea/8.314)(0.0048090535)
-7.551=(Ea/8.314)(0.0048090535))
-15,702.808=(Ea/8.314)
Ea= -1,192.24 J/mol

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

Temperature Problem - DrBob222, Saturday, February 2, 2013 at 3:05pm
I believe you have reversed the k1/T1 vs k2/T2. Also I think you are reporting too many significant figures
I would set it up as
ln(k2/k1) = (E/R)(1/T1 - 1/T2)
ln(0.235/1.23E-4) = (Ea/8.314)(1/301.15 - 1/352.15)
My answer is approximately 130,000

Answer this Question

Related Questions

Kinetics Problem I - A reaction has a rate constant of 1.23×10-4 s at 28 ...
Temperature - Q: A reaction rate has a constant of 1.23 x 10-4/s at 28 degrees C...
How to Plug these values into the Arrhenius Formul - A reaction has a rate ...
Finding the Activation Barrier - Q: A reaction rate has a constant of 1.23 x 10-...
Arrhenius Reaction - The activation energy of a certain reaction is 32.9 kJ/mol ...
Arrhenius Question - The activation energy of a certain reaction is 32.9 kJ/mol ...
Chemistry - A second order reaction has a rate constant of 8.7x10^-4/(M*s) at 30...
Chemistry - At a given temperature, the elementary reaction A <=> ...
science - If a temperature increase from 19.0 degrees C to 36.0 degrees C ...
Chemistry - Your Open Question Show me another » Chemistry help needed...

For Further Reading

First Name:
School Subject:
Answer: