Friday

August 28, 2015
Posted by **Devron** on Friday, February 1, 2013 at 10:15pm.

- Dr.Bob222 Chem -
**DrBob222**, Friday, February 1, 2013 at 10:31pmYes and no. What I said, or should have said, is that changing the temperature by 10 C (approximately) doubles the rate. The question asked for the temperature at which the rate was doubled. So yes, the reaction rate is constant but ONLY for that particular temperature. At any other temperature the rate will be different. The "rule of thumb" is that we double the rate for 10 C change. So increasing T from 20 to 70 we increase rate by approximately 2

^{5/sup> = 32 times faster. } - Dr.Bob222 Chem -
**Devron**, Friday, February 1, 2013 at 10:43pmOkay, I think I get what you are saying. So, let's say that I keep the concentration of reactants constant, but I increase the temperature or decrease the temperature for a reaction. Since I played around with the temperature (i.e., varying it) I changed the reaction rate, that I calculated for the reaction for the same reaction before I varied the temperature, therefore I changed the constant k, since I used the same concentration to determine k. Is there a specific equation that shows how the relationship works, or did you learn this in course work or while doing research?

- Dr.Bob222 Chem -
**Devron**, Friday, February 1, 2013 at 10:44pmSo increasing T from 20 to 70 we increase rate by approximately 25/sup> = 32 times faster.

This part here is why I am asking about an equation.

- Dr.Bob222 Chem -
**DrBob222**, Friday, February 1, 2013 at 11:13pmYes, it's the Arrhenius equation.

ln(k2/k1) = (Ea/R)(1/T1 - 1/T2). This give you a specific equation to determine what the new k will be at the new T. The 2^5 thing I quoted is a "rule of thumb" that is approximate. Try the equation giving values for k1 and T1 and Ea (R is 8.314 here) and use T2 that's 10+T1. Calculate k. It should be close to double k1. Or use ln (k2/k1) = ln 2 and solve for T2 for some given value of T1 and Ea. T2 should be close to 10 C.

- Dr.Bob222 Chem -
**Devron**, Friday, February 1, 2013 at 11:17pmGot it. I had to reread the original problem to see what went on. I haven't had use for some of these calculations and years, so some concepts are fuzzier than others.