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January 25, 2015

January 25, 2015

Posted by **Amy** on Sunday, February 3, 2013 at 3:26pm.

Given that the initial rate constant is 0.0160 s-1 at an initial temperature of 25 degrees C, what would the rate constant be at a temperature of 130 degrees C for the same reaction described in Part A?

I solved for the Ea, and I got 32,900 J.

ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)

ln(k2/0.0160) = (32,900/8.314)(1/298.15 - 1/314.55)

ln(k2/0.0160) = (3,957.180)(0.00174871)

ln(k2/0.0160) = (0.6919986)

(k2/0.0160) = e^(0.6919986)

(k2/0.0160) = 1.997704

k2 = 0.03196

The answer above is not correct.

Thanks!

- Rate Constant - My Work is Incorrect -
**DrBob222**, Sunday, February 3, 2013 at 4:00pmln(k2/k1) = (Ea/R)(1/T1 - 1/T2)

**The equation is OK**

ln(k2/0.0160) = (32,900/8.314)(1/298.15 - 1/314.55)**All of the substitutions look ok except for T2. 273.15 + 130 C = 403.15 and not 314.15.**

ln(k2/0.0160) = (3,957.180)(0.00174871)

ln(k2/0.0160) = (0.6919986)

(k2/0.0160) = e^(0.6919986)

(k2/0.0160) = 1.997704

k2 = 0.03196

**With the corrected T2 I obtained k2 = 0.5074.**

This is the same problem I did yesterday, I think, EXCEPT the problem then was to calculate T2 when k was doubled. This is just the reverse, so of course, you obtained 0.032 for k2 which is just double k1 when you substituted the T2 we calculated. If this k2 (0.5074 to the right number of significant figures) is not right, the problem may be the Ea. Since that came from another problem (at least it appears that way from your presentation) that could be the culprit. Hope this helps.

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