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### Topic: {activation energy}  (Read 1733 times)

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#### whirlofwings

• Regular Member
•   • Posts: 11
• Mole Snacks: +0/-0 ##### {activation energy}
« on: February 02, 2013, 02:09:53 AM »
Hi! I am having a lot of trouble with these two problems:

Q: The activation energy of a certain reaction is 32.9 kJ/mol At 25 degrees C the rate constant is 0.0160 units s-1 units. At what temperature in degrees Celsius would this reaction go twice as fast?

My work:
ln(0.0160/0.032) = (32,900/8.314)(1/T2 - 1/298.15)
-0.6931 = 3,957.180(1/T2 - 0.00335)
-0.6932 = 3,957.180(1/T2) - 13.2565
13.187T2 = 3,957.180
T2 = 300.08 K = 26.98 degrees C

The answer above is not correct.

Q: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)(0.00048090535)
-7.551=(Ea/8.314)(0.00048090535)
15,708.80800535=(Ea/8.314)
Ea = 13,0553.145
Ea=130.5531 J/mol
2,723 J/mol = 0.002723 kJ/mol

Thank you!

*MOD Edit*
title changed from: Can someone tell me what I am doing wrong with these chemistry problems? to more descriptive
« Last Edit: February 02, 2013, 09:33:49 AM by Arkcon »