March 28, 2024, 03:46:19 PM
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Topic: Help finding rate constant at T2 knowing T1, T2 and activation energy of reactio  (Read 1251 times)

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Offline flukeylukey

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Hello,
I have spent the past hour going over the same simple question and I can not work out where I am making a mistake. The question is:

"The rate constant k for a reaction is 4.70 ×10^-3 s^-1 at 25.0° C and the activation
energy is 33.6 kJ mol^-1. What is the rate constant at 65.0 °C?"

I am using the equation ln(k2/k1) = -Ea/R x (1/T2 - 1/T1).

I have change the units for temps to K and changed the given activation energy to J mol^-1. I am using the R value of 8.314 J mol^-1 K^-1. I can not see any issues with my units and I have gone over my working time and time again and found my K2 value to be 4.7 * 10^-3 each time. I don't understand, shouldn't the K2 value change from the K1 value due to the temp change?

I can take some photos of my working if that will help.
Thank you

Offline Alt+Tab

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Hello, you are correct about the fact that K2 changes due to temperature variation. Your result for K2 is pretty weird. In fact, I calculated here, and my result gave: ln k2 = 10.69 resulting in K2 = 43.91 * 10 ^ 3. My calculation is exposed to errors because I did it quickly, but it corresponded with the temperature analogy increasing the constant. Hope this helps.

Offline mjc123

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I get 2.3 x 10-2 s-1. There is no reason why K2 should be the same as K1; are you sure you're not just entering the numbers into a calculator in the wrong format, or something like that?
44 x 103 is very wrong (unless you misread K1 as 4 x 103). You could check your answer (ballpark) using the rule of thumb that a 10°C increase in temperature roughly doubles the rate. (This corresponds to Ea of about 50 kJ/mol, so the increase will be a bit less in this case.)

Offline Alt+Tab

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You are totally right. I re-read the correct values and K2 gave exactly 2.33 * 10 ^ -2. But I did not know about this rule. a good tip.

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