Rate and temperature calculation problem

[mourinho]

I having trouble with this question and the answer provided is not helping me either. If the rate of a certain reaction is increased by a factor of 10, the temperature increase is 3 degrees Celcius. What is the temperature increase from the same initial temperature if the rate is increased by a factor 0f 100? I know that I need to use the Arhenius equation but can't do the math?
The solution given is that if the if the rate is increased 10 times as much as previous one, the temperature increased will be twice as much. I can't comprehend that.

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[aldol16]

rate 2/rate 1 = k2/k1 = (e^-Ea/RT2)/(e^-Ea/RT1)
ln(rate 2/rate 1) = -Ea/RT2 - (-Ea/RT1)
ln(10) = -Ea/R*(1/T2-1/T1)

Are you given the initial or final temperature of the first reaction? A delta T of 3 K doesn't help you in this equation. If you are given the absolute temperature(s) of the first reaction, you can figure out what -Ea/R is and use that in the second equation where rate 2/rate 1 would equal 100 instead of 10.

 

[mourinho]

Actually the initial temp as well as the final temp is not given. The equation given is : 2.3lg(k2/k1) = (Ea/R) * [(T2-T1)/T1T2)].
So, I can't solve it using 2 equations since there are 3 unknowns. The answer given is 6degrees. The logic was that the log is proportional to the temperature increase. I am just baffled though because they ignored the temperature products at the denominator.

 

[aldol16]

Alright, so that gives me some information:

(from before) ln(k2/k1) = -Ea/R*(1/T2-1/T1)
2.3*log(k2/k1) = -Ea/R*(1/T2-1/T1)
2.3*log(k2/k1) = -Ea/R*((T1-T2)/T1*T2) = Ea/R*((T2-T1)/T1*T2)

This is the equation they have and it's a simple transformation from natural log to base 10 log. From here, the only conceivable thing I can think of is that you can assume T1*T2 is similar in both cases. You can make this assumption because the change in T is relatively small and thus the product of T1 and T2 will remain fairly constant. Imagine a reaction going from 298 K to 301 K. 298*301=89,698. Now this new case, it's going from 298 to 304 (given, you don't know this until you've solved which makes this assumption not very helpful although you're only increasing the rate by one order of magnitude so you might expect the delta T to not change by much, according to the rule of 6). 298*304=90,592. So if you take the product of T1 and T2 to be approximately constant over the range of interest, log(k2/k1) becomes dependent linearly on T2-T1. Therefore, since log(k2/k1) doubles from case 1 to case 2, you would also expect T2-T1 to double.

 

[mourinho]

Alright, without knowing the answer, I thought that the temperature increase would be higher when the rate is increased from 10 fold to 100 fold and their products could not be ignored. I hate questions like these which rely on assumptions!

 

[victorias]

Can we say that ln k is proportional to 1/T?

ln k1/ln k2 = (-Ea1/RT1)/(-Ea2/RT2)
R is a constant
Assuming Ea1 = Ea2
ln k1/ln k2 = (1/T1)/(1/T2)
ln k1/ln k2 = (T2)/(T1)
2.3/4.6 = T2/276K
T2 = 138K = -135 C

I am not sure if this is the correct way to think of this

 

[aldol16]

ln k1/ln k2 = (-Ea1/RT1)/(-Ea2/RT2)

Technically will work, but not with this question. The quantity lnk1/lnk2 does not help you with the information given. You know that k2/k1 is 10 or 100 but that doesn't tell you what lnk1/lnk2 will be.

 

[aldol16]

Alright, without knowing the answer, I thought that the temperature increase would be higher when the rate is increased from 10 fold to 100 fold and their products could not be ignored. I hate questions like these which rely on assumptions!

I agree, this is not a good question. But the temperature increase is actually higher when the rate is increased. The increase goes from +3 to +6 - an increase in the delta T. The product of T1 and T2, on the other hand, can be safely ignored as long as you're not working at the extreme, i.e. close to 0 K.

 

[victorias]

Technically will work, but not with this question. The quantity lnk1/lnk2 does not help you with the information given. You know that k2/k1 is 10 or 100 but that doesn't tell you what lnk1/lnk2 will be.

Isn't rate proportional to rate constant? so we know that the rate increases by 10x, given all else is constant, rate constant should also increase by 10x,

 

[aldol16]

Isn't rate proportional to rate constant? so we know that the rate increases by 10x, given all else is constant, rate constant should also increase by 10x,

Yes, rate is proportional to rate constant. But how does that help you here? You know that rate 2 > rate 1 by a factor of 10. Therefore, k2/k1 = 10. Now, how are you going to relate that to lnk2/lnk1?

Not only that, but you also arbitrarily used 3 centigrade as your temperature.