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Topic: Activation Enthalpy  (Read 6184 times)

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sci0x

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Activation Enthalpy
« on: January 08, 2006, 09:16:09 AM »
If activation enthalpy for a first order reaction is 76.7 kJ mol-1. How many times faster will the reaction occur at 50 degrees Celsius than 0 degrees Celsius?

Offline Borek

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Re:Activation Enthalpy
« Reply #1 on: January 08, 2006, 09:39:59 AM »
Arrhenius equation.

Unless I am wrong - it happens!
« Last Edit: January 08, 2006, 09:40:24 AM by Borek »
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sci0x

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Re:Activation Enthalpy
« Reply #2 on: January 08, 2006, 10:59:37 AM »
You mean this equeation  k=A*exp(-Ea/R*T) ?
How does this help?

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Re:Activation Enthalpy
« Reply #3 on: January 08, 2006, 11:19:10 AM »
What is k? What is Ea? WHat is goven? What do you have to find?
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Re:Activation Enthalpy
« Reply #4 on: January 08, 2006, 11:34:16 AM »
In this case, you might want to use version 2.0 of the arrhenius equation  ;)

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

You will want to find the ratio k2/k1.

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Re:Activation Enthalpy
« Reply #5 on: January 08, 2006, 11:42:49 AM »
In this case, you might want to use version 2.0 of the arrhenius equation  ;)

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

You will want to find the ratio k2/k1.

More confusing then helpfull.

As you have alredy solved the question, at least show sci0x that it is the same equation.
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Offline plu

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Re:Activation Enthalpy
« Reply #6 on: January 08, 2006, 11:59:29 AM »
More confusing then helpfull.

As you have alredy solved the question, at least show sci0x that it is the same equation.

Apologies.  I assumed he had covered both equations in class already.  To get the second form of the equation, you first take the ln of both sides.

(1)  ln k1 = ln A + (-Ea1/R)(1/T)
(2)  ln k2 = ln A + (-Ea2/R)(1/T)

Now, noting that A is equal in both equations, substracting (2) from (1) gives the second form of the Arrhenius equation  :D

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