15.61 (6th addition)

Arrhenius Equation:

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melodyzaki2E
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Joined: Fri Sep 28, 2018 12:18 am

15.61 (6th addition)

Postby melodyzaki2E » Fri Mar 15, 2019 2:50 pm

The rate constant of the first-order reaction 2 N2O(g)--->2 N2(g) + O2(g) is 0.76 1/s at 1000. K and 0.87 1/s at 1030. K.
Calculate the activation energy of the reaction.
How do you use Arrhenius here?

Maddy Mackenzie
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Re: 15.61 (6th addition)

Postby Maddy Mackenzie » Fri Mar 15, 2019 7:37 pm

You have to use the natural log of the Arrhenius equation: ln (K) = lnA - Ea/R (1/T). Then if you subtract, you can get the form: ln(K2/K1) = Ea/R (1/T2 - 1/T1). From this, you can now plug in the given values to get 39 kj per mol

Jeannine 1I
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Joined: Fri Sep 28, 2018 12:27 am

Re: 15.61 (6th addition)

Postby Jeannine 1I » Fri Mar 15, 2019 7:48 pm

Maddy Mackenzie wrote:You have to use the natural log of the Arrhenius equation: ln (K) = lnA - Ea/R (1/T). Then if you subtract, you can get the form: ln(K2/K1) = Ea/R (1/T2 - 1/T1). From this, you can now plug in the given values to get 39 kj per mol


Can you explain how to get the second form please? I'm having a difficult time understanding how to get that.

Maddy Mackenzie
Posts: 37
Joined: Fri Sep 28, 2018 12:22 am
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Re: 15.61 (6th addition)

Postby Maddy Mackenzie » Sat Mar 16, 2019 12:41 am

In the book what they have is: At temp T1: ln(k) = ln (A) - Ea/RT1 and at temp T2: ln (K2) = ln (A) - Ea/RT2
Then all they do subtract the first equation from the second equation: ln(k2) = ln (A) - Ea/RT2
- ln (K) = ln (A) - Ea/RT1
= ln(K2)-ln(K) = -Ea/RT2 -Ea/RT1
= ln(K2/K) = -Ea/R (1/T2 - 1/T1)
Thats how you get that form


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