Quiz prep 2 #11

Arrhenius Equation:

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Michael Lonsway 3O
Posts: 43
Joined: Wed Sep 21, 2016 2:57 pm

Quiz prep 2 #11

Postby Michael Lonsway 3O » Sat Feb 18, 2017 10:12 am

The rate constant for the reaction A ---> B is 25 /min at 298k and 35 /min at 350k. Calculate the value of the rate constant at 770k. Give your answer in /min.

I used a form of the Arrhenius equation to solve this and only got activation energy. What units are activation energy in and how can I find the rate constant at 770k with it?

Nerissa_Low_2F
Posts: 23
Joined: Fri Jul 15, 2016 3:00 am

Re: Quiz prep 2 #11

Postby Nerissa_Low_2F » Sat Feb 18, 2017 6:48 pm

I'm not sure how to find the rate constant here, but I believe the units for activation energy is kJ/mol. Also, I think we aren't being tested on the Arrhenius equation for the quiz.

NinaSheridan
Posts: 30
Joined: Wed Sep 21, 2016 2:58 pm

Re: Quiz prep 2 #11

Postby NinaSheridan » Fri Feb 24, 2017 1:11 am

Because we aren't being tested on it, should we still do the 20 or so problems on it? Is it likely to show up on future tests?

Cris Yuan 2I
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Re: Quiz prep 2 #11

Postby Cris Yuan 2I » Sat Feb 25, 2017 12:00 am

After you find the activation energy, plug in the value:
ln(k)=-Ea/(R700K)+ln(A)
and then use another equation:
ln(35)=-Ea/(R350K)+ln(A)
then subtract one from the other to get rid of A:
ln(k/35)=-Ea/(R700K)+Ea/(R350K)
k=35*e^(Ea(1/R350K-1/R700K)
and plug in the value of Ea here, and the rest are probably completed by the calculator.

I think the key to solving problems related to the Arrhenius equation is to get rid of certain valuables (k or A): in this case, subtracting one equation from the other gets rid of A and simplifies further calculation.

Amy_Shao_2D
Posts: 22
Joined: Fri Jul 15, 2016 3:00 am

Re: Quiz prep 2 #11

Postby Amy_Shao_2D » Sat Feb 25, 2017 12:24 am

They might show up on the final so I'd still do them even if they don't show up on quiz 2.


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