## Quiz prep 2 #11

Arrhenius Equation: $\ln k = - \frac{E_{a}}{RT} + \ln A$

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

### Quiz prep 2 #11

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

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

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
Posts: 12
Joined: Wed Sep 21, 2016 2:56 pm
Been upvoted: 1 time

### Re: Quiz prep 2 #11

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

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