## QUIZ 2 #11

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

alondra_1D
Posts: 17
Joined: Wed Sep 21, 2016 2:59 pm

### QUIZ 2 #11

The rate constant for the reaction $A\rightarrow B$ is 25 /min at 298K and 35/min at 350 K.
Calculate the value of the rate constant at 770K. Give your answer in min^-1

De we first use this equation, $ln\frac{k2}{k1}=\tfrac{Ea}{R}\cdot (\frac{1}{T1}-\frac{1}{T2})$ and get the Value of Ea and then use $lnK= A-\frac{Ea}{RT}$ to find the value of K at 770?

Thomas Hui 2J
Posts: 11
Joined: Wed Sep 21, 2016 2:55 pm

### Re: QUIZ 2 #11

What I did was first I used the Arrhenius equation to find the Activation Energy, E(a), which I found to be 5611 J/mol. However, I did not know how to solve for the rate constant because I do not know the value of A (rate of collisions) at 770 Kelvin...

Noor_Burney_1L
Posts: 13
Joined: Wed Sep 21, 2016 2:55 pm

### Re: QUIZ 2 #11

First, I defined k1, the rate constant, for 25 /min at 298K and k2, the other given rate constant, for 35 /min at 350 K. I then defined k3 as the rate constant value I am looking for. After using the Arrhenius equation to find the activation energy, which is 5611 J/mol, you can then plug in the value of the activation energy back into the Arrhenius equation, but this time you make k3 (rate constant you're looking for), as the variable to solve for. You can also choose to either plug in the information given for k1 or k2 in the equation to solve for k3. If we use the values given for k1 to solve for k3, it would look like this:

ln(k3/k1)=-Ea/R (1/T3 - 1/T1)

Next, you plug in 25 /min for k1 in the equation, 770K for T3, and 298K for T1. So the equation would now look like this:

ln(k3/25)=-5611/8.3145 (1/770 - 1/298)

The answer ends up to be 100 /min

Hope this helps!

nicoleclarke_lec1M
Posts: 25
Joined: Sat Jul 09, 2016 3:00 am

### Re: QUIZ 2 #11

Do we need to know how to do this for the quiz? I thought the material covered only up until page 73. Will we also need to know how to do #9?