## Practice Quiz winter 2015

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

Jacqueline Lee
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

### Practice Quiz winter 2015

The question goes "The rate constant for reaction A$\rightarrow$ B is 25min-1 at 298K and 35min-1 at 350K. Calculate the Value of the rate constant at 770K. Give your anser in min-1."

How would one approach the question?
I tried treating the given numbers as one finds slope, but the answer I got was 148min-1.

Because there was temperature given I assume I am supposed to be using the Arrhenius equation. Is this correct? How would I proceed from here when I do not have A or EA?

Saitiel1
Posts: 9
Joined: Fri Sep 25, 2015 3:00 am

### Re: Practice Quiz winter 2015

Hi,

To solve this question you're going to need to use the following equation:$ln\frac{k2}{k1}=\frac{Ea}{R}[\frac{1}{T1}-\frac{1}{T2}]$. We first need to find Activation energy(Ea) so we change the equation to $Ea=\frac{(R)(ln\frac{k1}{k2})}{\frac{1}{T1}-\frac{1}{T2}}$. You plug in the values given to you in the question to find Ea. Once you find Ea, you plug your values into $k3=k1*e^{(\frac{Ea}{R})*[\frac{1}{T1}-\frac{1}{T3}]}$. This should give you the final rate constant.

Hope this helps!

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