The reaction 2 N2O5¡2 N2O4 + O2 takes place at around
room temper...

Question

The reaction 2 N2O5¡2 N2O4 + O2 takes place at around
room temperature in solvents such as CCl4. The rate constant at
293 K is found to be 2.35 * 10 - 4 s - 1, and at 303 K the rate constant
is found to be 9.15 * 10 - 4 s - 1. Calculate the frequency
factor for the reaction.

Accepted Answer

Hi everyone for this problem. We're told the rate constant for the first order decomposition of hydrogen peroxide into water and oxygen is 7. times 10 to the negative 12 seconds inverse at 313 kelvin and 2.516 times 10 to the negative 10 seconds inverse at 343 kelvin. What is the value of the frequency factor for this reaction? So, what we're trying to solve for here is the frequency factor. And let's take a look at what were given in the problem. We're given to rate constants. So here's one. Here's the second and were given to temperatures. Okay. And from this given information, we want to solve for the frequency factor for this reaction. So we are going to want to first solve for the activation energy and we can solve for the activation energy using the rate constants at two different temperatures. And the equation that we can use in order to solve for activation energy. First is L N of K two over K one is equal to negative activation energy over R times one over T two minus one over T one. Okay. Where K two and K one. R R rate constants are is our constant as well and T two and T one our our temperatures and E A. Is our activation energy. So our goal here is to solve for our activation energy. Let's go ahead and plug in what we know based off of what was given in the problem. So we know our two K values. So we have L N f K two over K one. So we'll say that. All right it here. This would be our K one. This would be our T one. This would be K two and this would be T two. Okay, so we're going to go ahead and plug all that in. So L n f k two over K one is going to be 2.516 times 10 to the negative 10 seconds. In verse over K one 7.652 times 10 to the negative 12th seconds, inverse is equal to negative activation energy over R r S r gas constant, which is 8. jules over mall times kelvin. And this is going to be times one over T two minus one over T one. So this is one over 343 kelvin -1 over 313 Kelvin. Okay, so we can go ahead and simplify this a little bit. And when we do for the left side we're going to get 3.493 is equal to our activation energy times 3.361 times to the negative fifth. Okay, so let's go ahead. And softer activation energy here by dividing both sides of our equation by 3.361 times 10 to the negative five. Okay, So that means our activation energy is going to equal 3.493 divided by Let me clear that up. So 3.493 divided by 3.361 times 10 to the negative five. So now we get an activation energy of 1.39 times 10 to the fifth jewels over more. So now that we know the activation energy, we can use the activation energy and any of our K values to find the frequency factor which is A. And the equation that we're going to use to find our frequency factor is Ln of K is equal to Ln of a minus our activation energy over R. T. Okay, so this is going to be what we're going to use now. Let me just clear this up. This is what we're going to use now to solve for our frequency factor and our frequency factors represented by this letter A. Here and for our K we can use any of our K values. So let's go ahead and use K two. Okay, so when we plug in we're going to get R L N F K. So we're going to choose our K two. So Ln of 2.516 times 10 to the negative 10th seconds, inverse is equal to Ln of a minus our activation energy that we just solved for. So that's going to be 1. times 10 to the fifth jewels over mall. And this is going to be over R times T. So are are we said is 8. jewels over more times kelvin and R. T. Two. Because we're using K two, we need to use our T. Two. So this is going to be times T two which is 343 Calvin. Okay, so let's go ahead and simplify this here. So for the left side we're going to get negative two negative 22.10 is equal to Ln of a minus 36.43. So we can go ahead and simplify here. So our Ln of A is going to equal 36. -22.10. Okay, so let me just clear that here. So that's a minus sign. So 30 36.43 minus 22.10. So that means Ln of A is going to equal 14.33. We need to get rid of that L. N. And the way that we do that is by raising both sides by E. So we're going to get A is equal to e Raise to 14.33. Okay, so that gives us a final answer of a. Is equal to 1.67 times to the sixth seconds inverse. So this is our final answer. And this is going to be the value of our frequency factor for this reaction. That's the end of this problem. I hope this was helpful

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