In part b they want us toįind the activation energy, once again in kJ/mol. And so we've used all thatĭata that was given to us to calculate the activationĮnergy in kJ/mol. Activation energy is equal to 159 kJ/mol. Our answer needs to be in kJ/mol, so that's approximately 159 kJ/mol. And so we get an activation energy of, this would be 159205 approximately J/mol. So we can solve for the activation energy. Of the activation energy over the gas constant. So the slope is -19149, and that's equal to negative So to find the activation energy, we know that the slope m is equal to- Let me change colors here to emphasize. And so the slope of our line is equal to - 19149, so that's what we just calculated. And here are those five data points that we just inputted into the calculator. You can see that I have the natural log of the rate constant k on the y axis, and I have one over the And those five data points, I've actually graphed them down here. Alright, we're trying toįind the activation energy so we are interested in the slope. So we have, from our calculator, y is equal to, m was - 19149x and b was 30.989. So that's -19149, and then the y-intercept would be 30.989 here. We want a linear regression, so we hit this and we get ![]() Alright, so we have everything inputted now in our calculator. And then finally our last data point would be 0.00196 and then -6.536. Our third data point is when x is equal to 0.00204, and y is equal to - 8.079. So that's when x is equal to 0.00208, and y would be equal to -8.903. So when x is equal to 0.00213, y is equal to -9.757. So we go to Stat and we go to Edit, and we hit Enter twiceĪnd then start inputting. These different data points which we could put into the calculator to find the slope of this line. And if you took one over this temperature, you would get this value. Of this rate constant here, you would get this value. Temperature on the x axis, this would be your x axis here. This would be on the y axis, and then one over the Log of the rate constant on the y axis, so up here To the natural log of A which is your frequency factor. If you wanted to solveįor the frequency factor, the y-intercept is equal And so if you get the slope of this line, you can then solve for And the slope of that straight line m is equal to -Ea over R. ![]() The temperature on the x axis, you're going to get a straight line. Log of the rate constant on the y axis and one over And this is in the form of y=mx+b, right? So if you graph the natural Of the rate constant k is equal to -Ea over R where Ea is the activation energy and R is the gas constant, times one over the temperature plus the natural log of A, One way to do that is to remember one form of the Arrhenius equation we talked about in the previous video, which was the natural log And in part a, they want us to find the activation energy for We only have the rate constantsĪt different temperatures. Let's just say we don't have anything on the right side of the This is a first-order reaction and we have the different rate constants for this reaction atĭifferent temperatures. Have methyl isocyanide and it's going to turn into its isomer over here for our product. And so let's say our reaction is the isomerization of methyl isocyanide. Let's see how we can use the Arrhenius equation to find the activation energy for a reaction.
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