A reaction has a rate constant of 0.0117/s at 400.0 K and 0.689/s at 450.0 K. a. Determine the activation barrier for the reaction.

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Hi, everybody. Welcome back. Let's look at our next problem. It says a reaction has a rate constant of 0.0117 in verse seconds at 400.0. Kelvin and 0.689 inver seconds at 450.0. Kelvin determine the activation barrier for the reaction. A 122 kilojoules per mole. B 132 kilojoules per mole, C 224 kilojoules per mole or D 345 kilojoules per mole. So let's think about what are our known values we've been presented with. What are we looking for? What equation do we need to get there? So we have two different points in time. We have rate constants and temperatures. So if we take our first value, we have a rate constant K one and the temperature T one and we have a second K two and T two that correspond to each other. And we need the activation barrier or activation energy, which is the value E subscript A. Well, we know we have our erroneous equation that relates the rate to the temperature and includes the term E A that we could solve for. But more specifically, since we have two points in time, two rates two temperatures, we would want to use the two point form of this equation. So let's recall what that looks like that is written the natural log of K two divided by K one. That's in parenthesis, that K two over K one equals negative E A over capital R. And then in parentheses multiplied by one over one, divided by T two minus one, divided by T one enclose the parentheses. So that has the knowns, we have K ones and Ks and T and it has the E A term in there. So we can solve for that. We just need to remember what our capital R is, which is the gas constant. Now, the only trick here is that gas constant can be written in different ways with different units. So we need to pick the right one. For our situation, we're dealing with energy, we're going to want jewels. So we're going to use the version 8.314 jules, Puru Kelvin. There's of course, our other way of expressing the gas constant, which is with the units of liters atmospheres per mole Kelvin. And that would be if we're working with volumes of gas. But we want Juuls here, we're working with energy. So now let's go ahead and plug in our numbers to our equation note because this caused me some trouble when I was working through this problem myself, you will sometimes see this equation written differently with the temperature term written as one over T one minus one over T two. The key here is that our right side of the equation has a negative sign in front of that E A or R factor. Sometimes this equation is written with no negative sign there but the subtraction order reversed because they've multiplied the subtraction term by negative one. This sometimes makes it a little easier because having that negative sign right next to the equal sign. It's very easy to lose it in your calculations. So you can write it that way. You'll sometimes see it that way. Note it's a negative sign. You have to keep track of which does get a little tricky. So I'd like to write, put some parentheses around that negative E A over art. It just helps to kind of keep that negative sign there. Keep you focused on it. So we'll fill in our terms here. So Ln and then in parentheses, RK 20.689 inverse seconds divided by 0.0117 inverse seconds. That's going to equal open parentheses. Negative E subscript A divided by our gas constant 8.314 Joules per mole, Kelvin. And that's going to be multiplied by one over. Now, T two, I have 450.0 Kelvin minus one over T 1, 400.0 Kelvin. Now note, here's another part where it can be easy to trip yourself up. This is subtraction of fractions. You can't just subtract 450 minus 400. You have to convert them to a common denominator or you can change them into decimals. So either way you like, you can either convert both of these numbers to decimals and do a simple subtraction or you can keep it as a fraction. However, a little cautionary tale, if you use decimals, that will then be multiplied by E A. If you use fractions, remember that your number will be in the denominator and keep track of that. When you combine this whole term, don't end up putting the wrong number on top, the wrong number on bottom. So let's go down to the next step. So now we have the natural log, we'll do that division natural log of 58.888 nine. Another quick note here, when I'm thinking about significant figures in general, you don't want to round in your intermediate steps, you might make rounding errors and affect your final answer. However, looking at my problem, I see I'm going to have three sig figs, three significant figures in my answer because of that 0.689 in verse seconds term. And rather than carrying out all my intermediate terms out to really large numbers of decimal points as long as I, you know, here, I have five significant figures three decimal points. Give myself a little extra to reduce rounding errors. You don't actually have to write out. Uh you know, you don't have to have no rounding at all. Basically, you can make things a little more manageable for yourself. So now I have negative E A again, being careful not to lose that negative over 8.314 joules per mole. Kelvin. I'll use the fractional um method here. I will uh I have converted them into a common denominator and then reduced my result. And I end up with negative 1/3600 Kelvin. But again, you could have converted them to decimals. The other thing when you convert to decimals, don't forget those Kelvins are on the bottom. They're in the denominator. Then your units would need to be inverse Kelvins. Otherwise your units won't work out in the end. So now I can simplify this still further. My natural log of 58.889 will be four point 07 57. Note that there are no units on the side, they canceled out with that division. Now I'm going to get rid of my negative here because I have negative E A and the numerator on the side, negative one on the numerator on the other term, excuse me, not side term. So those two negatives are multiplied together and become positive. So negative has gone away. So IE a divided by now, I've m I'm multiplying the denominators times each other. So 8.314 joules per mole. Kelvin multiplied by 3600. Kelvin, my Kelvins will cancel out here. OK. Inverse Kelvins and my gas constant. And I end up with the very large number of 29,000 930 0.4 jules per mole in the denominator. So now to isolate ea I just multiply both sides by that denominator and I'm going to end up with. And now I've gotten my final answer. I will come to round it off to three significant figures and get 1.22 times 10 to the fifth jules promo. It's going to equal E A. The only thing left to do when I look to compare this to my answers, my answers are expressed in kill joules per mole. So I do need to convert that. So I'm just going to put my conversion factor over on the left here multiply this, I want to cancel out jewels. So in the denominator, I have 1000 jewels in the numerator, 1 kg jewel. So that will equal 1.22 times 10 to the second power kilojoules per mole or 122. So when I look at my answers, I see choice A is 122 kilojoules per mole. One other little side note, having these multiple choice options means we have the luxury. If we made an error, an error in our math along the way and we ended up with some really wacky, you know, much bigger, much smaller number. My multiple choice options in this case, at least give me a sort of range. And that would tell me I made a mistake in my math somewhere. We don't always have that luxury. But now we've got our answer here. We've determined the activation barrier when we were given rate constant temperature at two different points, using that two point form of our IOUs equation. And our answer is choice a 122 kilojoules per mole. See you in the next video.

A reaction has a rate constant of 0.0117/s at 400.0 K and 0.689/s at 450.0 K. a. Determine the activation barrier for the reaction.

Verified Solution

Key Concepts

Arrhenius Equation

Activation Energy (Ea)

Temperature Dependence of Reaction Rates