We're given that there is a relationship between k and temperature for a reaction that can be determined by this plot. The graphic information says that when the natural log of k is plotted versus 1 over temperature in kelvin, temperature's in kelvin, we'll get a straight line and it gives us the equation for that line. Why was it the natural log of K? X is one over temperature. Now this comes from the equation, that the natural log of K is equal- E sub a over R times one over temperature plus the natural log of A, where A is our frequency factor. So when we line this up, we see that this is a slope and the slope equals- E sub a over R. So if it's going to have a straight line, it's going to have a negative slope and the value of that slope, which is the same as- E sub a over R, is equal to- 20400. The y intercept up here, which is equal to the natural log of the frequency factor, is equal to 32. So we're going to use that information to answer these two parts. Part one, determine the activation energy. Since the slope, which is a- E sub a over R, is equal to a -20400, the units of this, by the way, are in Kelvin. Then we can solve for activation energy by plugging in R. I'm going to multiply both sides by R and change the sign. So activation energy equals 20400 K times R, which is 8.314 joules per mol Kelvin. This is going to give me an activation energy of 1.70 times 10 to the third joules per mol. Typicall,y activation energy is reported in kilojoules per mol, so that will be 170 kilojoules per mol. That's part one. Part two is asking us to determine the frequency factor. Well, that comes from the intersect. Natural log of A is equal to 32. So we're doing part two. If we take e to both sides, we'll get A. And A is equal to e to the 32, which is 7.9, times 10 to the 13th. And that is the value of the frequency factor.