As you read through this problem, you see that information is given about the plot. They plotted one over concentration versus time in seconds and that produced a straight line. From that information alone, I can know that this reaction is second order. The integrated rate law for a second order reaction is 1 over A is equal to Kt plus 1 over A null. That's y=mx+b. So this, the fact that you get a straight line and that this equation of a line tells me, it's second order. It also tells me that k is equal to the slope, so it's 0.296. So, we have just draw that out and let me draw that out over here where I have a little bit more room. We've got along the y-axis, 1 over concentration along the x-axis time. It has a straight line for its graph. The slope of this line is equal to 0.0296 and let's get units for that. Slope is change in y over change in x. So the units of y are 1 over molarity, a concentration unit. The units for x was in seconds. So, this gives me 1 over molarity times seconds for the slope. I mean, for the rate of constant k and that's always a unit for second order reaction. So now, we've answered part two. Part three tells me that the initial concentration of A is 0.1. So we can use our equation to calculate A at that time, because what we want to know is what percent reacted after five seconds. So if we knew how much was remaining, we'll be able to get the percent that reacted. So, we will plug in our k of 0.0296. We will put in our time of 5 seconds and we will put 1 over our initial concentration. This will give me 1 over A. If we multiply this out and add this number here, we get a value of 1 over A equals 10.148. So A would be the reciprocal of that, which is 0.0985. Now, that is not the answer to the question. The question says, what percent of A will react? So, let's figure out how much actually reacted. Well, if we start with point one and we finished with 0.0985. We will have a value of 0.001 molar that reacted. So, so much we start with. So, much we finish with. We subtract that, this is how much had to have reacted. So, what percent reacted? That percent reacted is 0.001 and I can only know that's the 1 decimal place, because I can only keep 3. I mean, one significant figure, because I can only keep three places to the right of the decimal point according to the rules of addition and subtraction in significant figures. So, the percent reacted would be the amount that reacted over the amount that I started with times 100 and that will give me 1% reacted. Part four ask, will this be the same percentage that would remain if the reaction began with a whole lot more? We began with ten molar, instead? Well, for second order reactions, the rate does depend very heavily and it changes more rapidly when you have more higher concentration. This would not be true for first order. But for second order, it is true, but let's let the numbers work for themselves. Let's figure out the amount of A that would remain, if we start it with a different quantity. Now the rate cost is not going to change, And the time isn't changing, but what is changing is this number here. We're going to start with 10 molar or 10.0 molar. Now when you multiply that out and solve for 1 over A, you get a very different number here and you get 0.248. And that's one over molar. So, A would be the reciprocal of that number and it would be 4.03. So, this is how much would remain. So if we're going to calculate the percent that reacted, we need to know first how much reacted. So, we started with 10, we ended with 4.03. So, that is 6.0 molar that reacted. And if 6.0 molar reacted and we started with 10 molar. To get the percent remaining, it's going to be 60% or the percent reacted would be 60%. So, you see that a much higher percentage reacts when you start with a greater quantity. This is typical for second order kinetics.