0:01

As you read through this problem, you see that information is given about the plot.

Â They plotted one over concentration versus time in seconds and

Â that produced a straight line.

Â From that information alone, I can know that this reaction is second order.

Â 0:19

The integrated rate law for a second order

Â reaction is 1 over A is equal to Kt plus 1 over A null.

Â That's y=mx+b.

Â So this, the fact that you get a straight line and

Â that this equation of a line tells me, it's second order.

Â It also tells me that k is equal to

Â the slope, so it's 0.296.

Â So, we have just draw that out and

Â let me draw that out over here where I have a little bit more room.

Â We've got along the y-axis,

Â 1 over concentration along the x-axis time.

Â It has a straight line for its graph.

Â The slope of this line is equal to 0.0296 and let's get units for that.

Â 1:21

Slope is change in y over change in x.

Â So the units of y are 1 over molarity, a concentration unit.

Â The units for x was in seconds.

Â So, this gives me 1 over molarity times seconds for the slope.

Â I mean, for the rate of constant k and that's always a unit for

Â second order reaction.

Â So now, we've answered part two.

Â Part three tells me that the initial concentration of A is 0.1.

Â So we can use our equation to calculate A at that time,

Â because what we want to know is what percent reacted after five seconds.

Â So if we knew how much was remaining,

Â we'll be able to get the percent that reacted.

Â So, we will plug in our k of 0.0296.

Â We will put in our time of 5 seconds and

Â we will put 1 over our initial concentration.

Â This will give me 1 over A.

Â If we multiply this out and add this number here,

Â we get a value of 1 over A equals 10.148.

Â So A would be the reciprocal of that,

Â which is 0.0985.

Â Now, that is not the answer to the question.

Â The question says, what percent of A will react?

Â So, let's figure out how much actually reacted.

Â Well, if we start with point one and

Â we finished with 0.0985.

Â We will have a value

Â of 0.001 molar

Â that reacted.

Â So, so much we start with.

Â So, much we finish with.

Â We subtract that, this is how much had to have reacted.

Â So, what percent reacted?

Â That percent reacted is 0.001 and I can only know

Â that's the 1 decimal place, because I can only keep 3.

Â I mean, one significant figure, because I can only keep three places to the right

Â of the decimal point according to the rules of addition and

Â subtraction in significant figures.

Â So, the percent reacted would be the amount that reacted

Â over the amount that I started with times 100 and

Â that will give me 1% reacted.

Â 4:33

But for second order, it is true, but let's let the numbers work for themselves.

Â Let's figure out the amount of A that would remain,

Â if we start it with a different quantity.

Â Now the rate cost is not going to change, And

Â the time isn't changing, but what is changing is this number here.

Â We're going to start with 10 molar or 10.0 molar.

Â Now when you multiply that out and solve for 1 over A,

Â you get a very different number here and you get 0.248.

Â 5:11

And that's one over molar.

Â So, A would be the reciprocal of that number and

Â it would be 4.03.

Â So, this is how much would remain.

Â So if we're going to calculate the percent that reacted,

Â we need to know first how much reacted.

Â So, we started with 10, we ended with 4.03.

Â So, that is 6.0 molar that reacted.

Â And if 6.0 molar reacted and we started with 10 molar.

Â