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### Course: Chemistry archive > Unit 11

Lesson 1: 2015 AP Chemistry free response questions- 2015 AP Chemistry free response 1a
- 2015 AP Chemistry free response 1b and c
- 2015 AP Chemistry free response 1d
- 2015 AP Chemistry free response 1e
- 2015 AP Chemistry free response 2a (part 1 of 2)
- 2015 AP Chemistry free response 2a (part 2/2) and b
- 2015 AP Chemistry free response 2c
- 2015 AP Chemistry free response 2d and e
- 2015 AP Chemistry free response 2f
- 2015 AP Chemistry free response 3a
- 2015 AP Chemistry free response 3b
- 2015 AP Chemistry free response 3c
- 2015 AP Chemistry free response 3d
- 2015 AP Chemistry free response 3e
- 2015 AP Chemistry free response 3f
- 2015 AP Chemistry free response 4
- 2015 AP Chemistry free response 5
- 2015 AP Chemistry free response 5a: Finding order of reaction
- 2015 AP Chemistry free response 6
- 2015 AP Chemistry free response 7

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# 2015 AP Chemistry free response 5a: Finding order of reaction

Finding the order of reaction based on graphs of absorbance, ln(absorbance) and 1/(absorbance) vs. time for kinetics of bleaching food coloring. An alternative method of solving 2015 AP Chemistry free response 5a.

## Want to join the conversation?

- why should the graph be linear?(2 votes)
- So the reaction is first order which means it will follow first order kinetic equations like the integrated rate law that they gave: ln([A]t) - ln([A]0) = -kt. You can rearrange the equation into: ln([A]t) = -kt + ln([A]0), which fits the slope-intercept equation y = mx + b where y is ln([A]t), x is t, m (the slope) is -k, and b (the y-intercept) is ln([A]0). Now if you graph the ln([A]t) as the y coordinate versus t as the x coordinate and they form a straight line, then the reaction is first order. And what's more you could find the rate constant k from the slope and initial concentration of A from the y-intercept.

So looking at the graphs they gave, graph 1 has [A] vs. t which follows the integrated rate of a 0th order reaction. But since the curve is not linear we can say that the data does not fit the 0th order model and that the reaction is not 0th order.

Graph 2 has ln[A] vs. t which follows the integrated rate law of a first order reaction as I've shown. Now since the data here is a nice and linear we can say that it fits much better using a first order model and that the reaction is first order.

Graph 3 has 1/[A] vs. t which would be following the integrated rate law of a second order reaction. And since it's not linear, it is similar to the 0th order model as to why it doesn't work.

It's pretty common to see if you're able to fit data onto a straight line in chemistry because it's a clear way to visualize if your model is predicting how the data should behave. I'm unsure as to why they didn't just solve it by looking at whether the graphs were linear or not, it's a much faster way to solve these problems especially in a test setting.

Hope that helps.(2 votes)

## Video transcript

- [Voiceover] Blue food
coloring can be oxidized by household bleach, which
contains hypochlorite ion, or OCI-, to form colorless products, as represented by the equation above. So we have this equation where
we have blue food coloring, which has this chemical formula, and it's reacting with
bleach, or hypochlorite, and it's making colorless products. So we're getting this color
change from blue to colorless. A student used a spectrophotometer set at a wavelength of 625 nanometers to study the absorbance
of the food coloring over time during the bleaching process. In the study, bleach is
present in large excess, and I will underline that
'cause that's important. It's present in large excess
so that the concentration of hypochlorite is essentially constant throughout the reaction. The student used data from the study to generate the graphs below. So we'd like to answer, in part a, based on the graphs
above, what is the order of the reaction with respect
to the blue food coloring? So we have a lot of information here. We have these three graphs. If we write absorbance as A, we're looking at the concentration of A over time. We're looking at the ln of A over time, the natural log, and we are also, lastly, looking at one over the
concentration A over time. So, we would like to
know, what is the order of the reaction with respect to our blue? My first instinct when I see this problem, is actually to be a little bit panicky because, hey we have a
lot of information here, so how do we deal with it? We also have a little bit more information for this problem. We also have the equation sheet. So this is the part of the equation sheet that's relevant for this problem. So it's all of the equations
that have to do with kinetics, which is looking at reaction rates, so that is what we're looking at here. So we have three possible things that might be helpful for this problem. Maybe not, right? So we have this first
equation which gives us the first order integrated rate law. We have the second
equation, which gives us the second order integrated rate law, and we have this third equation which is actually an
equation for the half-life. So it's saying half-life is .693 divided by k, and that is the half-life for a first order reaction. And the important thing to remember here, which will come up again in a second, is that this is a constant,
because k is a constant, so our half-life is also a constant for a first order reaction. So that's the information we have. Let's look back at our graphs. We're gonna start by looking
at just the first graph. It turns out we can get
a lot of information just by looking at the
concentration of A over time. So if we look at this first graph, and we look at the concentration of our blue stuff going down over time, we can see immediately
that this is not linear. So the fact that it's not linear immediately tells us it's not zero order. If it was zero order, we would expect to see a straight line,
which I will draw in pink, with a dotted line because that's not actually what we're seeing. So it's not zero order. We can also get some more information by looking at this one graph. We know it's not zero order. Can we tell from this first
graph if it's first order? And the answer is, we actually can. If you remember from looking
at the equations sheet, we said that first order reactions, so for first order reactions, we know, and I'd say this is maybe one of the most important things
about first order reactions, is that half-life is constant. So that means that the time it takes for the concentration to go down to half of what it was,
is the same no matter what your initial concentration is, no matter what point in your reaction you're at. So that's something we
can actually investigate just by looking at this first graph. We can see how much time it takes to get to half the concentration, and see if that's something
that changes over time. So for example, if we
start with .8 right here, half of .8 is .4, so we can see how much time it takes to get from .8 to .4, which is right here, and if we go down we can
see it takes 20 seconds to have the concentration starting at .8 absorbance units. We can look at the half-life
now to go from .4 to .2. So if we go from .4 to .2,
and I will change colors to make this look more obvious. We'll do this in orange. So going from .4 to .2
takes another 20 seconds. And so we can do this again and again. We could go from .2 to .1. We'll just see the
continuation of a pattern here, which is that, we can see that
it always takes 20 seconds to halve the concentration. And so the fact that half-life is constant based on Graph I, immediately tells us that our reaction is first order because our half-life is constant. We don't even have to look
at any of the other graphs. There is another way you
could do this problem. You might remember, based on
the equations that we have, or based on just equations you know off the top of your head. So one reason we know, and
one reason we could say this is first order, is
because t 1/2 is constant. The other reason why we
know this is first order is, when a reaction is first
order in your reactant, you know that graphing ln of your reaction versus time, gives you a straight line, and we can see from Graph II that this is the case. So we have these two pieces of evidence, from Graph I and Graph II, that tell us that this particular reaction
is actually first order with respect to the blue food coloring.