An AP Chemistry classic experiment for concentration changes over time is determining the rate law of fading Crystal Violet.
An AP Chemistry classic experiment for concentration changes over time is determining the rate law of fading Crystal Violet. (1:07) If the reaction is zero order, the plot of our concentration of our reactant vs time is linear. (2:51) If the reaction is first order with respect to our crystal violet, a plot of the natural log of the concentration of the reactant vs time will be linear. (3:17) If the reaction is second order with respect to crystal violet, a plot of 1 over concentration vs time will be linear. (3:45) We can use the graph and determine k as follows: for zeroth and first order k equals - slope and for second order reactions k = slope. (5:03) Integrated rate laws can be used to calculate the concentration after a specific amount of time or to determine how long a reaction has to run to get a specific concentration. A specific application of the rate laws is half-life. In AP Chemistry, half-life is always a first-order reaction. (6:15)
Question: If you have a reaction with a half-life of 4 days and an initial concentration of 0.1M, what fraction will be left after 20 days? (8:30)
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Hi and welcome to the APsolute Recap: Chemistry Edition. Today’s episode will recap Concentration Changes Over Time
Lets Zoom out:
Unit 5 - Kinetics
Topic - 5.3 Concentration Changes over time
Big idea - Transformation
In episode 23 we introduced Rate Laws and explained how you can use experimental data to describe how the rate of a reaction is determined by the concentration of each of the reactants. This method uses the idea that we can compare the initial rates of a reaction to determine the order with respect to each reactant. The order of a reaction can also be inferred from a graph showing the concentration of a reactant vs the time.
Let’s zoom in:
An AP Chemistry classic experiment for this topic is determining the rate law of the fading of Crystal Violet. In this experiment, a solution of crystal violet, which, you might have guessed, is a dark violet color, is mixed with sodium hydroxide. Crystal Violet will react with sodium hydroxide and the color will slowly fade. Since we have a colored solution and we want to know the concentration, we can use… the Beer-Lambert Law and a calibration curve. As the beautiful violet color fades, the measured absorbance will decrease. Okay, yeah, sure, great. But how does that relate to the order of the reaction?
We can use a graphical solution. If the reaction is zero order, that means that the reaction is independent of the concentration of the reactants. Graphically, we would see that the plot of our concentration for reactant vs time is linear. If the reaction is first order with respect to our crystal violet, a plot of the natural log of the concentration of the reactant vs time will be linear. Or, if the reaction is second order with respect to crystal violet, a plot of 1 over concentration vs time will be linear. So what do you have to do with your experimental data? You plot three graphs: concentration vs time; the ln of the concentration vs time and the reciprocal of concentration vs time. Then see which plot is closest to linear by calculating the R2. And by “calculating” I mean to hit the right buttons in your graphing program! Pretty straight forward!
We can even take an additional step and use the graph to determine the rate constant for the reaction. For zeroth order, we know that rate = k. And we know that rate is the negative change in concentration of our reactant over change in time. The change in concentration over time is represented by our slope! The rate constant k for our reaction is therefore the negative slope! For first order reactions, we know that the rate is equal to k times the concentration of our reactant to the first power. We’ve already discussed that for first order reactions we have a linear plot for the ln of the concentration of our reactant. Therefore, using some calculus, we get our integrated rate law. No worries, you don’t have to do calculus! It is actually given on the equation sheet and we’ll unpack that in a second. But using the integrated rate law, we can derive that, again, k equals the negative slope. For second order reactions we know that the rate is k times the concentration of reactant to the second power. Again, using calculus, we can derive an integrated rate law and see that k is equal to the slope of our plot.
As promised, what does it look like on the equation sheet? Let’s briefly unpack it for zeroth order: “Concentration of A at time t” means the concentration of A at a certain time point. “Concentration of A zero” means the initial concentration. By subtracting these two, we get the CHANGE in concentration, our delta concentration. On the equation sheet this is equal to negative kt. Which again makes sense, because if you divide by t, you will have what we discussed: k = change in concentration over change in t, aka rate. These integrated rate laws can be really useful - like if we have to calculate the concentration after a specific amount of time or to determine how long a reaction has to run to get a specific concentration.
One side note: Do you get confused determining when k is equal to the positive or negative slope? Just have in mind: k is never negative. Therefore, if the slope is negative, k has to be equal to the negative slope and if the slope is positive, k has to be equal to the slope.
A specific application of the rate laws is half-life, like with radioactive decay. Half-life is defined as the amount of time it takes a substance to be reduced to half of its initial concentration. We can have reactions that follow zero order, first order or second order, BUT in AP Chemistry, all half-life processes will be FIRST order. No exceptions! Now a short segment for our math-inclined listeners: since this is a first order reaction and the concentration at half-life is exactly half of the initial concentration, we can use the integrated rate law to get: ln of initial concentration of A divided by half of the initial concentration is equal to kt(½). That means: ln of 2 is equal to kt(½) and therefore half life is equal to 0.693/k. If you just skipped the math segment, let me repeat: half life is equal to 0.693/k and this is the equation on the equation sheet. You can use it to calculate k or, if you know k, to calculate half-life.
To recap:
An AP Chemistry classic experiment for concentration changes over time is determining the rate law of fading Crystal Violet. If the reaction is zero order, the plot of our concentration of our reactant vs time is linear. If the reaction is first order with respect to our crystal violet, a plot of the natural log of the concentration of the reactant vs time will be linear. If the reaction is second order with respect to crystal violet, a plot of 1 over concentration vs time will be linear. We can use the graph and determine k as follows: for zeroth and first order k equals - slope and for second order reactions k = slope. Integrated rate laws can be used to calculate the concentration after a specific amount of time or to determine how long a reaction has to run to get a specific concentration. A specific application of the rate laws is half-life. In AP Chemistry, half-life is always a first-order reaction.
Coming up next on the APsolute RecAP Chemistry Edition: Unit 5 Selected FRQs
Today’s Question of the day is about half-life.
If you have a reaction with a half-life of 4 days and an initial concentration of 0.1M, what fraction will be left after 20 days?