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If, in Example 4, one molecule of the product C is formed from one molecule of the reactant A and one molecule of the reactant B, and the initial concentrations of A and B have a common value $[A] = [B] = a moles/L,$ then $[C] = a^2kt/(akt + 1)$where $k$ is a constant.(a) Find the rate of reaction at time $t.$(b) Show that if $x = [C],$ then $\frac {dx}{dt} = k(a - x)^2$(c) What happens to the concentration as $t \to \infty?$(d) What happens to the rate of reaction as $t \to \infty?$(e) What do the result of parts (c) and (d) mean in practical terms?

a) $\frac{a^{2} k}{(a k t+1)^{2}}$b) $k(a-x)^{2}=\frac{d x}{d t}$c) concentration of the product becomes equal to initial concentration of the reactantsd) As time t increases and approaches infinity, the rate of reaction decreases and approaches zeroe) See explanation for answer.

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University of Nottingham

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to start this problem, we have the concentration of sea, and we're going to find the rate of reaction, which means we're going to find the derivative of this. So the derivative of the concentration of sea with respect to time. So because we have a quotient here, we're going to use the quotient rule. So we have a bottom a k T plus one times the derivative of the top, a squared K minus the top, a squared Katie, times the derivative of the bottom, a k over the bottom squared a k T plus one squared. Okay, we're going to want to simplify this. So let's start by distributing a squared K, and then we'll combine our like terms. So we have a cubed K squared T plus a squared K minus will multiply these together and we get a cube K squared T, and that is over a K T plus one quantity squared. Notice that we have opposite terms here that we can cancel. So let's look at wheat at what we have now. We have that the derivative of the concentration. So the rate of reaction is a squared K over a K T plus one quantity squared. Okay, we'll move on to part B in part B. We're replacing the concentration of sea with an X, and we're going to show that the derivative of that is equal to K times a minus X quantity squared. So that would mean that we're showing that this quantity here is equivalent to the answer we just found for part A. Now remember that the concentration of C was given earlier in the problem as a squared K T over a K T plus one. So the first thing we can do is substitute that quantity in for X and then show that these two are equivalent. So we have DX DT equals K times a quantity a minus a squared Katie over a K T plus one quantity squared. Okay, Our job is to just simplify that until we get it to look like the previous derivative. All right, so what I'm going to do is get a common denominator with this first term A and the second term. So we're going to multiply a by a K T plus one over itself. So we're going to have K times, a times a quantity, a k T plus one over a K T plus one minus a squared Katie over a K T plus one quantity squared. So now we have a common denominator so we can go ahead and combine those fractions. Subtract there numerator. We have k times a quantity while I'm subtracting the numerator is I'm also going to distribute the A So if k times a quantity a squared Katie plus a minus a skit a squared Katie tongue twister over a k T plus one and that's all squared. Now look at what happens when we cancel the A square Katie and the minus a squared Katie. So all we have left for the numerator inside the fraction is a and notice that that will be squared. So that means we have que tons a squared over a k T plus one squared. And that is exactly what we were hoping to show. That matches our answer from part A. And now, for part C, we want to know what happens to the concentration when t goes toward infinity. So let's look at the limit as t approaches infinity of the concentration and the concentration was a squared Katie over a k t plus one. So one of the methods we have for dealing with the limit that involves infinity is we can divide the top and bottom by t divide the top and bottom by the highest power of tea that you see in the denominator. And that happens to be t. So really, what we're doing here is just multiplying by a special form of one. And we get the limit as he goes to infinity of a squared K over a K plus one over tea. Now it's imagine t going to infinity. So the numerator is just a number and the denominator is a number plus one over infinity and one over infinity is essentially zero. And so this part here is basically zero. So what we're getting at then is the limit is going to be whatever a square K over a K is, and we can definitely reduce that. Let's divide the top and bottom by a and by K and all we have left is a. So as time goes toward infinity, the concentration is approaching a value of A on what was a It was the initial concentration of the reactant and for Part D. We want to know what's happening to the rate of react in rate of reaction as time approaches infinity. And so let's take a limit of that derivative we found as T approaches infinity. So that would be the limit of a squared okay, over a K T plus one quantity squared. All right, so this time there's no tea on the top. So let's just imagine what happens if t goes toward infinity. That makes the denominator exceedingly huge. We have basically infinity on the denominator, and any number divided by infinity is just going to head toward zero. So what's happening over time is that the rate of reaction is going toward zero. In other words, the reaction is ending. That's pretty much what party is asking about is what these two limits tell us. They tell us that over time the concentration is approaching. The initial concentration of the reactant concentration of see is approaching the initial concentration of A and B and is telling us that as time goes on, the reaction is slowing down to nothing. The reaction is ending

Oregon State University
Catherine R.

Missouri State University

Anna Marie V.

Campbell University

Kayleah T.

Harvey Mudd College

Samuel H.

University of Nottingham

Lectures

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