Concentration–time data for the conversion of A and B to
D are l...

Question

Concentration–time data for the conversion of A and B to D are listed in the following table. (c) What is the rate law?

Accepted Answer

Hello everyone in this video be until the X and Y are converted to Z. And in this table here this is the concentration time data. And we're being asked this problem to determine the rate law. So what we can take away from this part of the problem here is that X and Y are reacting so start materials and then Z. Is our product. And because there's a product we do not included in the rate law. Also from this table here where we can take away is that for a is not going to be consumed in a reaction. And therefore it can be concluded that a callus will be we'll just aid in a faster reaction rate. So we know first that the equation for a rate, his heart rate is equal to KR constant, multiplied by the concentration of X. Raised to the power of M multiplied by the concentration of Y. Raised to power of N. Multiplied by the concentration of a razor power of P. Now to determine the initial rates. So now we're doing initial right here we go ahead and use the equation. That rate is equal to negative delta. The concentration of X. Or the change of the concentration of X over the change of time. So delta T. Or we can just simplify this to also be the concentration of X. Or they find a concentration of X minus the initial concentration of X. On the same thing here. Just delta T. So this here and this is the same thing. But we just want to go ahead and be a little bit more specific as to what we're looking forward for delta concentration of X. Alright, so I'm not taking this kind of equation here to apply to our experiments. So again we're giving experiment 123 and four. So we'll be doing these calculations using all these values here to whatever we're being asked for. So Let's start for experiment one. So not for experiment one the rate is equal to because I forget that it's negative delta concentration of X over the change of time. So for our final value that's 2.7 molars and our initial is 3.0 molars. So again, we're just taking the final minus the initial as we said here and before they change in time, that's just 30 seconds since we're going from 0 to 30 seconds. So once you put that into the calculator, we see that the value that we get for the rate is one times 10 to the negative two molars per second. Doing this exact same process. For experiments too, we can see here that the rate is equal to again negative delta concentration of X over the change in time. So now for the or the final value rather for experiment two, That's 5.4 molar initial is 6.0 Molar. And for they change in time it's also 30 seconds as well. So once you put that into a calculator, receive we get the value of two times 10 to the -2 molars per second. Now scrolling down for more space here. So we'll do it for experiments three. So we have that rate is equal to negative delta X. Over delta T. So for our final value in experiment three we have 2.4 molar. And the initial value is three molars. And that's going to be all divided by 30 seconds because that's their change in time. So once you put that into a calculator, we also get two times 10 to the - molars per second. Now finally we can go ahead and do this calculations for experiment. For so rate is equal to negative delta change in X over the change of T. So time. So we have negative 2.7 molars minus three molars. And of course our changing times also is just 30 seconds. Once you put that in the calculator we get one times 10 to the negative two molars per second. Now we're gonna go ahead and just scroll down and we'll see that for experiment one and 2, the concentration of x does double. And the initial rate also doubles This leads us to know that our M. is equal to one Enough. For experiments one and 3, the concentration of a doubles and the initial rates also doubles. Which we can find out that the value of P is equal to one. Now, lastly for experiments one and 4, the concentration of why here doubles. And the initial rates remains the same, which means that our value of N, then is equal to zero. So they find an equation for our rate. Here we have rate is equal to K. For our constant multiplied by the concentration of X, multiplied by the concentration of A. So this right here is going to be my final answer for this problem.