Consider the following hypothetical aqueous reaction:
A1aq2S B1aq...

Question

Consider the following hypothetical aqueous reaction: A1aq2S B1aq2. A flask is charged with 0.065 mol of A in a total volume of 100.0 mL. The following data are collected: Time (min) 0 10 20 30 40 Moles of A 0.065 0.051 0.042 0.036 0.031 (a) Calculate the number of moles of B at each time in the table, assuming that there are no molecules of B at time zero and that A cleanly converts to B with no intermediates.

Accepted Answer

hi everyone for this problem it reads given a hypothetical acquis reaction, C yields D. A solution of 0.78 moles of C in a total volume of 200 mL was made in a flask, the moles of C for every 10 seconds was recorded. What is the amount and moles of D? And the solution for each time recorded in the following data with the assumption that there is only see initially and no molecule of D and that the reaction does not have an intermediate. Okay, so our goal here is to figure out the amount Of moles of D and the solution for each time recorded. Okay, so this is what we want to do and we want to do that based off of what is given in the data set above. Okay. And so the first thing we want to do is look at our multiple ratio. Okay, so we have a 1-1 mole ratio. And what that means is for every one mole of C, which is a reactant for every one mole of C consumed, one mole of D is produced as a product. Okay, and so with that the way that we'll be able to find the moles of D and we'll write it here, the mole of D is going to equal the mole of C, initial minus the moles of C final. Okay, so here it's going to look like this. So this is our initial so for time equals zero seconds, the initial moles of C is 00.78 and the final moles of C is 0.62 at 10 seconds. Okay. And so we're going to follow this pattern for initial minus final and then we're going to calculate at that particular time what is the mole of D? Okay, so let's go ahead and scroll down and we'll write it like this. Okay, So for T equals zero. Okay, Are more of D is also going to equal zero. Okay, because we have we don't have any product yet at T equals zero. Alright, so we'll go ahead and write moles of D and just copy it over here. So at T equals zero. We only have a reactant. Okay, that's moles of C and we have zero product. So our molds of D is going to equal zero at time equals zero. Now for T equals seconds. Okay, for T equals 10 seconds. Our mole of D. Remember we said we're going to take the mole of C initial minus the mold of C final. So our mole of C initial is 0.78 minus moles of C. Final is 0.62. So this is our first arrow here are 0.78 is the initial and our 0.62 is the final four time equals 10 seconds. So once we do this calculation we get 0. for moles of D at 10 seconds. Okay, so let's go ahead and write that up here. So zero point 016 is the most of D at 10 seconds. All right, so let's move it up. So it's not confusing. So that's at 10 at time equals 10 seconds. Alright, So for a t equals 20 seconds are more of D is now going to equal the moles of C. Initial. So the initial is gonna be the .062. So for time equals 20 seconds, the initial is . -152. Okay, because the .062 is the initial and the .052 is the final. So once we do that calculation, we get 0.01 For moles of D at 20 seconds. Alright, so for time equals 30 seconds, the moles of D is going to equal our initial, Which is 0.05, 2 - the final 0.045. And we get most of D at T equals 30 seconds is 0.07. Excuse me, 0.7. And lastly at t equals 40 seconds are moles of D is going to be the initial minus. Final initial is 400. 0. 0.039. And this gives us modes of D at 40 seconds is equal to 0.006. Okay, so this is it. This is the amount of moles of D. And the solution for each time recorded from the following data set. Okay, so that is it for this problem. I hope this was helpful.

Watch next