The equilibrium constant for the dimerization of acetic acid in b...

Question

The equilibrium constant for the dimerization of acetic acid in benzene solution is 1.51 * 102 at 25 °C.
2 CH3CO2H ∆ 1CH3CO2H22 Kc = 1.51 * 102 at 25 °C
(b) What is the osmotic pressure of the solution at 25 °C?

Accepted Answer

hi everyone for this problem. It reads at 298 kelvin, the equilibrium constant when a forms a dime er B and a benzene solution is 2.4 times 10 to the two determine the osmotic pressure of the solution at 298 kelvin if 2980.1 moles of pure A is dissolved in enough benzene to produce 500 mL of solution. So the question that we want to answer here is the osmotic pressure. Okay, and there's an equation that we can use to solve this in that equation is the osmotic pressure is going to equal M. R. T. Okay. And let's go ahead and define this. So osmotic pressure is equal to M is going to be the equilibrium concentration of A plus the equilibrium concentration of B. And this is going to be times the gas constant R times temperature. Okay, so gas constant R is equal to 0.8206 leaders, atmosphere over mole Kelvin & T. is temperature and Calvin. And we were given the temperature and the problem we're told the temperature is 298 Kelvin. So what we need to do in order to solve for the osmotic pressure is we need to find the equilibrium concentration of A and B. So what we're going to need to do is we're going to need to create an ice table for the the dye amortization of A to B. So let's go ahead and create our ice table to do this. We need to rewrite the equation. So we have to a yields be okay. And for our ice table we have our eyes on the left side and remember I represent our initial concentration. So let's take a look at the problem. We have .1 moles of pure a. And the volume is 500 ml. Okay? So we know that the volume of solution Is equal to ml and we know that the moles Will signify this with N is equal to 0.100. So remember concentration is most over leader. So let's go ahead and convert this volume two leaders. Okay, so this is going to become 0.500 leaders. So that means our concentration of a. When we put it in moles over leader we have 0.100 over 0.500. And this gives us 0.2 molar. Okay, so this is what we're gonna put here. So 0.2 and then we have no product. So for B this is going to be zero. So what this means then is our our reaction is going to move in the forward direction. So for our change in concentration, which is the second row of the ice table, our concentration of reactant is decreasing. So we have minus and we have two moles and then we write X. Okay? And then for our products, our concentration of products is increasing. So this becomes A. Plus and we only have one mold. So this is just X. So when we combine the two rows we get 0. minus two X. For the equilibrium role and just X. So let's go ahead and write our K. C expression. Our equilibrium constant expression and our equilibrium constant expression is the concentration of products over the concentration of reactant. So based off of our reaction, our product is B. So this becomes concentration of B. Over concentration of our reactant are reacting is A. And we have two moles of that so that too now becomes an exponent. Okay, so now what we do with this K C. X. Or equilibrium constant expression is we plug in the values that we know based off of the K. C. That was given and based off of what's in the equilibrium role of our ice table. So our case our equilibrium constant was given in the problem and we're told so let's go ahead and write this here because this is our given um list of givens, so r K C is equal to 2.40 times 10 to the two. That's what we see right here. Alright, so that means K C is 2.40 times 10 to the two is going to equal concentration of products over concentration of reactant. So now we're going to plug this in for a concentration of products over concentration of reactant. So for our products concentration of B this is represented by X. And the equilibrium over ice table. And for reactant we have 0.2 minus two X. And we have two moles of it. So we need to raise that to the second power. Alright, so let's take a look at whether or not X is negligible. We can do this by looking at the value of K. C. Since our equilibrium constant is greater than one, That means we're going to use the quadratic formula to solve for X. And this is because the initial if the initial concentration over the KC is less than 100, then we use the quadratic formula. Alright, so we're going to simplify this equation so we know that we need to so KC is greater than one. Use quadratic formula to solve for X. Okay, so let's go ahead and simplify let's get rid of our fraction. Okay, so when we simplify this we get 2.40 times 10 to the second is equal to x over 0.4 minus 0. X plus four X squared. Okay, so let's go ahead and simplify this will multiply both sides by multiply these two by each other to get rid of the fraction. And when we do that we get 9.6 minus X plus 960 X squared is equal to X. And we want to bring everything over to one side and set it equal to zero. So when we do that, what we get is 960 X squared minus 193 X plus 9.6 is equal to zero. So now we can use the quadratic formula to solve for X. And let's just recall that the quadratic formula is X. Is equal to negative B plus or minus the square root of B squared minus four. A. C over To a. Okay, so based off of our equation here, we'll see that a is equal to B is equal to -193 and C is equal to 9.6. So let's go ahead and plug these values into our quadratic our quadratic equation. So X is going to equal negative negative 1 93 plus and minus the square root of B squared which is negative 1 93 squared minus four times A. Which is 9 60 times C which is 9.6. And this is all over two times A. So over two times 960. So what we're going to get when we saw for this is two values for X. For the first one we're going to get X is equal to nine 9.0301 Times 10 to the -2 and 1.1074 times 10 to the negative one. So only one of these values is going to work for us and we're going to disregard the second X. So 1.74 times 10 to the negative one. Since it will lead to a negative concentration, which is impossible. Okay, so what this means then is we can now solve for the concentrations of B and A. Based off of our ice table. So we know based off our ice table the concentration of be at equilibrium is equal to X. Okay. And we just solve for X. So that means the concentration of B at equilibrium is 9.301 times 10 to the negative to moller. And the concentration of a at equilibrium is equal to 0.200 minus or the way we wrote it in the ice table is 0.2 minus two X. So let's go ahead and plug this X in. So we get 0.2 minus two times X. Which is 9.301 times 10 to the negative two. Okay, so when we saw for this we'll get the concentration of A. At equilibrium Is equal to 1.939, 8 times 10 to the negative to moller. These were the two pieces of information were missing up above to solve for the osmotic pressure. Okay, so recall up here that we wrote the equation to solve for the osmotic pressure. So let's just go ahead and rewrite that down here. Okay. So he said the equation to solve for the osmotic pressure is equal to the concentration of B at equilibrium. Plus excuse me, let me rewrite that the way that we wrote it up above the osmotic pressure is equal to the concentration of A. At equilibrium plus the concentration of be at equilibrium times are times team. So now we have everything we need. So let's just go ahead and plug all that in. So we have the concentration of A which is 1.9398 times 10 to the negative two molar plus the concentration of B which is 9.301 times 10 to the negative to moller. And this is times are where we said our is our gas constant which we rode up above 0.8 to 06 leaders. Atmosphere over mole Calvin. And the temperature which we brought up above as well was given and that's 298 Calvin. So let's go ahead and solve. And when we do we'll get the osmotic pressure is equal to 2.68 atmospheres. And this is going to be our final answer. That is it for this problem. I hope this was helpful

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