Acetic acid tends to form dimers, (CH 3 CO 2 H 2 ), because of hydrogen bonding: The equilibrium constant Kc for this reaction is 1.51⨉10 2 in benzene solution but only 3.7⨉10 -2 in water solution. (b) Calculate the ratio of dimers to monomers for 0.100 M acetic acid in water.

Hi everyone for this problem. It reads formic acid. The simplest car bulk silic acid exists as a dime er and benzene solution due to hydrogen bonding for this reaction, the equilibrium constant is 1.5 times 10 to the two in the benzene solution, determine the ratio of di MERS two monomers for a 20.3 molar formic acid and benzene. So we want to determine the ratio of di MERS two monomers. Okay, so let's go ahead and start off by creating an ice table for the reaction. So let's write out our reaction. Okay? And we're going to create an ice table For the I. Row of our table. We're going to put the initial concentration here and we're told in the problem that there is .3 molars of formic acid. Okay, so this is going to go here. So 0.300 and we have no product. So what that means is our reaction is shifting in the four direction and we're going to lose concentration of reactant and gain concentration of products are change then is going to be minus for the reactant and we have two moles of that. So it's going to be minus two X. And we're gaining products. So this is going to be plus and we have one mole of this. So plus X. So we're gonna combine R. I N. C bro for equilibrium row. So this becomes zero point 300 minus two X. And this becomes X. So now let's go ahead and write out our equilibrium constant expression and that is K. C. Is equal to the concentration of products over the concentration of react ints so looking at our reaction, what this is is our concentration of products is going to be the following. And this is over our concentration of reactant. So we have our reactant and because there's two moles of it that two moles now becomes the exponent. Alright, so when we do this and we replace what's in the equilibrium row of our ice table, we already know what the value is for the equilibrium constant. It's 1.1, it's 1.50. So K C. Which is 1.50 times 10 to the two is going to equal the concentration of products over concentration of reactant. We're copying what we see here and the hero of our equilibrium or in the hero of our rice table. So this becomes the concentration of X. Over the concentration of 0.300 minus two X. Okay, so now that that this is going to be our equation. Alright, so let's take a look at the value for K. C. R. Value for K. C is greater than one. All right, So we're going to right here since K C. Is greater than one, the concentration of the product over K C Will be less than 500. So what that means is we're going to need to use the quadratic formula to software X. So let's go ahead and simplify our equation. Okay, we're going to simplify this equation right here. So when we simplify this equation, what we get is 1.50 times 10 to the two is equal to X over 0.9 minus 1. X plus four X squared. Okay, so let's go ahead and simplify this out by by getting rid of the fraction. So we're going to get 13.5 minus 180 X plus 600 X squared is equal to X. And we want to bring X over to one side and set this equal to zero. So when we do that, what we get is 600 X squared minus 181 X plus 13.5 equals zero. So now we can use the quadratic formula to solve for X. Alright. And let's recall that the quadratic formula is X is equal to negative B plus and minus the square root of B squared minus four A. C over to a. Okay. And here based off of our equation, this equation here we're gonna say that A is equal to 600 B is equal to negative 1 81 and C is equal to 13.5. Alright, so we have everything we need to plug in to our quadratic formula. So let's go ahead and do that. So what we're going to get is X is equal to negative And our number four B is already negative. So this is negative, 181 plus and minus the square root of B squared which is negative 81 squared minus four times 600 times 13. over two times 600. Okay, so when we saw for this we're going to get two values for X. The first one we're going to get is x is equal to 1.6667 times to the negative one And 1. times 10 to the -1. Okay, so we're going to discard the first X which is 1.667 times 10 to the negative one. Since it leads to a negative equilibrium concentration, which is impossible. Okay, so we're going to get rid of that. So what that means then is X. Yes. So what this means then is the concentration at equilibrium is going to equal this the value for the second X 1. times 10 to the negative one. Okay. So when we look at our equilibrium row of the ice table, what that tells us then is the concentration at equilibrium is equal to 0.300 minus two X. And we know what the value of X is now. So we can go ahead and plug this in. So this becomes 0.300 minus two times the X. We just solve for which is 1.3500 times to the negative one. So this equals 3. Times 10 to the -2. Okay, and we're almost done. Just make some room. Okay, So solving for the ratio of dime er, to monitor, what we're going to do is dime er over. Monomer Is going to equal that value for X. So we have 1.3500 times 10 to the negative one and this is over What we just solve for, which is 3.00 times 10 to the -2. So this gives us 4.50. Okay, so our answer then for this problem is 4.50 is the ratio for our timer to monitor that. Is it for this problem? I hope this was helpful.

Ch.15 - Chemical Equilibrium

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McMurry 8th Edition

Ch.15 - Chemical Equilibrium

Problem 154b

Chapter 15, Problem 154b

Acetic acid tends to form dimers, (CH₃CO₂H₂), because of hydrogen bonding: The equilibrium constant Kc for this reaction is 1.51⨉10² in benzene solution but only 3.7⨉10^-2 in water solution. (b) Calculate the ratio of dimers to monomers for 0.100 M acetic acid in water.

Verified step by step guidance

Identify the equilibrium reaction for the dimerization of acetic acid: 2 \text{CH}_3\text{CO}_2\text{H} \rightleftharpoons (\text{CH}_3\text{CO}_2\text{H})_2.

Write the expression for the equilibrium constant, K_c, for the reaction: K_c = \frac{[(\text{CH}_3\text{CO}_2\text{H})_2]}{[\text{CH}_3\text{CO}_2\text{H}]^2}.

Substitute the given K_c value for water (3.7 \times 10^{-2}) into the equilibrium expression.

Let x be the concentration of dimers at equilibrium. Then, the concentration of monomers at equilibrium is (0.100 - 2x) M.

Solve the equation 3.7 \times 10^{-2} = \frac{x}{(0.100 - 2x)^2} to find the ratio of dimers to monomers, \frac{x}{0.100 - 2x}.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hydrogen Bonding

Hydrogen bonding is a type of intermolecular force that occurs when a hydrogen atom covalently bonded to a highly electronegative atom, such as oxygen or nitrogen, experiences an attraction to another electronegative atom. In the case of acetic acid, the hydrogen bonds between molecules lead to the formation of dimers, which significantly affects the concentration of monomers and dimers in solution.

Equilibrium Constant (Kc)

The equilibrium constant (Kc) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. In this context, Kc indicates the extent to which acetic acid forms dimers in different solvents, with higher values suggesting a greater tendency to form dimers, as seen in benzene compared to water.

Concentration and Ratio Calculations

Concentration refers to the amount of a substance in a given volume of solution, typically expressed in molarity (M). To calculate the ratio of dimers to monomers in a solution of acetic acid, one must use the equilibrium constant and the initial concentration to determine the concentrations of both species at equilibrium, allowing for the calculation of their ratio.