Consider the reaction: A(g) ⇌ B(g) + C(g) Find the equilibrium concentrations of A, B, and C for each value of K_c. Assume that the initial concentration of A in each case is 1.0 M and that the reaction mixture initially contains no products. Make any appropriate simplifying assumptions. b. K_c = 0.010

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Accepted Answer

hi everyone for this problem. It reads given the following equilibrium reaction and we're also given the equilibrium constant, given that the initial concentration of D is 1.50 moller and no products are present, calculate the equilibrium concentrations of D. E and F. Alright, so our goal here is to calculate the equilibrium concentrations of D. E and F. So what we can do here is we can calculate the equilibrium concentrations from the given initial concentration of D. So let's go ahead and create an ice table here by rewriting our reaction. So our reaction is D. Is that equilibrium? And then we have E plus F. Okay, so this is our reaction. We're going to write out ice on the side and I'm going to draw this line here just to separate Our react mints from our products and so our eye represents our initial concentration and so they tell us that our initial concentration of D is 1.50 and we have no products. So that means our concentration is going to be zero. So in order for us to reach equilibrium, our reaction is going to want to shift towards the right to produce products. And so that means our change is going to be, we're going to have our reactant is consumed, so this is going to be minus and then we're going to have our products created. So this is going to be a plus. And so now we need to look at our number of moles for each, so for everything, we just have one mole and so we're just going to have X represent our change. And so we have minus X plus X and plus X. Okay, our last row is E. And that's our equilibrium concentrations. So we're going to add up our first two rows. So we have 1.50 minus X. X. And X. Okay, so now that our ice table is complete, we want to write out our equilibrium expression and that expression is our equilibrium constant is going to equal our concentration of products over our concentration of reactant. Each raise to its Tokyo metric coefficient. So because our reaction is proceeding to the right here are two products are E and F. Okay, so we're going to copy down what we have in our equilibrium row for this part. So we have our two products are E and F. So we have X times X. And they're both raised to the first. Everything is going to be raised to the first power because we only have one mole of each and our our reactant is D and that is represented by 1.50 minus x. Okay, so we do know what the value of K C is here because they tell us that it is 0.450. So we're going to set this equal to what we just wrote out. So our K C which is 0.450 is going to be set equal to our concentration of products over our concentration of reactant. So we have X squared, we're just simplifying the right side is over 1.50 minus x. Okay so our goal here is to solve for X. Now one thing that we can do before we move on is to check if X is negligible. If X is negligible then we can ignore this X at the bottom and we can eliminate it. But first we need to check to see if it's negligible. First and the way that we do that is by dividing our initial concentration by our K. C value. So our initial concentration is point, our initial concentration is 1.50 and our K C value is 0.450. Now if this value is greater than 500 then X is negligible. But if it's less than 500 then X is not negligible. So this value is going to be less than 500. So that means X is not negligible. Okay so we tried so that means we're not going to ignore that X in the denominator. So let's go ahead and solve, we're going to isolate X squared by multiplying both sides of our equation by 1.50 minus x. Okay So we have 0. times 1.50 minus X is going to equal X squared. So let's go ahead and foil out the left side. So we have 0. sorry nope that is 0.675 minus 0.450 X. Is equal to X squared. So now we want to isolate our our set, everything equal to zero. So then we're going to get X squared plus 0.450 X minus 0.675 is equal to zero. So now what we can do to solve for R. X in this case is use the quadratic equation. Okay, so when we use the quadratic equation, we're going to get two values for X. And Our first value is going to be 0.6268. And our second value is going to be negative 1.768. So we can only have one correct answer for X here and concentrations cannot be negative. So we're going to go ahead and ignore this second X. So we know that our concentration is going to be 10.6268. Okay, this is a correction here, this should be . 68. So now that we know what our X. Is, we can plug it in for X. And the equilibrium row of our table to find out the concentration for everything in our reaction. So let's go ahead and do that. Our concentration for D. We have is 1.50 minus X. So we know what X. Is. So let's go ahead and plug that in. So 1.50 minus 0.6268 gives us a concentration and equilibrium concentration of 0.873 molar. Our concentration for E Is just represented by X. And so that is going to be an equilibrium concentration of 0.6-7 molar. And our equilibrium concentration for F is going to be the same thing, it is represented by X, which is 0.6 to seven moller. So our final answer is our equilibrium concentration for D is 0.873 moller. Our equilibrium concentration for E is 0.6 to seven moller, and our equilibrium concentration for F is 0.6 to seven moller. That's the end of this problem, I hope this was helpful.

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Key Concepts

Chemical Equilibrium

Equilibrium Constant (Kc)

ICE Table (Initial, Change, Equilibrium)