For the reaction shown here, K c = 0.513 at 500 K. N 2 O 4 (g) ⇌ 2 NO 2 (g) If a reaction vessel initially contains an N 2 O 4 concentration of 0.0500 M at 500 K, what are the equilibrium concentrations of N 2 O 4 and NO 2 at 500 K?

hi everyone for this problem, we're told to consider the following hypothetical equilibrium reaction. We need to determine the concentration of A. And B. Once equilibrium is established, assuming you start only with 2.50 molar of A. So we're going to calculate the equilibrium concentrations from the given initial concentration and equilibrium constant using an ice table. So let's go ahead and rewrite out our equation. So we have to a and we have to be okay, so we ride out ice on the side and I'm going to draw this line here to separate our products from our reactant, since we are told we have 2.5 molar of A. So this is what we start off with initially, this is what our I wrote of our ice table represents. And so at the very beginning of the reaction, we have 2.50 moles of A. And we have zero moles of of B. So the reaction is currently not at equilibrium because we have that zero mol of product. So our reaction is going to shift in the forward direction in order to reach equilibrium. So what that means is our reactant is going to be consumed and our products is going to be created. So, in terms of R C ro for our ice table, our change in concentration for our reactant is going to be negative and our change in concentration of reactant is going to be positive. So now we need to pay attention to our stoke eom a tree. So we have two moles of A. So we're going to have minus two X. And we also have two moles of B. So this is going to be plus two X. Ok. And r E is our equilibrium concentration, which is what we're trying to solve for. So we're going to add up our first two rows and just write it out here. So we have 2.50 minus two X. And then we just have to X here. Alright, so we need to solve for X in order to figure out what our equilibrium concentration is going to be for both A and B. For us to solve for X. We're going to need to write out our equilibrium expression and that expression is K C is equal to our concentration of products over our concentration of reactant by doing this, we're going to be able to solve for X. And the problem, they tell us what our K C value is, it's 5.64. So let's go ahead and plug that in here. Our concentration of products, we see here that it's two X. Okay, so we have two X. But we need to raise it to its Tokyo metric coefficient because we have two moles of B. We need to raise this to the second power for our concentration of reactant, we have 2.50 minus two X. Okay, we're copying everything from our equilibrium row here for this concentration of products and concentration of reactant. So we have 2.50 minus two X. And we need to pay attention to the stoke eo metric coefficient because there's a two there, we need to raise this to the second power. Okay, so our goal here is to solve for X and by solving for X will be able to plug it into our equilibrium row and figure out what our concentrations are. So what we can do here because our rights, everything is square a ble here, we can take the square root of both sides of our equation. And when we do that we get rid of the squares on the right side. So we're going to take the square root of both sides. And when we do that we get a simplification of two point 3749 is equal to two X over 2. minus two X. Okay, so we're going to multiply both sides by 2.50 minus two X. So we have 2. times 2.50 minus two X is equal to two X. So our goal is to solve for X. So let's go ahead and foil out. When we foil out we get 5. minus 4. X is equal to two X. We can divide both sides by let's simplify some more. So we get 5.9471 is equal to 6. 97 X. So we can divide both sides by 6.7497. And when we do that we get X. Is equal to 0.881. Okay, so now that we know what X is, we can plug in X to our equilibrium row of our ice table here to solve for the equilibrium concentration of A. And B. So let's go ahead and do a first. So are equilibrium concentration of A says that it is 2. minus two X. So we know what X is. We just solved for it. Let's bring it up here. We said X is equal to zero point 881. Okay, so that means that our concentration of A is going to equal 2. minus two times 0.881. Which gives us a concentration of 0. Moeller for A. And our equilibrium Rose says that our equilibrium concentration for B is going to equal to X. So we have two times 0. and this gives us a equilibrium concentration of 1.76 moller for B. So these are going to be our final answer. The concentration of A is 0.740 moller and our equilibrium concentration of B is 1.76 smaller. Okay, so that is the end of this problem. I hope this was helpful

Ch.16 - Chemical Equilibrium

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Ch.16 - Chemical Equilibrium

Problem 53

Chapter 16, Problem 53

For the reaction shown here, K_c = 0.513 at 500 K. N₂O₄(g) ⇌ 2 NO₂(g) If a reaction vessel initially contains an N₂O₄concentration of 0.0500 M at 500 K, what are the equilibrium concentrations of N₂O₄and NO₂ at 500 K?

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Set up the initial concentrations of the reactants and products. For N2O4, it is 0.0500 M, and for NO2, it is 0 M since it has not yet formed.

Let x be the change in concentration of N2O4 that reacts to reach equilibrium. The concentration of N2O4 at equilibrium will then be (0.0500 - x) M.

Since the reaction produces 2 moles of NO2 for every 1 mole of N2O4 that reacts, the concentration of NO2 at equilibrium will be (0 + 2x) M.

Write the expression for the equilibrium constant, Kc, in terms of the concentrations of the reactants and products at equilibrium. For the reaction N2O4(g) ⇌ 2NO2(g), Kc = \( \frac{[NO2]^2}{[N2O4]} \).

Substitute the equilibrium concentrations into the Kc expression and solve for x. Use the quadratic formula if necessary. Once x is found, substitute back to find the concentrations of N2O4 and NO2 at equilibrium.

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Key Concepts

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

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. For the reaction N2O4 ⇌ 2NO2, Kc = [NO2]^2 / [N2O4]. A Kc value less than 1 indicates that at equilibrium, reactants are favored, while a value greater than 1 indicates that products are favored.

ICE Table

An ICE table (Initial, Change, Equilibrium) is a tool used to organize the concentrations of reactants and products at different stages of a chemical reaction. It helps in calculating the changes in concentration as the system reaches equilibrium. By filling in initial concentrations, the changes that occur during the reaction, and the equilibrium concentrations, one can solve for unknown values.

Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust itself to counteract the change and restore a new equilibrium. This principle is useful in predicting how changes in concentration, pressure, or temperature will affect the position of equilibrium in a reaction, such as the one involving N2O4 and NO2.

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Le Chatelier's Principle

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Textbook Question

Silver sulfate dissolves in water according to the reaction: Ag₂SO₄(s) ⇌ 2 Ag⁺(aq) + SO₄^2-(aq) K_c = 1.1⨉10^-5 at 298 K. A 1.5-L solution contains 6.55 g of dissolved silver sulfate. If additional solid silver sulfate is added to the solution, will it dissolve?

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Textbook Question

Consider the reaction and the associated equilibrium constant: aA(g) ⇌ bB(g) Kc = 4.0 Find the equilibrium concentrations of A and B for each value of a and b. Assume that the initial concentration of A in each case is 1.0 M and that no B is present at the beginning of the reaction. c. a=1;b=2

Consider the reaction and the associated equilibrium constant: aA(g) + bB(g) ⇌ cC(g) with Kc = 5.0. Find the equilibrium concentrations of A, B, and C for the given values of a, b, and c. Assume that the initial concentrations of A and B are each 1.0 M and that no product is present at the beginning of the reaction. Calculate the equations for x without solving them for the following cases: a) a=1, b=1, c=2; b) a=1, b=1, c=1; c) a=2, b=1, c=1.

Textbook Question

For the reaction shown here, K_c = 255 at 1000 K. CO(g) + Cl₂(g) ⇌ COCl₂(g) If a reaction mixture initially contains a CO concentration of 0.1500 M and a Cl₂ concentration of 0.175 M at 1000 K, what are the equilibrium concentrations of CO, Cl₂, and COCl₂ at 1000 K?

Consider the reaction: NiO(s) + CO(g) ⇌ Ni(s) + CO2(g) Kc = 4.0 * 10^3 at 1500 K. If a mixture of solid nickel(II) oxide and 0.20 M carbon monoxide comes to equilibrium at 1500 K, what is the equilibrium concentration of CO2?

Open Question

Consider the reaction: CO(g) + H2O(g) ⇌ CO2(g) + H2(g) Kc = 102 at 500 K. If a reaction mixture initially contains 0.110 M CO and 0.110 M H2O, what will the equilibrium concentration of each of the reactants and products be?

For the reaction shown here, Kc = 0.513 at 500 K. N2O4(g) ⇌ 2 NO2(g) If a reaction vessel initially contains an N2O4 concentration of 0.0500 M at 500 K, what are the equilibrium concentrations of N2O4 and NO2 at 500 K?

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Key Concepts

Equilibrium Constant (Kc)

ICE Table

Le Chatelier's Principle

For the reaction shown here, K_c = 0.513 at 500 K. N₂O₄(g) ⇌ 2 NO₂(g) If a reaction vessel initially contains an N₂O₄concentration of 0.0500 M at 500 K, what are the equilibrium concentrations of N₂O₄and NO₂ at 500 K?