At 2000°C, the equilibrium constant for the reaction 2 NO(𝑔) ⇌ N 2 (𝑔) + O 2 (𝑔) is 𝐾 𝑐 = 2.4×10 3 . If the initial concentration of NO is 0.175 M, what are the equilibrium concentrations of NO, N 2 , and O 2 ?

Hello everyone today. We are being given the following question and asked to solve for it. So it says the value of K. C. For the reaction is 0.771 at 6 50 degrees Celsius and a two liter reaction vessel three moles of hydrogen gas and carbon dioxide are added and allowed to reach equilibrium at 6 50 degrees Celsius. We are then being asked to calculate the mass of water vapor present at equilibrium. So the first thing we want to do for this chemical reaction is construct our K. C. Expression. In our case the expression is going to involve our products, the concentration of products in the numerator and the concentration of our reactant and the denominator for Casey expressions, we only want to include gasses, not solids or liquids. So with that in mind for our products for this reaction we have a carbon monoxide molecule C. O. And we also have a second product which is water. So we're gonna write H 20. Times that first product. And then in the denominator we have our reactant. So we have hydrogen gas being multiplied by carbon dioxide. And so we have our K. C. Expression written for us. The next thing we wanna do is we want to construct an ice table ice chart. And so in doing that we have to write out our reaction first. So we have this hydrogen gas combining with this carbon dioxide gas and this is an irreversible reaction denoted by the barbed arrow. The double headed arrow. And this is going to form carbon monoxide in the gas form as well as water in the gas form. And so as part of our ice chart, we're going to have I C. E. The I stands for initial that you see as the change and the E. Is the equilibrium. So our initial concentrations of our hydrogen gas is one 0. Mueller. As is the case with our C. 02. How do we know this well? Polarity is calculated by moles over leaders. And so we were given three moles of both H2 and carbon dioxide. So therefore We can take that three moles and divide that by our two leaders. Given to us in the question as well to give us our 1.5 verity. And so that's where that 1.5 comes from. Our initial concentrations for carbon monoxide and water are going to be of course zero. Since we are taking away from our reactant, sweet to know our changes negative X or minus X. And the fire products is going to be plus X and plus X. The E. Is going to be essentially the I subtracted by the sea. So in this case we will have 1.5 minus X. 1.5 minus X. For our products, we're just going to have X and X. And so now we can plug in for R K C expression. Rkc expression. Or we said Casey was equal to 0.771. We're going to equal that to our variables that we just found. And so for our products we have two exes which when combined form X squared and we have two reactions which are the same. And when we combine those, that's going to be 1.5 minus X squared. Our next up is going to be to take the square root of both sides. And when we do that We get 0.8781 is equal to X over 1.5 -1. And then we can simply solve from X. Here. When we solve for X, we get that X Is equal to 0.7013 Molar which is also going to be equal to our concentration of our water at equilibrium. And so now that we have our concentration of water equilibrium, we can now solve for the mass of the water vapor. So that's gonna be our fourth and final step. So we're going to say that the mass of this water, it's going to be equal to the polarity that we just sold for. So 0.7013 moles per one liter. We're going to multiply that by the initial volume that we had our two leaders and then to get rid of this mold or these molds. We're going to multiply this by the molar mass of water which is 18.2 g per one mall our units are going to cancel out And we will be left with 25.3 g of water. I hope this helped, and until next time.

Ch.15 - Chemical Equilibrium

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Brown 14th Edition

Ch.15 - Chemical Equilibrium

Problem 51

Chapter 15, Problem 51

At 2000°C, the equilibrium constant for the reaction 2 NO(𝑔) ⇌ N₂(𝑔) + O₂(𝑔) is 𝐾_𝑐= 2.4×10³. If the initial concentration of NO is 0.175 M, what are the equilibrium concentrations of NO, N₂, and O₂?

Verified step by step guidance

Write the balanced chemical equation: 2 \text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g).

Set up an ICE (Initial, Change, Equilibrium) table to track the concentrations of each species. Initially, [NO] = 0.175 \text{ M}, and [N_2] = [O_2] = 0 \text{ M}.

Define the change in concentration for NO as -2x, for N_2 as +x, and for O_2 as +x, where x is the change in concentration at equilibrium.

Express the equilibrium concentrations in terms of x: [NO] = 0.175 - 2x, [N_2] = x, [O_2] = x.

Substitute these expressions into the equilibrium constant expression: K_c = \frac{[N_2][O_2]}{[NO]^2} = \frac{x \cdot x}{(0.175 - 2x)^2} = 2.4 \times 10^3, and solve for x.

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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 2 NO(g) ⇌ N2(g) + O2(g), Kc = [N2][O2]/[NO]^2. A larger Kc value indicates that products are favored at equilibrium, while a smaller Kc suggests that reactants are favored.

ICE Table

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

Stoichiometry

Stoichiometry involves the quantitative relationships between the amounts of reactants and products in a chemical reaction, based on the balanced chemical equation. In the reaction 2 NO(g) ⇌ N2(g) + O2(g), stoichiometry allows us to relate the change in concentration of NO to the formation of N2 and O2, using the coefficients from the balanced equation to determine how much of each species is present at equilibrium.