The following equilibria were attained at 823 K: CoO(s) + H 2 (g) → Co(s) + H 2 O(g) K c = 67 CoO(s) + CO(g) → Co(s) + CO 2 (g) K c = 490 Based on these equilibria, calculate the value of 𝐾 𝑐 for H 2 (𝑔) + CO 2 (𝑔) ⇌ CO(𝑔) + H 2 O(𝑔) at 823 K.

Welcome back, everyone. At 298 Kelvin, the equilibria given below are observed. What is the equilibrium constant KQ for the following reaction? Now, if we look at the target reaction, we are given hso two solid reacting with so gas to produce hso gas and so two gas, and specifically, we're given two equations. If we manipulate them, we can combine them in a specific way such that we get the target equation. So essentially what we're going to do is just carefully analyze the target equation. And immediately what we notice is that one of our reactants is HSOT, which is actually a product in the first equilibrium. So what we're going to do is just reverse equation number one, if we reverse it, we obtain HS two gas reacts with N no gas to produce hso gas and no two gas. Let's recall that whenever we reverse an equilibrium equation, we are essentially taking the inverts of the equilibrium constant. So the equilibrium constant for this part would become a one divided by K one, right? Because we're essentially taking the inverts or the reciprocal value of the original equilibrium constant. Looking at the second reaction, we can essentially notice that the target equation has so on the reactant side and the second equation has so on the reactant side as well, meaning we are not going to change anything. So let's just rewrite it. So reacts with no two to produce sot and I know with an equilibrium constant of K to what we're going to notice is that if we add them up, we're going to first of all eliminate no two because no two appears on opposite sides of the two equations, right? And we're going to get ho two as our first species in the sun plus. If so, we can also notice that we can eliminate no, since it appears on opposite sides of the two equations and the remaining species are hso which is consistent with the target equation plus so two, which is also consistent with the target equation. What we have shown here is that if we reverse the first equilibrium and if we add the two equations, we simply obtain the expression of the target equation. Therefore, if we add two equations, then the equilibrium constant of the resultant equation is simply the product of the individual equilibrium constants. Now, what does that mean? Well, the equilibrium constant for the first equation was one divided by K, one in the equilibrium constant. For the second step was K two. If we add them up, we obtain an equation with an equilibrium constant, that is a product of one divided by K one ant K two or in a simplified form K two divided by K one. So we simply want to substitute those results were specifically those values K two would be 1.4 multiplied by it. Sense the power of negative 11th. And we want to divide that by K one 9.6 multiplied by sense the power of negative 12. And if we evaluate the result, we get an equilibrium constant of 1.5. That's our final answer. Thank you for watching.

Ch.15 - Chemical Equilibrium

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Brown 15th Edition

Ch.15 - Chemical Equilibrium

Problem 27

Chapter 15, Problem 27

The following equilibria were attained at 823 K:
CoO(s) + H₂(g) → Co(s) + H₂O(g) K_c = 67
CoO(s) + CO(g) → Co(s) + CO₂(g) K_c = 490
Based on these equilibria, calculate the value of 𝐾_𝑐 for H₂(𝑔) + CO₂(𝑔) ⇌ CO(𝑔) + H₂O(𝑔) at 823 K.

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Identify the given reactions and their equilibrium constants: Reaction 1: CoO(s) + H_2(g) → Co(s) + H_2O(g) with K_{c1} = 67, Reaction 2: CoO(s) + CO(g) → Co(s) + CO_2(g) with K_{c2} = 490.

Write the target reaction: H_2(g) + CO_2(g) ⇌ CO(g) + H_2O(g).

Recognize that the target reaction can be derived by reversing Reaction 2 and adding it to Reaction 1.

Reverse Reaction 2: CO(s) + CO_2(g) → CoO(s) + CO(g), which changes the equilibrium constant to 1/K_{c2}.

Add the reversed Reaction 2 to Reaction 1 to obtain the target reaction, and calculate the equilibrium constant for the target reaction as K_{c} = K_{c1} * (1/K_{c2}).

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Key Concepts

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

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. The equilibrium constant (Kc) quantifies the ratio of product concentrations to reactant concentrations at equilibrium, providing insight into the position of the equilibrium. Understanding this concept is crucial for manipulating and calculating Kc values in various reactions.

Equilibrium Constant (Kc)

The equilibrium constant (Kc) is a numerical value that expresses the relationship between the concentrations of products and reactants at equilibrium for a given reaction at a specific temperature. It is calculated using the formula Kc = [products]^[coefficients] / [reactants]^[coefficients]. The value of Kc indicates whether the products or reactants are favored in the equilibrium state, which is essential for solving equilibrium-related problems.

Manipulating Equilibrium Expressions

Manipulating equilibrium expressions involves using the properties of Kc to derive new equilibrium constants from known ones. For example, if a reaction is reversed, the new Kc is the reciprocal of the original. Additionally, if reactions are added together, the Kc for the overall reaction is the product of the individual Kc values. This concept is vital for calculating Kc for complex reactions based on simpler, known equilibria.

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Consider the following equilibrium: 2 H₂(𝑔) + S₂( 𝑔) ⇌ 2 H₂S(𝑔) 𝐾_𝑐= 1.08×10⁷ at 700°C (c) Calculate the value of 𝐾𝑐 if you rewrote the equation H₂(𝑔) + 1/2 S₂( 𝑔) ⇌ H₂S(𝑔)

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At 1000 K, 𝐾_𝑝= 1.85 for the reaction SO₂(𝑔) + 12 O₂(𝑔) ⇌ SO₃(𝑔) (c) What is the value of 𝐾𝑐 for the reaction in part (b)?

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Consider the following equilibrium, for which at 𝐾_𝑝= 0.0752 at 480°C: 2 Cl₂(𝑔) + 2 H₂O(𝑔) ⇌ 4 HCl(𝑔) + O₂(𝑔) (a) What is the value of 𝐾_𝑝 for the reaction 4 HCl(𝑔) + O₂(𝑔) ⇌ 2 Cl₂(𝑔) + 2 H₂O(𝑔)?

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Consider the equilibrium N₂(𝑔) + O₂(𝑔) + Br₂(𝑔) ⇌ 2 NOBr(𝑔) Calculate the equilibrium constant 𝐾𝑝 for this reaction, given the following information at 298 K:

2 NO(𝑔) + Br₂(𝑔) ⇌ 2 NOBr(𝑔) 𝐾_𝑐= 2.02

NO(𝑔) ⇌ N₂(𝑔) + O₂(𝑔) 𝐾_𝑐= 2.1×10³⁰

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Textbook Question

The equilibrium 2 NO(𝑔) + Cl₂(𝑔) ⇌ 2 NOCl(𝑔) is established at 500.0 K. An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for NO, Cl₂, and NOCl, respectively. (a) Calculate 𝐾_𝑝 for this reaction at 500.0 K.

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Textbook Question

The equilibrium 2 NO(𝑔) + Cl₂(𝑔) ⇌ 2 NOCl(𝑔) is established at 500.0 K. An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for NO, Cl₂, and NOCl, respectively. (b) If the vessel has a volume of 5.00 L, calculate Kc at this temperature.

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The following equilibria were attained at 823 K:CoO(s) + H2(g) → Co(s) + H2O(g) Kc = 67CoO(s) + CO(g) → Co(s) + CO2(g) Kc = 490Based on these equilibria, calculate the value of 𝐾𝑐 for H2(𝑔) + CO2(𝑔) ⇌ CO(𝑔) + H2O(𝑔) at 823 K.

Verified Solution

Key Concepts

Chemical Equilibrium

Equilibrium Constant (Kc)

Manipulating Equilibrium Expressions

The following equilibria were attained at 823 K:
CoO(s) + H₂(g) → Co(s) + H₂O(g) K_c = 67
CoO(s) + CO(g) → Co(s) + CO₂(g) K_c = 490
Based on these equilibria, calculate the value of 𝐾_𝑐 for H₂(𝑔) + CO₂(𝑔) ⇌ CO(𝑔) + H₂O(𝑔) at 823 K.