The following equilibria were measured at 823 K: CoO(s) + H 2 (g) ⇌ Co(s) + H 2 O(g) K c = 67 H 2 (g) + CO 2 (g) ⇌ CO(g) + H 2 O(g) Kc = 0.14 (a) Use these equilibria to calculate the equilibrium constant, K c , for the reaction CoO(s) + CO(g) ⇌ Co(s) + CO 2 (g) at 823 K.

everyone. So here at the equilibrium for the following reactions. Remember aspects on the given equilibrium. What would be the K. C. For the overall reaction were given to reactions with Casey values. We need to find a K. C. For the overall reaction. We need to look at the overall reaction for each of the reactors and products. And the two reactions here we have copper metal sulfide as a reactant. And it's a reality in this reaction as well with the same number of moles. So we don't have to do anything here and we have hydrogen gas which is in both reactions but it is not in the overall reaction at all. So we don't have to do anything here. Now we have copper which is a product and it's a product in this reaction as well with the same number of moles. They don't have hydrogen sulfide which is a product in both reactions but it is not in the overall reaction at all. Then we have carbon dot sulfide which is the reactant. That is a product in this reaction. So we need to reverse the reaction. So we're gonna have carbon mono sulfide because hydrogen sulfide and it's gonna be your carbon di sulfide plus hydrogen gas. Since we reversed the reaction for the K. C. Value, Gonna be equal to one Over the KC buyer for the second reaction which is 0.289. Can we get 3.46. And now we have carbon mono so far which is a reactor in the new equation and it's a reacting here as well. Now we're gonna add both the reactions and cancel out the common species. Get the overall reaction. So we have the first reaction which is copper wanna sulfide that's harsh and gas and then shows copper that's hydrogen sulfide. And then we have the reverse of the second reaction which is carbon mono sulfide, that's hydrogen. So fine building apartment dot sulfide. That's hydrogen gas for the K. C. Values. You have 78.9 for the first one And there are new K. c. value for the second which is 3.46. So you can cancel any common species and it's gonna be hydrogen gas. It has been so five and I left a copper model. Sophie as carbon model sulfide, you're in copper does carbon di sulfide. And now for the K. C. Value we need the most part the two K. C. We get K. C. Equals 273. Thanks for watching my video and I hope it was helpful

Ch.15 - Chemical Equilibrium

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Brown 14th Edition

Ch.15 - Chemical Equilibrium

Problem 96a

Chapter 15, Problem 96a

The following equilibria were measured at 823 K: CoO(s) + H₂(g) ⇌ Co(s) + H₂O(g) K_c = 67 H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g) Kc = 0.14 (a) Use these equilibria to calculate the equilibrium constant, K_c, for the reaction CoO(s) + CO(g) ⇌ Co(s) + CO₂(g) at 823 K.

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Identify the given reactions and their equilibrium constants. The first reaction is CoO(s) + H2(g) ⇌ Co(s) + H2O(g) with Kc = 67. The second reaction is H2(g) + CO2(g) ⇌ CO(g) + H2O(g) with Kc = 0.14.

Write the target reaction: CoO(s) + CO(g) ⇌ Co(s) + CO2(g). Notice that this reaction can be obtained by reversing the first given reaction and adding it to the second given reaction.

Reverse the first reaction to get Co(s) + H2O(g) ⇌ CoO(s) + H2(g). The equilibrium constant for the reversed reaction will be the reciprocal of the original, which is 1/67.

Add the reversed first reaction to the second reaction. This involves adding Co(s) + H2O(g) ⇌ CoO(s) + H2(g) to H2(g) + CO2(g) ⇌ CO(g) + H2O(g). The H2(g) and H2O(g) will cancel out, resulting in CoO(s) + CO(g) ⇌ Co(s) + CO2(g).

Calculate the equilibrium constant for the target reaction by multiplying the constants of the individual reactions used. The equilibrium constant for the target reaction, Kc, is (1/67) * 0.14.

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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. It is calculated using the formula Kc = [products]^[coefficients] / [reactants]^[coefficients]. A Kc value greater than 1 indicates that products are favored at equilibrium, while a value less than 1 suggests that reactants are favored.

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 crucial for predicting how changes in concentration, pressure, or temperature will affect the position of equilibrium in a chemical reaction.

Manipulating Equilibrium Expressions

When dealing with multiple equilibria, it is essential to manipulate the equilibrium expressions correctly to derive the desired Kc for a new reaction. This involves using the relationships between the given reactions, such as reversing a reaction (which inverts Kc) or adding reactions (which multiplies Kc values). Understanding how to combine these expressions is key to solving complex equilibrium problems.

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Equilibrium Constant Expressions

Related Practice

Textbook Question

Consider the hypothetical reaction A(𝑔) + 2 B(𝑔) ⇌ 2 C(𝑔), for which 𝐾_𝑐= 0.25 at a certain temperature. A 1.00-L reaction vessel is loaded with 1.00 mol of compound C, which is allowed to reach equilibrium. Let the variable x represent the number of mol/L of compound A present at equilibrium.

(d) The equation from part (c) is a cubic equation (one that has the form ax3 + bx2 + cx + d = 0). In general, cubic equations cannot be solved in closed form. However, you can estimate the solution by plotting the cubic equation in the allowed range of x that you specified in part (b). The point at which the cubic equation crosses the x-axis is the solution.

(e) From the plot in part (d), estimate the equilibrium concentrations of A, B, and C. (Hint: You can check the accuracy of your answer by substituting these concentrations into the equilibrium expression.)

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Textbook Question

At a temperature of 700 K, the forward and reverse rate constants for the reaction 2 HI(g) ⇌ H₂(g) + I₂(g) are k_f = 1.8×10⁻³⁰ M⁻¹s⁻¹ and k_r = 0.063 M⁻¹s⁻¹.

(a) What is the value of the equilibrium constant K_c at 700 K?

(b) Is the forward reaction endothermic or exothermic if the rate constants for the same reaction have values of k_f= 0.097 M⁻¹s⁻¹ and k_r= 2.6 M⁻¹s⁻¹ at 800 K?

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Open Question

Consider the reaction IO4- (aq) + 2 H2O (l) ⇌ H4IO6- (aq); Kc = 3.5 * 10^-2. If you start with 25.0 mL of a 0.905 M solution of NaIO4 and then dilute it with water to 500.0 mL, what is the concentration of H4IO6- at equilibrium?

Textbook Question

The following equilibria were measured at 823 K: CoO(s) + H₂(g) ⇌ Co(s) + H₂O(g) K_c = 67 H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g) Kc = 0.14 (d) If the reaction vessel from part (c) is heated to 823 K and allowed to come to equilibrium, how much CoO(s) remains?

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Textbook Question

The phase diagram for SO₂ is shown here. (d) At which of the three points marked in red does SO₂(g) most closely approach ideal-gas behavior?

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Textbook Question

The phase diagram for SO₂ is shown here. (e) At which of the three red points does SO₂(g) behave least ideally?

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The following equilibria were measured at 823 K: CoO(s) + H2(g) ⇌ Co(s) + H2O(g) Kc = 67 H2(g) + CO2(g) ⇌ CO(g) + H2O(g) Kc = 0.14 (a) Use these equilibria to calculate the equilibrium constant, Kc, for the reaction CoO(s) + CO(g) ⇌ Co(s) + CO2(g) at 823 K.

Verified Solution

Key Concepts

Equilibrium Constant (Kc)

Le Chatelier's Principle

Manipulating Equilibrium Expressions

The following equilibria were measured at 823 K: CoO(s) + H₂(g) ⇌ Co(s) + H₂O(g) K_c = 67 H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g) Kc = 0.14 (a) Use these equilibria to calculate the equilibrium constant, K_c, for the reaction CoO(s) + CO(g) ⇌ Co(s) + CO₂(g) at 823 K.