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 equilibrium constant 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

All textbooks

Brown 14th 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 equilibrium constant for H₂(g) + CO₂(g) → CO(g) + H₂O(g) at 823 K.

Verified step by step guidance

Identify the given reactions and their equilibrium constants: \( \text{CoO(s) + H}_2\text{(g) } \rightarrow \text{ Co(s) + H}_2\text{O(g) } \) with \( K_c = 67 \) and \( \text{CoO(s) + CO(g) } \rightarrow \text{ Co(s) + CO}_2\text{(g) } \) with \( K_c = 490 \).

Write the target reaction: \( \text{H}_2\text{(g) + CO}_2\text{(g) } \rightarrow \text{ CO(g) + H}_2\text{O(g) } \).

Recognize that the target reaction is the reverse of the sum of the given reactions. Therefore, reverse the first reaction and add it to the second reaction.

Calculate the equilibrium constant for the reverse of the first reaction: \( K_{c, \text{reverse}} = \frac{1}{67} \).

Use the relationship \( K_{c, \text{target}} = K_{c, \text{reverse}} \times K_{c, \text{second}} \) to find the equilibrium constant for the target reaction.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

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 extent of the reaction. Understanding this concept is crucial for manipulating and calculating equilibrium constants in various reactions.

Equilibrium Constant (Kc)

The equilibrium constant (Kc) is a dimensionless number that expresses the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their coefficients in the balanced equation. For reactions involving gases, Kc can be calculated using partial pressures. In this question, Kc values for two different reactions are provided, which can be used to derive the Kc for a third reaction through manipulation of the given equilibria.

Manipulating Equilibrium Expressions

To find the equilibrium constant for a new reaction based on known equilibria, one can manipulate the given reactions by reversing them or adding them together. When a reaction is reversed, the equilibrium constant is inverted, and when reactions are added, their equilibrium constants are multiplied. This principle allows for the calculation of Kc for complex reactions by using simpler, known equilibria.