Problem 18
Question
Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? (a) \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) at \(1300 \mathrm{~K} K_{c}=0.57\) (b) \(2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\) at \(900 \mathrm{~K} K_{p}=0.0572\)
Step-by-Step Solution
Verified Answer
For reaction (a) with \(K_c=0.57\) at \(1300 \mathrm{~K}\), and reaction (b) with \(K_p=0.0572\) at \(900 \mathrm{~K}\), both equilibrium constants are less than \(1\), indicating that both reactions favor the formation of reactants and lie to the left.
1Step 1: Reaction (a) analysis
We are given the reaction: \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) at \(1300 \mathrm{~K}\) with \(K_c=0.57\). Since the given equilibrium constant, \(K_c\), is less than \(1\), the reaction favors the formation of reactants. Therefore, the reaction lies to the left.
2Step 2: Reaction (b) analysis
We are given the reaction: \(2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\) at \(900 \mathrm{~K}\) with \(K_p=0.0572\). Since the given equilibrium constant, \(K_p\), is less than \(1\), the reaction favors the formation of reactants. Therefore, the reaction lies to the left.
Based on the analysis of the given equilibrium constants, both reactions (a) and (b) lie to the left, favoring formation of reactants.
Key Concepts
Equilibrium ConstantReaction DirectionGaseous Reactions
Equilibrium Constant
In chemistry, the equilibrium constant is a numerical value that helps us understand the position of a chemical reaction at equilibrium. Represented as either \(K_c\) or \(K_p\), depending on whether the concentration or pressure is used, this constant informs us about the relative amounts of products and reactants at equilibrium.
An equilibrium constant greater than 1 suggests that, at equilibrium, products are favored over reactants. This means the reaction lies to the right. Conversely, if the equilibrium constant is less than 1, reactants are favored, and the reaction lies to the left. This indicates more reactants are present than products when the reaction reaches equilibrium.
In the exercises given, both reactions have equilibrium constants less than 1, implying they lie to the left. This means they favor the formation of reactants over products.
An equilibrium constant greater than 1 suggests that, at equilibrium, products are favored over reactants. This means the reaction lies to the right. Conversely, if the equilibrium constant is less than 1, reactants are favored, and the reaction lies to the left. This indicates more reactants are present than products when the reaction reaches equilibrium.
In the exercises given, both reactions have equilibrium constants less than 1, implying they lie to the left. This means they favor the formation of reactants over products.
Reaction Direction
For any given reaction, knowing whether it lies to the right or left is essential in predicting the dominant species at equilibrium. The direction of a reaction is determined by its equilibrium constant:
- If \( K > 1 \), the reaction proceeds to the right, favoring product formation.
- If \( K < 1 \), the reaction proceeds to the left, favoring reactant formation.
Gaseous Reactions
Gaseous reactions involve reactants and products in the gas phase. Their behavior is often described by partial pressures, especially when using the equilibrium constant \(K_p\). The exercises given are excellent examples of gaseous reactions and demonstrate how equilibrium concepts are applied to them.
In such reactions, changes in conditions like temperature and pressure can significantly affect the position of equilibrium. For instance, increasing pressure may favor the side of the equation with fewer moles of gas. However, the equilibrium constant itself is only temperature-dependent and remains unchanged by shifts in pressure or concentration.
Gaseous reactions require a careful balance, making knowledge of equilibrium principles vital for effectively managing and predicting their outcomes. Understanding these principles allows chemists and students alike to manipulate conditions favorably for desired results.
In such reactions, changes in conditions like temperature and pressure can significantly affect the position of equilibrium. For instance, increasing pressure may favor the side of the equation with fewer moles of gas. However, the equilibrium constant itself is only temperature-dependent and remains unchanged by shifts in pressure or concentration.
Gaseous reactions require a careful balance, making knowledge of equilibrium principles vital for effectively managing and predicting their outcomes. Understanding these principles allows chemists and students alike to manipulate conditions favorably for desired results.
Other exercises in this chapter
Problem 15
Write the expression for \(K_{c}\) for the following reactions. \(\operatorname{In}\) each case indicate whether the reaction is homogeneous or heterogeneous. (
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Which of the following statements are true and which are false? (a) The equilibrium constant can never be a negative number. (b) In reactions that we draw with
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Which of the following statements are true and which are false? (a) For the reaction \(2 \mathrm{~A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{A}_{2} \mathrm{
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