Problem 192
Question
The equilibrium constant for the reaction, \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) At temperature \(\mathrm{T}\) is \(4 \times 10^{-4} .\) The value of \(\mathrm{K}_{c}\) for the reaction \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons 1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})\) at the same temperature is (a) \(4 \times 10^{-6}\) (b) \(2.5 \times 10^{2}\) (c) \(0.02\) (d) 50
Step-by-Step Solution
Verified Answer
The correct answer is (d) 50.
1Step 1: Identify the Given Reaction Equilibrium Constant
The equilibrium constant for the reaction \( \mathrm{N}_{2} (\mathrm{~g}) + \mathrm{O}_{2} (\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO} (\mathrm{g}) \) is given as \( K_c = 4 \times 10^{-4} \) at a certain temperature \( T \).
2Step 2: Recognize the Reverse Reaction
The problem asks for the equilibrium constant \( K_c \) for the reverse reaction: \( \mathrm{NO} (\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2} (\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2} (\mathrm{~g}) \).
3Step 3: Express Relationship between Forward and Reverse Reaction Constants
The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. Therefore, \( K'_c = \frac{1}{K_c} \), where \( K'_c \) is the equilibrium constant for the desired reverse reaction.
4Step 4: Calculate the Equilibrium Constant for the Reverse Reaction
Apply the reciprocal relationship to find \( K'_c \):\[K'_c = \frac{1}{K_c} = \frac{1}{4 \times 10^{-4}}\]
5Step 5: Simplify the Expression for \( K'_c \)
Calculate \( K'_c \) by simplifying the expression:\[K'_c = 2.5 \times 10^{3}\]
6Step 6: Adjust for Stoichiometry of the Reaction
The equation \( \mathrm{NO} (\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2} (\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2} (\mathrm{~g}) \) implies that the equilibrium constant should be adjusted for the stoichiometry by taking the square root of \( K'_c \):\[K_{c,\text{adjusted}} = \sqrt{2.5 \times 10^{3}} = 50\]
7Step 7: Confirm the Correct Answer
The correct value of \( K_c \) for the reverse reaction \( \mathrm{NO} (\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2} (\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2} (\mathrm{~g}) \) is 50, which matches option (d).
Key Concepts
Reversible ReactionsStoichiometryReaction Kinetics
Reversible Reactions
Reversible reactions are an essential concept in chemistry. These are reactions that proceed in both directions - forward and backward. This means that products can revert to reactants and vice-versa, which is what happens in the dynamic state of equilibrium. In the presented exercise, the reaction between nitrogen (\(\mathrm{N}_{2}\)) and oxygen (\(\mathrm{O}_{2}\)) forming nitrogen monoxide (NO) is a reversible one. As such, it is represented with a double arrow (\(\rightleftharpoons\)), indicating that both the forward and reverse reactions are occurring simultaneously in a closed system.
Understanding the equilibrium constant \(K_c\) helps us predict the extent of these reversible reactions. For the forward reaction \(\mathrm{N}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\), \(K_c\) is given. To find \(K_c\) for the reverse reaction (\(\mathrm{NO} \rightleftharpoons \frac{1}{2} \mathrm{N}_{2} + \frac{1}{2} \mathrm{O}_{2}\)), the reciprocal value of the forward \(K_c\) is calculated, since reversing the reaction inverts the equilibrium condition. This principle illustrates how the same reaction under the same conditions can exhibit different dynamics depending on its directionality.
Reversible reactions are significant in many real-world processes, including biochemical pathways and industrial synthesis applications, where control over the equilibrium position can optimize yields and efficiencies.
Understanding the equilibrium constant \(K_c\) helps us predict the extent of these reversible reactions. For the forward reaction \(\mathrm{N}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\), \(K_c\) is given. To find \(K_c\) for the reverse reaction (\(\mathrm{NO} \rightleftharpoons \frac{1}{2} \mathrm{N}_{2} + \frac{1}{2} \mathrm{O}_{2}\)), the reciprocal value of the forward \(K_c\) is calculated, since reversing the reaction inverts the equilibrium condition. This principle illustrates how the same reaction under the same conditions can exhibit different dynamics depending on its directionality.
Reversible reactions are significant in many real-world processes, including biochemical pathways and industrial synthesis applications, where control over the equilibrium position can optimize yields and efficiencies.
Stoichiometry
Stoichiometry involves the quantitative relationships between the reactants and products in a chemical reaction. Accurate stoichiometric calculations are crucial for understanding and manipulating reaction conditions. In our exercise, stoichiometry plays a pivotal role in adjusting the equilibrium constant of the reverse reaction.
The original reaction \(\mathrm{N}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) \) has a 1:1:2 stoichiometry ratio. When considering the reverse reaction \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_2(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g}) \), the stoichiometry is 1:0.5:0.5. In this case, because the stoichiometric coefficients change, the equilibrium constant needs an adjustment.
Simply calculating the reciprocal is not enough. It must account for the stoichiometric coefficients by taking a square root, reflecting the fact that halving each reactant for the reverse reaction also influences the concentration terms in the equilibrium law. This adjustment leads to the correct equilibrium constant for the reaction as it accounts for the fractional stoichiometry, ensuring accurate modeling of the chemical system's behavior.
Understanding stoichiometry in equilibrium is vital for chemists. It allows accurate predictions of concentrations necessary for processes like titration, pharmacology, and reaction scaling in the industry.
The original reaction \(\mathrm{N}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) \) has a 1:1:2 stoichiometry ratio. When considering the reverse reaction \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_2(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g}) \), the stoichiometry is 1:0.5:0.5. In this case, because the stoichiometric coefficients change, the equilibrium constant needs an adjustment.
Simply calculating the reciprocal is not enough. It must account for the stoichiometric coefficients by taking a square root, reflecting the fact that halving each reactant for the reverse reaction also influences the concentration terms in the equilibrium law. This adjustment leads to the correct equilibrium constant for the reaction as it accounts for the fractional stoichiometry, ensuring accurate modeling of the chemical system's behavior.
Understanding stoichiometry in equilibrium is vital for chemists. It allows accurate predictions of concentrations necessary for processes like titration, pharmacology, and reaction scaling in the industry.
Reaction Kinetics
Reaction kinetics studies how fast a reaction occurs and what factors influence this rate. Though not directly solved in the given exercise, understanding kinetics can provide a complete picture of reversible reactions.
Kinetics and equilibrium go hand-in-hand. For reversible reactions like \(\mathrm{N}_2 + \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} \), reaction kinetics would analyze the collision frequencies, energy states, and how temperature impacts both forward and reverse reaction rates. This is integral because while the equilibrium constant ( \(K_c\)) tells us about the position of equilibrium, kinetics reveals how swiftly equilibrium is achieved.
In many reversible reactions, catalysts might be involved, which can lower activation energy barriers, allowing equilibrium to be reached faster without affecting \(K_c\). Understanding kinetics enables chemists to optimize reactions and conditions for production, especially in industrial applications where speed and yield are crucial.
Kinetics also influences the consideration of reaction pathways, any temporary intermediates formed, and the potential energy profiles of the reactants and products. By studying these, one can engineer reactions to favor desired products quickly, efficiently, and safely.
Kinetics and equilibrium go hand-in-hand. For reversible reactions like \(\mathrm{N}_2 + \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} \), reaction kinetics would analyze the collision frequencies, energy states, and how temperature impacts both forward and reverse reaction rates. This is integral because while the equilibrium constant ( \(K_c\)) tells us about the position of equilibrium, kinetics reveals how swiftly equilibrium is achieved.
In many reversible reactions, catalysts might be involved, which can lower activation energy barriers, allowing equilibrium to be reached faster without affecting \(K_c\). Understanding kinetics enables chemists to optimize reactions and conditions for production, especially in industrial applications where speed and yield are crucial.
Kinetics also influences the consideration of reaction pathways, any temporary intermediates formed, and the potential energy profiles of the reactants and products. By studying these, one can engineer reactions to favor desired products quickly, efficiently, and safely.
Other exercises in this chapter
Problem 189
For the reaction equilibrium, \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})=2 \mathrm{NO}(\mathrm{g})\) the concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) a
View solution Problem 191
What is the equilibrium expression for the reaction \(\mathrm{P}_{4}(\mathrm{~s})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{P}_{4} \mathrm{O}_{10
View solution Problem 193
Consider an endothermic reaction \(\mathrm{X} \longrightarrow \mathrm{Y}\) with the activation energies \(E_{b}\) and \(E_{f}\) for the backward and forward rea
View solution Problem 194
For the reaction, \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) \(\left(\mathrm{K}_{c}=1.8 \times
View solution