Problem 81
Question
If \(2 \mathrm{NO} \rightleftharpoons \mathrm{N}_{2}+\mathrm{O}_{2}\) \(\mathrm{K}_{\mathrm{c}}=2.5 \times 10^{30}\) \(\mathrm{NO}+\frac{1}{2} \mathrm{Br}_{2} \rightleftharpoons \mathrm{NOBr}\) \(\mathrm{K}_{c_{2}}=1.6\) find \(K_{c}\) for the reaction given below \(\frac{1}{2} \mathrm{~N}_{8}+\frac{1}{2} \mathrm{O}_{8}+\frac{1}{2} \mathrm{Br}_{\mathrm{g}} \rightleftharpoons \mathrm{NOBr}\) (a) \(1.01 \times 10^{-15}\) (b) \(2.02 \times 10^{-15}\) (c) \(1.01 \times 10^{30}\) (d) \(2.02 \times 10^{15}\)
Step-by-Step Solution
Verified Answer
To calculate \(K_c\) for the target reaction, balance and combine reactions using given \(K_c\) values.
1Step 1: Analyze the Given Reactions
First, understand the given reactions and how they relate to the target reaction. We have two equilibrium reactions provided: 1. \(2 \text{NO} \rightleftharpoons \text{N}_{2} + \text{O}_{2}\) with \(K_c = 2.5 \times 10^{30}\).2. \(\text{NO} + \frac{1}{2} \text{Br}_2 \rightleftharpoons \text{NOBr}\) with \(K_{c_2} = 1.6\).The target reaction is \(\frac{1}{2} \text{N}_2 + \frac{1}{2} \text{O}_2 + \frac{1}{2} \text{Br}_2 \rightleftharpoons \text{NOBr}\).
Key Concepts
Equilibrium Constant (Kc)Reaction StoichiometryEquilibrium Reactions
Equilibrium Constant (Kc)
The equilibrium constant, denoted by \( K_c \), is a vital concept in chemistry that expresses the ratio of the concentrations of products to reactants at equilibrium for a particular chemical reaction. Understanding \( K_c \) helps us grasp how far a reaction proceeds before reaching a state where the forward and reverse reaction rates equalize. For a general reaction like \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K_c \) can be expressed as:\[K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]
This equation shows that \( K_c \) depends on the concentrations of the respective reactants and products raised to the power of their stoichiometric coefficients. A larger \( K_c \) value indicates a reaction that favors products at equilibrium, while a smaller value suggests reactants are more favored.
When determining \( K_c \) for a target reaction, it's crucial to understand how \( K_c \) values from related reactions might be used, especially when reactions are combined or modified.
This equation shows that \( K_c \) depends on the concentrations of the respective reactants and products raised to the power of their stoichiometric coefficients. A larger \( K_c \) value indicates a reaction that favors products at equilibrium, while a smaller value suggests reactants are more favored.
When determining \( K_c \) for a target reaction, it's crucial to understand how \( K_c \) values from related reactions might be used, especially when reactions are combined or modified.
Reaction Stoichiometry
Reaction stoichiometry involves understanding the quantitative relationship between reactants and products in a chemical equation. It is essential for calculating equilibrium constants and manipulating chemical reactions to obtain the desired \( K_c \).
Given reactions can often be rearranged or multiplied by factors to assist in finding the \( K_c \) for a different reaction. For example, if a reaction is halved, the resulting \( K_c \) needs to be taken to the power of the factor by which the reaction is changed. In the context of the exercise, adjusting the given reactions to synthesize the target reaction is where stoichiometry becomes crucial.
- Stoichiometry lays the foundation for your calculations by ensuring that all elements are balanced according to their coefficients in the chemical equation.
- Any changes to coefficients, concentration, or the reaction equation can alter the equilibrium constant and must be taken into account.
Given reactions can often be rearranged or multiplied by factors to assist in finding the \( K_c \) for a different reaction. For example, if a reaction is halved, the resulting \( K_c \) needs to be taken to the power of the factor by which the reaction is changed. In the context of the exercise, adjusting the given reactions to synthesize the target reaction is where stoichiometry becomes crucial.
Equilibrium Reactions
Equilibrium reactions are those in which the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time, not because the reactions cease, but because they occur at the same rate.
Recognizing the interconnection of individual equilibria and how their collective interaction leads to an overall equilibrium aids in solving complex reaction problems as seen in the exercise.
- The idea of dynamic equilibrium is fundamental. It implies motion and transformation, although at a macroscopic scale, composition doesn't appear to change.
- In the given exercise, understanding the individual equilibrium reactions is key to determining how they collectively form the target reaction and what the resulting \( K_c \) would be.
- Equilibrium reactions further highlight the dynamic nature of chemical processes, where adjustments to concentration, temperature, or pressure can shift the equilibrium position as described by Le Chatelier's Principle.
Recognizing the interconnection of individual equilibria and how their collective interaction leads to an overall equilibrium aids in solving complex reaction problems as seen in the exercise.
Other exercises in this chapter
Problem 77
In which of the following gaseous reaction, the value of \(K_{p}\) is less than \(K_{c}\) ? (a) \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCI}_{3}+\mathrm{C
View solution Problem 80
The ratio of \(\mathrm{Kp} / \mathrm{Kc}\) for the reaction \(\mathrm{SO}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(
View solution Problem 84
\(9.2\) grams of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) is taken in a closed one litre vessel and heated till the following equilibrium is reached \(\ma
View solution Problem 85
Equilibrium constant for the reaction \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}
View solution