Problem 2
Question
For a reversible reaction, if the concentrations of the reactants are doubled, at constant temperature the equilibrium constant will be (a) one-fourth (b) halved (c) doubled (d) the same
Step-by-Step Solution
Verified Answer
The equilibrium constant remains the same when reactant concentrations are doubled; the answer is (d) the same.
1Step 1: Understand the Equilibrium Constant
The equilibrium constant \( K \) for a reversible reaction is a ratio of the concentration of products to reactants, each raised to the power of their respective stoichiometric coefficients in the balanced chemical equation. This constant depends only on temperature for a given reaction.
2Step 2: Recall the Impact of Concentration Changes
The equilibrium constant itself, \( K \), does not change with the concentration of reactants or products. Changes in concentrations affect the position of equilibrium (i.e., whether the reaction shifts to the right or left), but not the value of \( K \).
3Step 3: Analyze the Given Scenario
In this problem, the concentration of reactants is doubled. However, since the equilibrium constant is solely dependent on temperature and not on the concentration, doubling the reactant concentration will not affect the value of \( K \).
4Step 4: Conclusion
Since the equilibrium constant \( K \) remains the same with changes in concentration at constant temperature, the correct answer is (d) the same.
Key Concepts
Reversible ReactionsImpact of Concentration ChangesTemperature Dependence of Equilibrium Constant
Reversible Reactions
In chemistry, reversible reactions are those that can proceed in both forward and reverse directions. This typically occurs when the reactants form products, and those products can re-form reactants under the same conditions. An equilibrium state is reached when the rate of the forward reaction equals the rate of the reverse reaction.
At equilibrium, the concentration of reactants and products remains constant, although they are not necessarily equal. This balance can be described by the equilibrium constant, represented as \( K \). In a reversible reaction, \( K \) is calculated using the concentrations of the products and reactants, each raised to the power of their coefficients from the balanced equation.
Understanding reversible reactions can illuminate why changes in certain conditions affect chemical systems. They highlight the dynamic nature of chemical equilibrium, where reactions are always ongoing, even if their net change appears static.
At equilibrium, the concentration of reactants and products remains constant, although they are not necessarily equal. This balance can be described by the equilibrium constant, represented as \( K \). In a reversible reaction, \( K \) is calculated using the concentrations of the products and reactants, each raised to the power of their coefficients from the balanced equation.
Understanding reversible reactions can illuminate why changes in certain conditions affect chemical systems. They highlight the dynamic nature of chemical equilibrium, where reactions are always ongoing, even if their net change appears static.
Impact of Concentration Changes
While considering concentration changes in reversible reactions, it's essential to note that these changes do not impact the equilibrium constant \( K \). According to Le Chatelier’s Principle, adding or removing reactants or products shifts the position of equilibrium to counteract the change.
If concentration of reactants in a reversible reaction is increased, the system will shift towards the products to re-establish equilibrium. Similarly, if product concentration increases, the system may shift towards the reactants. These shifts change the reaction quotient but leave \( K \) unchanged.
If concentration of reactants in a reversible reaction is increased, the system will shift towards the products to re-establish equilibrium. Similarly, if product concentration increases, the system may shift towards the reactants. These shifts change the reaction quotient but leave \( K \) unchanged.
- Adding more reactants typically shifts the equilibrium to the right, favoring products.
- Adding more products shifts the equilibrium to the left, favoring reactants.
- The equilibrium constant \( K \) remains constant unless the temperature changes.
Temperature Dependence of Equilibrium Constant
The equilibrium constant \( K \) for a reaction is inherently temperature-dependent. Unlike concentration changes, variations in temperature can alter the value of \( K \). This occurs because temperature changes affect the kinetic energy of molecules, potentially altering reaction rates.
An increase in temperature typically favors the endothermic reaction direction, increasing \( K \) for endothermic reactions. Conversely, a decrease in temperature favors exothermic reactions, potentially decreasing \( K \).
An increase in temperature typically favors the endothermic reaction direction, increasing \( K \) for endothermic reactions. Conversely, a decrease in temperature favors exothermic reactions, potentially decreasing \( K \).
- For endothermic reactions, raising the temperature increases \( K \).
- For exothermic reactions, raising the temperature decreases \( K \).
Other exercises in this chapter
Problem 1
The reaction \(\mathrm{PCl}_{5}(\mathrm{~g})=\mathrm{PCl}_{3}(\mathrm{~s})+\mathrm{Cl}_{2}(\mathrm{~g})\) is an example of (a) backward reaction (b) forward rea
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The value of \(K_{p}\) in the reaction \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) is (a) \(\mat
View solution Problem 4
In the reaction \(\mathrm{BaO}_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})\) \(\Delta \mathrm{H}=\) tve. In equilib
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If equilibrium constant for the reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) is \(\mathrm{K}_{\mathrm{c}}\), then the equil
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