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.
1Step 1: Understand the Equilibrium Constant
The equilibrium constant, denoted as \( K_{eq} \), for a reversible reaction at a given temperature, is a ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. It is given by the expression: \[K_{eq} = \frac{{[C]^c [D]^d}}{{[A]^a [B]^b}}\]where \([A]\), \([B]\), \([C]\), and \([D]\) are the molar concentrations of the reactants and products, and \(a\), \(b\), \(c\), and \(d\) are their respective stoichiometric coefficients.
2Step 2: Note the effect of changing reactant concentrations
When concentrations of the reactants are changed, the reaction will shift according to Le Chatelier's principle to re-establish equilibrium, but the equilibrium constant itself remains unchanged because it is only affected by temperature, not by the concentrations of reactants or products.
3Step 3: Answer the question based on the given understanding
Since the equilibrium constant \( K_{eq} \) is unaffected by changing concentrations at constant temperature, it remains the same regardless of whether the reactants' concentrations are doubled or reduced. The correct choice that reflects this is (d) the same.
Key Concepts
Reversible ReactionsLe Chatelier's PrincipleEffect of Concentration Changes on Equilibrium
Reversible Reactions
Reversible reactions are chemical reactions where the conversion of reactants to products and the conversion of products back to reactants occur simultaneously. In other words, these reactions can proceed in both forward and backward directions. For example, consider the reaction:
- Reactants \( A \) and \( B \) can combine to form products \( C \) and \( D \).
- Similarly, products \( C \) and \( D \) can react to regenerate reactants \( A \) and \( B \).
Le Chatelier's Principle
Le Chatelier's principle is a crucial concept in chemistry that predicts how a system at equilibrium will respond to changes in conditions such as concentration, temperature, or pressure. It states that if an external change is applied to a system at equilibrium, the system adjusts itself to counteract the effect of the change and a new equilibrium is established.
This principle can be summarized as follows:
- If the concentration of reactants or products is changed, the system shifts the equilibrium position to either consume or produce more of the substance whose concentration was altered.
- For temperature changes, increasing temperature will favor the endothermic direction, while decreasing temperature will favor the exothermic direction.
- Pressure changes affect only gaseous reactions where a change in the number of moles of gas occurs.
Effect of Concentration Changes on Equilibrium
Changes in the concentration of reactants or products do not alter the equilibrium constant; instead, they cause the equilibrium position to shift. This is an application of Le Chatelier's principle. Let's say the concentration of a reactant is increased:
- The system will temporarily experience an excess of reactants.
- This will shift the equilibrium towards the products side to reduce the excess reactants, forming more products.
- The system will adjust by shifting the equilibrium towards the product side to make up for the decreased concentration.
Other exercises in this chapter
Problem 1
The reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~s})+\mathrm{Cl}_{2}(\mathrm{~g})\) is an example of (a) backward react
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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 \(\mathrm{BaO}_{2}(\mathrm{~s}) \longrightarrow \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})\) \(\Delta \mathrm{H}=\) tve. In equilibriu
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If equilibrium constant for the reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) is \(\mathrm{K}_{e}\), then the equilibrium co
View solution