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In which of the following reactions, the concentration of reactant is equal to concentration of product at equilibrium \((\mathrm{K}=\) equilibrium constant \()\) : (a) \(\mathrm{A} \rightleftharpoons \mathrm{B} ; \mathrm{K}=0.01\) (b) \(\mathrm{R} \rightleftharpoons \mathrm{P} ; \mathrm{K}=1\) (c) \(\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}=10\) (d) \(\mathrm{L} \rightleftharpoons \mathrm{J} ;=0.025\)

Short Answer

Expert verified
Reaction (b) has \( K = 1 \), so it's the correct answer.

Step by step solution

01

Understand Equilibrium Constant

The equilibrium constant, K, is calculated from the concentrations of products and reactants at equilibrium: \( K = \frac{[\text{Products}]}{[\text{Reactants}]} \). If \( K = 1 \), it implies that the concentration of reactants equals the concentration of products at equilibrium, as the ratio becomes \( 1 = \frac{[\text{Products}]}{[\text{Reactants}]} \).
02

Analyze Each Reaction

Review the given equilibrium constants for each reaction:- (a) K = 0.01,- (b) K = 1,- (c) K = 10,- (d) K = 0.025.Among these, only reaction (b) has \( K = 1 \), indicating that the concentration of reactant \( R \) equals the concentration of product \( P \) at equilibrium.
03

Determine the Correct Reaction

Since only \( K = 1 \) meets the condition where the concentrations of reactants and products are equal at equilibrium, reaction (b) \( \mathrm{R} \rightleftharpoons \mathrm{P} \) is the only one satisfying this criteria.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
In chemical equilibrium, the equilibrium constant, denoted by \( K \), is a crucial value that helps describe the balance between reactants and products in a reversible chemical reaction. Understanding the concept of \( K \) is essential since it outlines whether the reactants or products are favored in a reaction at equilibrium.
The equilibrium constant is calculated using the following formula:
  • \( K = \frac{[\text{Products}]}{[\text{Reactants}]} \)
This equation shows that \( K \) is essentially the ratio of the concentration of products to the concentration of reactants when the reaction reaches equilibrium. Critical to understanding this is the fact that if \( K = 1 \), it indicates a perfect balance, meaning the concentration of reactants is equal to the concentration of products. This scenario is essential because it implies neither side is favored, and the system is truly at equilibrium.
Conversely, if \( K \) is less than 1, reactants are favored, meaning there are more reactants than products at equilibrium. If \( K \) is greater than 1, products are favored, reflecting a higher concentration of products compared to reactants.
Overall, the equilibrium constant helps chemists predict the direction and extent of a chemical reaction under given conditions.
Reactant Concentration
The concentration of reactants in a chemical reaction plays a significant role in determining the dynamics of equilibrium. Reactant concentration, represented by brackets in chemical equations (e.g., \([\text{Reactant}]\)), is the amount of substance present in a given volume of solution.
In the context of equilibrium, the concentration of reactants is critical because it tops the denominator in the equilibrium constant expression:
  • \( K = \frac{[\text{Products}]}{[\text{Reactants}]} \)
When a chemical reaction reaches equilibrium, the concentrations of reactants and products remain constant, even though they are involved in ongoing forward and reverse reactions. If the equilibrium constant \( K \) is equal to 1, it reflects a unique state where the concentration of reactants is equal to that of the products. This balance leads to no observable change in concentrations over time.
A deeper understanding of reactant concentration allows chemists to manipulate conditions such as temperature, pressure, or concentration to achieve desired reaction outcomes in laboratory or industrial settings.
Product Concentration
Product concentration, much like reactant concentration, is foundational in understanding chemical equilibrium. The concentration of products is represented using brackets (e.g., \([\text{Product}]\)) and describes the amount of product present in a solution at equilibrium.
In the equation for the equilibrium constant:
  • \( K = \frac{[\text{Products}]}{[\text{Reactants}]} \)
Product concentration is the numerator, influencing the value of \( K \). A larger product concentration means a higher \( K \) value, indicating that products are favored in the equilibrium mixture, whereas a lower product concentration would suggest reactants are more prevalent.
Just like with reactants, when \( K = 1 \), it indicates an ideal balance between product and reactant concentrations. This equilibrium state highlights the importance of product concentration in predicting and observing the results of chemical reactions.
Ultimately, by understanding how product concentration fits into the equilibrium constant concept, chemists can better control reaction outcomes and conditions.

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Most popular questions from this chapter

The equilibrium constants \(\mathrm{K}_{\mathrm{P}_{1}}\) and \(\mathrm{K}_{\mathrm{P}_{2}}\) for the reactions \(\mathrm{X}\) \(\rightleftharpoons 2 \mathrm{Y}\) and \(\mathrm{Z} \rightleftharpoons \mathrm{P}+\mathrm{Q}\), respectively are in the ratio of \(1: 9\). If the degree of dissociation of \(X\) and \(Z\) be equal then the ratio of total pressure at these equilibria is: (a) \(1: 36\) (b) \(1: 1\) (c) \(1: 3\) (d) \(1: 9\)

In which of the following gaseous reaction, \(\mathrm{K}_{\mathrm{p}}\) and \(\mathrm{K}_{\mathrm{c}}\) have the same values: (a) \(2 \mathrm{Hl} \rightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}\) (b) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) (c) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) (d) \(\mathrm{PCI}_{5} \rightleftharpoons \mathrm{PCI}_{3}+\mathrm{Cl}_{2}^{3}\)

\(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) in the above reaction \(K_{p}\) and \(K_{c}\) are related as: (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})^{-1}\) (c) \(K_{c}=K_{p} \times(R T)^{2}\) (d) \(K_{p}=K_{c} \times(R T)^{-2}\)

For the reaction \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) \(\mathrm{g}\) \(\mathrm{g}\) If we start with 2 mol. \(\mathrm{SO}_{2}\) and \(1 \mathrm{~mol} . \mathrm{O}_{2}\) in \(1 \mathrm{~L}\) flask, the mixture needs \(0.4 \mathrm{~mol} \mathrm{MnO}_{4}^{-}\) in acidic medium for the complete oxidation of \(\mathrm{SO}_{2}\). The value of \(\mathrm{K}_{\mathrm{c}}\) is: (a) \(1 / 2\) (b) 2 (c) 1 (d) \(0.6\)

One mole of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) at 300 is kept in a closed container under one atmosphere. It is heated to 600 when \(20 \%\) by mass of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) decomposes to \(\mathrm{NO}_{2}\) (g). The resultant pressure is: (a) \(1.2 \mathrm{~atm}\) (b) \(2.4 \mathrm{~atm}\) (c) \(2.0 \mathrm{~atm}\) (d) \(1.0 \mathrm{~atm}\)

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