Problem 206
Question
The equilibrium constant at \(298 \mathrm{~K}\) for a reaction \(\mathrm{A}+\) \(\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\) is \(100 .\) If the initial concentration of all the four species were \(1 \mathrm{M}\) each, then equilibrium concentration of \(\mathrm{D}\left(\right.\) in \(\left.\mathrm{mol} \mathrm{L}^{-1}\right)\) will be: \(\quad[\mathbf{2 0 1 6}]\)(a) \(0.818\) (b) \(1.818\) (c) \(1.182\) (d) \(0.182\)
Step-by-Step Solution
Verified Answer
The equilibrium concentration of \( \mathrm{D} \) is \(1.818 \text{ M}\).
1Step 1: Write the expression for the equilibrium constant
For the reaction \( \mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D} \), the equilibrium constant \( K_c \) is given by:\[ K_c = \frac{[C][D]}{[A][B]}\]Given that \( K_c = 100 \).
2Step 2: Set up the equilibrium concentration changes
Assuming that at equilibrium the change in the concentration of \( \mathrm{A} \) and \( \mathrm{B} \) is \( x \), then the concentration of \( \mathrm{C} \) and \( \mathrm{D} \) will also change by \( +x \).Initial concentrations: \([A]_0 = [B]_0 = [C]_0 = [D]_0 = 1 \text{ M}\).At equilibrium:\[[A] = [B] = 1 - x, \quad [C] = [D] = 1 + x\]
3Step 3: Substitute concentrations into the equilibrium constant expression
Substitute the equilibrium concentrations into the equilibrium expression:\[ K_c = \frac{(1+x)^2}{(1-x)^2} = 100\]
4Step 4: Solve for x
Start by taking the square root of both sides:\[ \sqrt{\frac{(1+x)^2}{(1-x)^2}} = \sqrt{100} \\frac{1+x}{1-x} = 10\]Cross-multiply to solve for \( x \):\[1+x = 10(1-x) \ 1+x = 10 - 10x \ 11x = 9 \ x = \frac{9}{11} = 0.818\]
5Step 5: Find the equilibrium concentration of D
The equilibrium concentration of \( D \) is given by \( [D] = 1 + x \).Substituting the value of \( x \):\[[D] = 1 + 0.818 = 1.818 \text{ M}\]
Key Concepts
Chemical EquilibriumEquilibrium ConcentrationsEquilibrium Constant Expression
Chemical Equilibrium
In chemistry, chemical equilibrium refers to a state in which the concentrations of reactants and products remain constant over time. This doesn't mean that the reactant and product concentrations are equal, rather that their rates of formation and consumption are balanced. At this point, the reaction proceeds in both forward and reverse directions at the same rate.
When a reaction reaches equilibrium, no further changes in concentration occur, and the system's macroscopic properties, such as temperature, pressure, and concentrations, have stabilized. Understanding chemical equilibrium is crucial because it allows for the prediction and manipulation of reactions, especially those that are reversible.
When a reaction reaches equilibrium, no further changes in concentration occur, and the system's macroscopic properties, such as temperature, pressure, and concentrations, have stabilized. Understanding chemical equilibrium is crucial because it allows for the prediction and manipulation of reactions, especially those that are reversible.
- The reaction can shift back and forth between reactants and products, responding to changes in conditions, such as concentration, temperature, or pressure.
- Equilibrium does not mean that the reactants and products are in equal concentrations; it means that their formation rates are equal.
- Factors such as temperature and pressure can affect the position of equilibrium, according to Le Chatelier's principle.
Equilibrium Concentrations
The concentrations of reactants and products at the equilibrium state are termed equilibrium concentrations. These values are essential for evaluating the extent to which a reaction has proceeded. In a chemical reaction, the equilibrium concentrations depend on various factors: initial concentrations, the equilibrium constant, and any changes made to the reaction's conditions.
At equilibrium, the concentrations of the substances involved in the reaction remain constant. To find the equilibrium concentrations, one must set up an equilibrium table (often referred to as an ICE table, for Initial, Change, Equilibrium) where the concentrations are calculated systematically.
At equilibrium, the concentrations of the substances involved in the reaction remain constant. To find the equilibrium concentrations, one must set up an equilibrium table (often referred to as an ICE table, for Initial, Change, Equilibrium) where the concentrations are calculated systematically.
- The initial concentrations are noted as the starting point.
- Changes in concentrations are represented by variables, often with an algebraic expression describing their relationship.
- Equilibrium concentrations are then determined by solving these equations, often using the equilibrium constant.
Equilibrium Constant Expression
The equilibrium constant, denoted as \( K_c \), is a numerical value that characterizes the position of equilibrium for a reversible reaction at a specific temperature. It is a fundamental concept in chemical equilibrium, representing the ratio of the product concentrations to the reactant concentrations, each raised to the power of their stoichiometric coefficients in the balanced chemical equation.
The expression for the equilibrium constant for a reaction \( aA + bB \rightleftharpoons cC + dD \) is written as:\[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]This helps in quantifying the relationship between the reactants and products in a chemical equilibrium system.
The expression for the equilibrium constant for a reaction \( aA + bB \rightleftharpoons cC + dD \) is written as:\[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]This helps in quantifying the relationship between the reactants and products in a chemical equilibrium system.
- If \( K_c \) is large (much greater than 1), the equilibrium favors the products.
- If \( K_c \) is small (much less than 1), the equilibrium favors the reactants.
- The magnitude of \( K_c \) can predict the extent of a reaction but not the speed at which equilibrium is reached.
Other exercises in this chapter
Problem 204
For the reaction \(\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})\) if \(\mathrm{K}_{\math
View solution Problem 205
The following reaction is performed at \(298 \mathrm{~K}\). $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathr
View solution Problem 203
The equilibrium constant \((\mathrm{K})\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g}) \quad
View solution