91Ó°ÊÓ

Consider the hypothetical reaction \(\mathrm{A}(g)+2 \mathrm{~B}(g) \rightleftharpoons\) \(2 \mathrm{C}(g),\) for which \(K_{c}=0.25\) at a certain temperature. \(\mathrm{A}\) 1.00-L reaction vessel is loaded with \(1.00 \mathrm{~mol}\) of compound C, which is allowed to reach equilibrium. Let the variable \(x\) represent the number of mol/L of compound A present at equilibrium. (a) In terms of \(x\), what are the equilibrium concentrations of compounds \(\mathrm{B}\) and \(\mathrm{C}\) ? (b) What limits must be placed on the value of \(x\) so that all concentrations are positive? (c) By putting the equilibrium concentrations (in terms of \(x\) ) into the equilibriumconstant expression, derive an equation that can be solved for \(x .(\mathbf{d})\) The equation from part \((\mathrm{c})\) is a cubic equation (one that has the form \(\left.a x^{3}+b x^{2}+c x+d=0\right)\). In general, cubic equations cannot be solved in closed form. However, you can estimate the solution by plotting the cubic equation in the allowed range of \(x\) that you specified in part (b). The point at which the cubic equation crosses the \(x\) -axis is the solution. \((\mathbf{e})\) From the plot in part (d), estimate the equilibrium concentrations of \(\mathrm{A}, \mathrm{B},\) and C. (Hint: You can check the accuracy of your answer by substituting these concentrations into the equilibrium expression.)

Step by step solution

01

Initial Setup and Equilibrium Concentrations in Terms of x

Given the balanced chemical equation: \( \mathrm{A} + 2 \mathrm{~B} \rightleftharpoons 2 \mathrm{C} \), and initial moles of C as 1.00 mol in 1.00 L, the initial concentration of \( \mathrm{C} \) is 1 M.Define \( x \) as the concentration of \( \mathrm{A} \) at equilibrium.The changes in concentration will be:\[ \mathrm{A}: 0 + x = x \\mathrm{~B}: 0 + 2x = 2x \\mathrm{~C}: 1.00 - 2x \]

02

Identify Equilibrium Concentrations of B and C

Using the results from Step 1, determine the equilibrium concentrations:- The equilibrium concentration of \( \mathrm{B} \) is \(2x\).- The equilibrium concentration of \( \mathrm{C} \) is \(1.00 - 2x\).

03

Determine Limits for Positive Concentrations

To ensure all concentrations remain positive at equilibrium, set the inequalities:\[ x > 0 \] (A must be positive)\[ 2x > 0 \] (B must be positive)\[ 1.00 - 2x > 0 \] (C must be positive)Solving \( 1.00 - 2x > 0 \) gives \( x < 0.50 \).Thus, \( 0 < x < 0.50 \).

04

Formulate Equilibrium Expression in Terms of x

The equilibrium constant expression is:\[ K_c = \frac{[\mathrm{C}]^2}{[\mathrm{A}][\mathrm{B}]^2} \]Substitute equilibrium concentrations:\[ 0.25 = \frac{(1.00 - 2x)^2}{x(2x)^2} \]This equation simplifies to:\[ 0.25 = \frac{(1.00 - 2x)^2}{8x^3} \]

05

Derive and Simplify the Cubic Equation

Multiply through by \(8x^3\) to eliminate the fraction:\[ 0.25 \, \times \, 8x^3 = (1.00 - 2x)^2 \]Simplify to obtain:\[ 2x^3 = (1.00 - 2x)^2 \]Further expansion gives:\[ 2x^3 = 1.00 - 4x + 4x^2 \]Rearranging terms we have:\[ 2x^3 - 4x^2 + 4x - 1.00 = 0 \] This is the cubic equation to solve.

06

Estimate Solution Range and Plot

The cubic function \( 2x^3 - 4x^2 + 4x - 1.00 = 0 \) should be plotted within the range determined: \( 0 < x < 0.50 \).The intersection points of the plot with the \( x \)-axis will approximate the solutions.

07

Conclusion from Plot and Equilibrium Concentrations

Estimate \( x \approx 0.20 \) from the plot.Thus, equilibrium concentrations are:- \( [\mathrm{A}] = 0.20 \)- \( [\mathrm{B}] = 2 \times 0.20 = 0.40 \)- \( [\mathrm{C}] = 1.00 - 2 \times 0.20 = 0.60 \)Check these values satisfy the original equilibrium expression: \( \frac{(0.60)^2}{(0.20)(0.40)^2} = 0.25 \), which holds true.

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Key Concepts

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

Equilibrium Constant

The equilibrium constant, often denoted as \( K_c \), is a vital concept in the study of chemical equilibrium. It provides a quantitative measure that expresses the ratio of the concentrations of the products to the reactants at equilibrium, raised to the power of their stoichiometric coefficients. This ratio is crucial in predicting the direction and extent of a chemical reaction.

For the reaction \( ext{A} + 2 ext{~B} \rightleftharpoons 2 ext{C} \), the equilibrium constant expression is written as:

• \( K_c = \frac{[ ext{C}]^2}{[ ext{A}][ ext{B}]^2} \)

Understanding \( K_c \) helps determine whether the products or reactants are favored at equilibrium. If \( K_c \) is much larger than 1, products are favored. If \( K_c \) is much smaller than 1, reactants are favored. In our example, \( K_c = 0.25 \) suggests that, at equilibrium, reactants are more prevalent than products.

Reaction Stoichiometry

Reaction stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. This relationship is defined by the coefficients of the balanced chemical equation.

For our example, the balanced equation \( ext{A} + 2 ext{~B} \rightleftharpoons 2 ext{C} \) indicates that 1 mole of \( ext{A} \) reacts with 2 moles of \( ext{B} \) to form 2 moles of \( ext{C} \). Thus, stoichiometry aids in predicting how changes in the amount of one substance affect another:

• If \( x \) moles of \( ext{A} \) are formed, \( 2x \) moles of \( ext{B} \) are likewise formed, while \( (1.00 - 2x) \) moles of \( ext{C} \) are consumed.

This relationship is crucial in concentration calculations and determines the feasibility of a reaction under given conditions.

Cubic Equations

Cubic equations are algebraic expressions of the form \( ax^3 + bx^2 + cx + d = 0 \). These equations arise naturally in equilibrium problems, especially when deriving expressions involving concentrations. In our case, the equilibrium constant expression leads to:

• \( 2x^3 - 4x^2 + 4x - 1.00 = 0 \)

Solving cubic equations typically requires numerical methods or graphing, as closed-form solutions are complex and not always feasible.

In practice, estimating the solution involves plotting the equation and identifying where it crosses the x-axis. This graphing approach helps approximate solutions, particularly in specific bounded ranges, like the determined \( 0 < x < 0.50 \) for this exercise.

Concentration Calculations

Concentration calculations are central to studying chemical equilibria. These involve determining the amount of each species at equilibrium by considering initial amounts, changes due to reaction progress, and the extent of reaction.

Initially, the reaction setup involves 1.00 mol of \( ext{C} \) in a 1.00-L container, equating to a concentration of 1.00 M for \( ext{C} \). The change in concentration when equilibrium is established can be detailed as:

• \([\text{A}] = x\)
• \([\text{B}] = 2x\)
• \([\text{C}] = 1.00 - 2x\)

To ensure all concentrations remain positive, certain constraints (

• \(0 < x < 0.50\)

) were applied. This logical approach ensures metals remain feasible and highlights the relevance of careful calculations in chemical equilibrium analysis.