Question

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. A \(1.00-\mathrm{L}\) reaction vessel is loaded with \(1.00 \mathrm{~mol}\) of compound \(\mathrm{C}\), which is allowed to reach equilibrium. Let the variable \(x\) represent the number of \(\mathrm{mol} / \mathrm{L}\) 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 equilibrium-constant expression, derive an equation that can be solved for \(x\). (d) The equation from part (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. (e) From the plot in part (d), estimate the equilibrium concentrations of \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\).

Short answer

The equilibrium concentrations of compounds A, B, and C can be represented as A: \(x\), B: \(2(1-x)\), and C: \(1+2x\). The condition for positive concentrations is \(0 < x < 1\). The equation to be solved for x is given by \(0.25 = \frac{(1+2x)^2}{x(2-2x)^2}\), which simplifies to a cubic equation: \(x^3 - 3x^2 + 2x - 0.25 = 0\). By plotting this equation, we estimate the value of x to be approximately 0.17. Hence, the equilibrium concentrations of A, B, and C are approximately 0.17 mol/L, 1.66 mol/L, and 1.34 mol/L, respectively.

Step-by-step solution

(a) Equilibrium concentrations of A, B, and C

Since the initial volume of the reaction vessel is 1 L, we can treat moles as concentrations throughout the problem. Let the initial number of moles of A, B, and C be A_initial, B_initial, and C_initial, respectively. The given reaction is: \(\mathrm{A}(g)+2 \mathrm{~B}(g) \rightleftharpoons 2 \mathrm{C}(g)\). Initially, we have 1 mol of C, and we can assume no A and B are present. So, A_initial = 0 mol, B_initial = 0 mol, and C_initial = 1 mol. Let x be the moles of A formed at equilibrium. Since we follow stoichiometry, for every mole of A generated, we consume 2 moles of B and generate 2 moles of C. Therefore, the equilibrium concentrations of A, B, and C can be written as: - A: \(x\) - B: \(2(1-x)\) - C: \(1+2x\)

(b) Limits on the value of x

For all concentrations to remain positive, \(x > 0\), because the concentration of A cannot be negative. Similarly, the concentration of B must also be positive so, \(1 - x > 0\), which implies \(x < 1\). Hence, \(0 < x < 1\).

(c) Derive the equation in terms of x

The equilibrium constant, \(K_c\), is given by the ratio of the product of concentrations of the products to the product of concentrations of the reactants. For the given reaction, \(K_c = \frac{[C]^2}{[A][B]^2} = 0.25\) Now, substitute the expressions for equilibrium concentrations from step (a) into the equation: \(0.25 = \frac{(1+2x)^2}{x(2-2x)^2}\) We need to solve this equation for x.

(d) Cubic equation and plot

We can simplify the equation derived in step (c) to get a cubic equation: \(x^3 - 3x^2 + 2x - 0.25 = 0\) The allowed range of x is \(0 < x < 1\). We can plot this cubic equation within this range to find the value of x where the equation crosses the x-axis.

(e) Estimate equilibrium concentrations

By plotting the cubic equation in the allowed range of x, the point at which it crosses the x-axis is found to be approximately \(x ≈ 0.17\). Now, we can substitute this value of x into the expressions for equilibrium concentrations from step (a) to estimate them: A: \(x \approx 0.17 \ \mathrm{mol/L}\) B: \(2(1 - 0.17) \approx 1.66 \ \mathrm{mol/L}\) C: \(1 + 2(0.17) \approx 1.34 \ \mathrm{mol/L}\) Hence, the equilibrium concentrations of A, B, and C are approximately 0.17 mol/L, 1.66 mol/L, and 1.34 mol/L, respectively.

Key concepts

Equilibrium Constant

In a chemical reaction, the equilibrium constant (\( K_c \)) is a crucial concept that helps predict the proportions of reactants and products at equilibrium. It is a numerical value that represents the ratio of the products' concentrations to the reactants' concentrations, each raised to the power of their respective coefficients in the balanced chemical equation. Essentially, \( K_c \) gives us insight into the position of equilibrium—whether the products or reactants are favored when the reaction is at equilibrium.
- For reactions favoring products, \( K_c \) is greater than 1.
- For reactions favoring reactants, \( K_c \) is less than 1.
In our exercise, the equilibrium constant \( K_c \) is given as 0.25, indicating that at equilibrium, the reactants are more prevalent than the products.

Reaction Stoichiometry

Reaction stoichiometry involves using the balanced chemical equation to determine the quantitative relationship between the amounts of reactants and products involved in a chemical reaction. In the equation \(A(g) + 2B(g) \rightleftharpoons 2C(g),\) stoichiometry tells us how many moles of each substance participate in the reaction.
- 1 mole of A reacts with 2 moles of B to produce 2 moles of C.
- The shift of \( x \) moles in the concentration of A at equilibrium affects the other compounds according to their stoichiometric ratios.
This principle aids in determining how the concentrations of other substances change as \( x \) changes, allowing us to derive their equilibrium concentrations accurately.

Cubic Equation

A cubic equation is a polynomial equation of degree three, typically in the form \( ax^3 + bx^2 + cx + d = 0 \). In chemical equilibrium problems, cubic equations can often arise when substituting equilibrium concentrations into the expression for the equilibrium constant. These equations are more complex than linear or quadratic equations because they can have up to three real roots.
In our specific example, after substituting equilibrium concentrations into the expression for \( K_c \), we obtain a cubic equation:\[x^3 - 3x^2 + 2x - 0.25 = 0\]
Solving cubic equations usually requires numerical methods or plotting, as they typically cannot be solved algebraically like quadratic equations.

Equilibrium Concentrations

Equilibrium concentrations refer to the amounts of reactants and products present at equilibrium. These concentrations are determined by considering both the initial conditions and how they change as the system reaches equilibrium.
In our problem, defining \( x \) as the equilibrium concentration of A helps derive expressions for B and C:

• A: \( x \) mol/L
• B: \( 2(1-x) \) mol/L
• C: \( 1+2x \) mol/L

These expressions incorporate reaction stoichiometry and are used to solve for \( x \) using the cubic equation. Once \( x \) is found, the concentrations can be accurately calculated. Plotting the cubic equation as mentioned helps pinpoint the exact value of \( x \), reaffirming these calculations with practical accuracy.