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 -L reaction vessel is loaded with 1.00 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 equilibrium-constant expression, derive an equation that can be solved for \(x .(\mathbf{d})\) The equation from part (c) is a cubic equation (one that has the form \(a x^{3}+b x^{2}+c x+d=0 )\) . 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 \(A, B,\) 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 conditions

Before the reaction reaches equilibrium, we have 1.00 mol of compound C in the 1 L reaction vessel. Since there was initially no A or B, we can assume that the mol/L of A at equilibrium will be x. Then, as the stoichiometry shows, the mols of B consumed per mole of A formed are 2 times the mols of A. Therefore, we have 2x equilibrium mol concentration of B. Now, for every mol of A formed, 2 mols of C are consumed, so we have (1.00 - 2x) mol of C left at equilibrium. #b. Ensuring all concentrations are positive#

02

Positive concentrations

In order to ensure that all concentrations are positive, the values of x should be in the range: 0 < x 0 < 2x 0 < (1.00 - 2x) From the last inequality, we get that x < 0.5. #c. Deriving the equation to solve for x#

03

Equilibrium constant expression

We can write the equilibrium constant expression for this reaction as follows: \(K_c = \frac{[C]^2}{[A][B]^2}\) Now substitute the equilibrium concentrations in terms of x into this expression: \(0.25=\frac{(1.00 - 2x)^2}{(x)(2x)^2}\) #d. Plotting the cubic equation#

04

Cubic equation

The equation derived above is a cubic equation in x, which cannot be solved in closed form. We can rewrite this equation as: \(a x^3 + b x^2 + c x + d = 0\) where a = 16, b = -64, c = 64, and d = -16, and plot this equation in the range 0 < x < 0.5. #e. Estimating the equilibrium concentrations#

05

Equilibrium concentrations

From the plot of the cubic equation, we can estimate where it intersects the x-axis. This estimate will give us the value of x at equilibrium. By substituting this value of x back into the expressions for the equilibrium concentrations of A, B, and C, we can find the concentrations of each species at equilibrium. To check our answer, we can plug these concentrations back into the equilibrium constant expression to make sure it equals the given value of 0.25.

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