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 \(\mathrm{C}\).

Step by step solution

01

(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: \(1+2x\)

02

(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, \(0 < x < 1\).

03

(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.

04

(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.

05

(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.

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