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Chapter 15: Problem 38

For the following reaction, \(K_{\mathrm{c}}=2.00\) at \(1000^{\circ} \mathrm{C}\) $$2 \operatorname{COF}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{CF}_{4}(\mathrm{g})$$ If a \(5.00 \mathrm{L}\) mixture contains \(0.145 \mathrm{mol} \mathrm{COF}_{2}, 0.262 \mathrm{mol}\) \(\mathrm{CO}_{2},\) and \(0.074 \mathrm{mol} \mathrm{CF}_{4}\) at a temperature of \(1000^{\circ} \mathrm{C}\) (a) Will the mixture be at equilibrium? (b) If the gases are not at equilibrium, in what direction will a net change occur? (c) How many moles of each gas will be present at equilibrium?

Short Answer

Expert verified
a) No, the mixture is not at equilibrium. b) The reaction will shift to the right to reach equilibrium. c) The moles of each gas at equilibrium can be calculated by finding 'x', plugging it into the equilibrium concentration, and then multiplying the concentration by 5.

Step by step solution

01

Calculate the Initial Concentrations

Firstly, calculate the initial concentrations of the reactants and products. Use the formula \(Concentration = moles \, of \, substance \, / \, volume \, of \, solution\). Thus, the initial concentrations are: \[ [COF_2] = 0.145 \, mol \, / \, 5.00 \, L = 0.029 \, M\] \[ [CO_2] = 0.262 \, mol \, / \, 5.00 \, L = 0.0524 \, M\] \[ [CF_4] = 0.074 \, mol \, / \, 5.00 \, L = 0.0148 \, M\]
02

Determine the Reaction Quotient (Q) and Its Relation to Kc

Next, calculate the initial reaction quotient (Q) using the formula: \[ Q = [Products]^{coefficients} \, / \, [Reactants]^{coefficients} \] \(Q = [CO_2][CF_4] \, / \, [COF_2]^2 \) Substituting the values into the formula, we get: \(Q = (0.0524 \times 0.0148) \, / \, (0.029)^2 = 1.12\) Compare the Q value with the given Kc. If Q > Kc, the reaction will shift to the left (reactants) to reach equilibrium. If Q < Kc, the reaction will shift to the right (products) to reach equilibrium. If Q = Kc, the reaction is at equilibrium. In this case, we have Q > Kc (1.12 > 2.00), indicating the reaction will shift to the right, toward the products. Hence, the reaction is not currently at equilibrium.
03

Set Up and Solve the Equilibrium Expression

Finally, set up and solve the equilibrium expression for the reaction: \[ Kc = [CO_2][CF_4] \, / \, [COF_2]^2 = 2.00 \] Since the reaction shifts right, consider a change in concentrations as \( x\). Then, the concentrations at equilibrium will be: \( [COF_2] = 0.029 - 2x \) \( [CO_2] = 0.0524 + x \) \( [CF_4] = 0.0148 + x \) Substitute these values back into the Kc expression. Solve for x (numerically if needed). Once x is determined, find the equilibrium concentrations by substitution. The solutions will provide the concentration, and for the moles, multiply the concentration by 5 (since the mixture volume is 5L).

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Most popular questions from this chapter

For the dissociation reaction \(2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{g})+\) \(\mathrm{S}_{2}(\mathrm{g}), K_{\mathrm{p}}=1.2 \times 10^{-2}\) at \(1065^{\circ} \mathrm{C} .\) For this same reaction at \(1000 \mathrm{K},\) (a) \(K_{\mathrm{c}}\) is less than \(K_{\mathrm{p}} ;\) (b) \(K_{\mathrm{c}}\) is greater than \(K_{\mathrm{p}} ;(\mathrm{c}) K_{\mathrm{c}}=K_{\mathrm{p}} ;\) (d) whether \(K_{\mathrm{c}}\) is less than, equal to, or greater than \(K_{\mathrm{p}}\) depends on the total gas pressure.

Determine \(K_{c}\) for the reaction $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NOBr}(\mathrm{g})$$ from the following information (at \(298 \mathrm{K}\) ). $$\begin{aligned} 2 \mathrm{NO}(\mathrm{g}) & \rightleftharpoons \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) K_{\mathrm{c}}=2.1 \times 10^{30} \\ \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{g}) & \rightleftharpoons \mathrm{NOBr}(\mathrm{g}) \quad K_{\mathrm{c}}=1.4 \end{aligned}$$

Can a mixture of \(2.2 \mathrm{mol} \mathrm{O}_{2}, 3.6 \mathrm{mol} \mathrm{SO}_{2},\) and \(1.8 \mathrm{mol}\) \(\mathrm{SO}_{3}\) be maintained indefinitely in a \(7.2 \mathrm{L}\) flask at a temperature at which \(K_{\mathrm{c}}=100\) in this reaction? Explain. $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) $$

Lead metal is added to \(0.100 \mathrm{M} \mathrm{Cr}^{3+}(\mathrm{aq}) .\) What are \(\left[\mathrm{Pb}^{2+}\right],\left[\mathrm{Cr}^{2+}\right],\) and \(\left[\mathrm{Cr}^{3+}\right]\) when equilibrium is established in the reaction? $$\begin{aligned} \mathrm{Pb}(\mathrm{s})+2 \mathrm{Cr}^{3+}(\mathrm{aq}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cr}^{2+}(\mathrm{aq}) & \\ K_{\mathrm{c}}=3.2 \times 10^{-10} & \end{aligned}$$

In the reversible reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons\) \(2 \mathrm{HI}(\mathrm{g}),\) an initial mixture contains \(2 \mathrm{mol} \mathrm{H}_{2}\) and 1 mol I \(_{2} .\) The amount of HI expected at equilibrium is (a) \(1 \mathrm{mol} ;\) (b) \(2 \mathrm{mol} ;\) (c) less than \(2 \mathrm{mol}\); (d) more than 2 mol but less than 4 mol.
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