Question

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

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

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: \[ [COF_2] = 0.145 \, mol \, / \, 5.00 \, L = 0.029 \, M\] \[ [CO_2] = 0.262 \, mol \, / \, 5.00 \, L = 0.0524 \, M\] \[ [CF_4] = 0.074 \, mol \, / \, 5.00 \, L = 0.0148 \, M\]

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.

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).

Key concepts

Reaction Quotient

The reaction quotient, denoted as \(Q\), is a tool used to assess the progress of a chemical reaction relative to its equilibrium state.
It helps indicate whether a reaction is moving toward equilibrium or if it has already reached equilibrium.

To calculate \(Q\), you use a formula similar to that of the equilibrium constant \(K_c\):

• For a general reaction \(aA + bB \rightleftharpoons cC + dD\), the reaction quotient \(Q\) is calculated as: \[ Q = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]
• Here, the square brackets \([ ]\) indicate the concentrations of the respective species, and \(a, b, c, d\) are their coefficients in the balanced chemical equation.

By comparing the value of \(Q\) to \(K_c\):- If \(Q = K_c\), the system is at equilibrium.- If \(Q < K_c\), the reaction will proceed in the forward direction (toward the products) to reach equilibrium.- If \(Q > K_c\), the reaction will proceed in the reverse direction (toward the reactants).

This serves as a valuable diagnostic tool in determining the direction in which a reaction should proceed to attain equilibrium.

Equilibrium Constant

The equilibrium constant, \(K_c\), is a value that expresses the ratio of product concentrations to reactant concentrations at equilibrium.
It is specific to a particular reaction at a given temperature.

In the context of the reaction \(2 \operatorname{COF}_{2} \rightleftharpoons \operatorname{CO}_{2} + \operatorname{CF}_{4}\), the equilibrium constant is given by:

• \[ K_c = \frac{[CO_2][CF_4]}{[COF_2]^2} \]
• Each concentration is raised to the power of its coefficient in the reaction equation.

For this reaction at \(1000^{\circ}C\), \(K_c = 2.00\).

The magnitude of \(K_c\) provides insight into the composition of the mixture at equilibrium:- A large \(K_c\) (much greater than 1) suggests a higher concentration of products at equilibrium.- A small \(K_c\) (much less than 1) shows the reactants are favored over the products.- A \(K_c\) around 1 indicates that neither reactants nor products are particularly favored at equilibrium.

Understanding \(K_c\) is crucial because it informs chemists about the expected composition of the system at equilibrium, helping them predict the extent and behavior of chemical reactions.

Concentration Calculations

Concentration calculations are fundamental in determining the concentrations of species involved in a reaction.
They are necessary when using the reaction quotient \(Q\) and the equilibrium constant \(K_c\).

For a given solution, concentration is calculated as:

• \[ \text{Concentration} = \frac{\text{moles of substance}}{\text{volume of solution}} \]

To proceed with equilibrium calculations, follow this process:

• Determine the initial concentrations of all reactants and products. For example, \([COF_2] = 0.029 \, M\), \([CO_2] = 0.0524 \, M\), and \([CF_4] = 0.0148 \, M\).
• Use these concentrations to compute the reaction quotient \(Q\) to decide if the system is at equilibrium or in which direction it will shift.
• If the system is not at equilibrium, predict the changes in concentration and set up the equilibrium expressions involving a variable \(x\) that represents these concentration changes.
• Substitute the equilibrium concentrations into the \(K_c\) expression and solve for \(x\).

Calculated concentrations give you the equilibrium state of each species. Multiply these concentrations by the volume of the mixture to convert them to moles, yielding a complete understanding of the system’s composition.