Question

For the following reaction at a certain temperature $$ \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g) $$ it is found that the equilibrium concentrations in a 5.00-L rigid container are \(\left[\mathrm{H}_{2}\right]=0.0500 M,\left[\mathrm{~F}_{2}\right]=0.0100 M\), and \([\mathrm{HF}]=\) \(0.400 \mathrm{M}\). If \(0.200 \mathrm{~mol} \mathrm{~F}_{2}\) is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

Short answer

The new equilibrium concentrations after adding \(0.200 \ \mathrm{mol} \ \mathrm{F}_2\) to the initial mixture are: \([H_2] = 0.0353 \ \mathrm{M}\), \([F_2] = 0.0353 \ \mathrm{M}\), and \([HF] = 0.4294 \ \mathrm{M}\).

Step-by-step solution

1. Write down the balanced chemical equation.

The balanced chemical equation is given: \[ \mathrm{H}_2(g) + \mathrm{F}_2(g) \rightleftharpoons 2 \mathrm{HF}(g) \]

2. Write the expression for the equilibrium constant (K).

Based on the balanced chemical equation, the expression for K is: \[ K = \frac{[\mathrm{HF}]^2}{[\mathrm{H}_2][\mathrm{F}_2]} \]

3. Calculate the initial value of K.

Use the initial equilibrium concentrations to calculate K: \[ K = \frac{[0.400 \ \mathrm{M}]^2}{[0.0500 \ \mathrm{M}][0.0100 \ \mathrm{M}]} = 320 \]

4. Write the reaction quotient (Q) with the initial concentrations after adding F₂.

Add 0.200 mol of F₂ to the 5.00 L container and calculate the new concentration of F₂: \[ [\mathrm{F}_2]_\mathrm{new} = 0.0100 \ \mathrm{M} + \frac{0.200 \ \mathrm{mol}}{5.00 \ \mathrm{L}} = 0.0500 \ \mathrm{M} \] Now, we can write the reaction quotient Q using the initial concentrations after adding F₂: \[ Q = \frac{[\mathrm{HF}]^2}{[\mathrm{H}_2][\mathrm{F}_2_\mathrm{new}]} \]

5. Determine the direction of the shift.

Compare the value of Q with the value of K to determine the direction of the shift: \[ Q = \frac{[0.400 \ \mathrm{M}]^2}{[0.0500 \ \mathrm{M}][0.0500 \ \mathrm{M}]} = 64 \] Since \( Q < K \), the reaction will shift to the right, towards the products, to reach the new equilibrium.

6. Define the change (x) and set up the equilibrium table.

Let the concentration of F₂ decrease by x. Then, following stoichiometry, the concentration of H₂ will also decrease by x, and the concentration of HF will increase by 2x: \[ \begin{array}{c|c|c|c} & [\mathrm{H}_2] & [\mathrm{F}_2] & [\mathrm{HF}] \\ \hline \text{Initial} & 0.0500-x & 0.0500-x & 0.400+2x \\ \text{Change} & -x & -x & 2x \\ \text{Equilibrium} & 0.0500 - x & 0.0500 - x & 0.400 + 2x \end{array} \]

7. Write an expression for K in terms of the change (x) and solve for x.

Use the change quantities and the calculated K to solve for x: \[ 320 = \frac{(0.400 + 2x)^2}{(0.0500 - x)^2} \] Solving for x, we find that \( x = 0.0147 \ \mathrm{M} \).

8. Calculate the new equilibrium concentrations.

Use the calculated value of x to determine the new equilibrium concentrations: \[ [\mathrm{H}_2]_\mathrm{eq} = 0.0500 - 0.0147 = 0.0353 \ \mathrm{M}\\ [\mathrm{F}_2]_\mathrm{eq} = 0.0500 - 0.0147 = 0.0353 \ \mathrm{M}\\ [\mathrm{HF}]_\mathrm{eq} = 0.400 + 2(0.0147) = 0.4294 \ \mathrm{M} \] The new equilibrium concentrations are: \([H_2] = 0.0353 \ \mathrm{M}\), \([F_2] = 0.0353 \ \mathrm{M}\), and \([HF] = 0.4294 \ \mathrm{M}\).

Key concepts

Equilibrium Constant (K)

In chemical equilibrium, the equilibrium constant, denoted as \( K \), is a value that expresses the ratio of products to reactants at equilibrium for a particular reaction at a given temperature. This constant is crucial because it provides a quantitative measure of the position of the equilibrium. The equation for \( K \) typically reflects the stoichiometry of the balanced chemical reaction.

For instance, consider the reaction \( \mathrm{H}_2(g) + \mathrm{F}_2(g) \rightleftharpoons 2 \mathrm{HF}(g) \). Here, the equilibrium constant \( K \) is expressed as:

• \( K = \frac{[\mathrm{HF}]^2}{[\mathrm{H}_2][\mathrm{F}_2]} \)

This formula signifies that at equilibrium, the concentration of \( \mathrm{HF} \), squared because of its coefficient in the balanced equation, is divided by the product of the concentrations of \( \mathrm{H}_2 \) and \( \mathrm{F}_2 \).

In practice, by calculating \( K \) using equilibrium concentrations, you can understand whether the products or reactants are favored under specific conditions.

Reaction Quotient (Q)

The reaction quotient, \( Q \), serves as an essential tool to predict the direction a reaction will proceed in order to reach equilibrium. It has the same form as the equilibrium constant expression, but uses the current concentrations or pressures rather than the equilibrium ones.

To find \( Q \), you use the formula:

• \( Q = \frac{[\mathrm{HF}]^2}{[\mathrm{H}_2][\mathrm{F}_2]} \)

If \( Q \) is compared to \( K \), it gives valuable information:

• If \( Q < K \), the reaction shifts right, forming more products.
• If \( Q > K \), the reaction shifts left, forming more reactants.
• If \( Q = K \), the system is at equilibrium.

In the given problem, after adding more \( \mathrm{F}_2 \), you calculate \( Q \) with the new concentrations and compare it to \( K \). This tells you the direction in which the reaction will shift to achieve the new equilibrium state.

Equilibrium Concentrations

Equilibrium concentrations refer to the concentrations of reactants and products in a chemical reaction when it reaches a state of equilibrium. At this point, the rates of the forward and reverse reactions are equal, thus stabilizing the concentrations.

To determine these concentrations after a change, such as adding a reactant, you set up an ICE table (Initial, Change, Equilibrium) to organize and calculate the shifts in concentrations. In the exercise, given the initial concentrations and changes due to stoichiometry (

• \( [\mathrm{H}_2] = 0.0500 - x \)
• \( [\mathrm{F}_2] = 0.0500 - x \)
• \( [\mathrm{HF}] = 0.400 + 2x \)

After solving for \( x \), you substitute back to find the equilibrium concentrations.

Therefore, after adjusting for the changes caused by adding \( \mathrm{F}_2 \), we find the new equilibrium concentrations:\[ [H_2] = 0.0353 \, \mathrm{M}, \ [F_2] = 0.0353 \, \mathrm{M}, \, \ [HF] = 0.4294 \, \mathrm{M}.\] These values indicate the stable state of the system after reaching the new equilibrium.