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Chapter 13: Problem 90

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

Expert verified
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

01

1. Write down the balanced chemical equation.

The balanced chemical equation is given: \[ \mathrm{H}_2(g) + \mathrm{F}_2(g) \rightleftharpoons 2 \mathrm{HF}(g) \]
02

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

Based on the balanced chemical equation, the expression for K is: \[ K = \frac{[\mathrm{HF}]^2}{[\mathrm{H}_2][\mathrm{F}_2]} \]
03

3. Calculate the initial value of K.

Use the initial equilibrium concentrations to calculate K: \[ K = \frac{[0.400 \ \mathrm{M}]^2}{[0.0500 \ \mathrm{M}][0.0100 \ \mathrm{M}]} = 320 \]
04

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: \[ Q = \frac{[\mathrm{HF}]^2}{[\mathrm{H}_2][\mathrm{F}_2_\mathrm{new}]} \]
05

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.
06

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: \[ \begin{array}{c|c|c|c} & [\mathrm{H}_2] & [\mathrm{F}_2] & [\mathrm{HF}] \\ \hline \text{Initial} & 0.0500-x & 0.0500-x & 0.400+2x \\ \text{Change} & -x & -x & 2x \\ \text{Equilibrium} & 0.0500 - x & 0.0500 - x & 0.400 + 2x \end{array} \]
07

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 \( x = 0.0147 \ \mathrm{M} \).
08

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 \([HF] = 0.4294 \ \mathrm{M}\).

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Most popular questions from this chapter

As we have attempted to lessen our dependence on fossil fuels, the demand for biofuels, such as ethanol, which is produced by the fermentation of the sugars found in corn, has increased. Using Le Châtelier's principle, predict which way the equilibrium will shift during the fermentation of sugar for each of the following changes. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(a q) $$ a. when the concentration of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is increased b. when the concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is decreased c. when \(\mathrm{CO}_{2}\) gas is added to the solution d. when the volume of water in the solution is doubled

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A sample of iron(II) sulfate was heated in an evacuated container to \(920 \mathrm{~K}\), where the following reactions occurred: $$ \begin{aligned} 2 \mathrm{FeSO}_{4}(s) & \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{3}(g)+\mathrm{SO}_{2}(g) \\ \mathrm{SO}_{3}(g) & \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \end{aligned} $$ After equilibrium was reached, the total pressure was \(0.836\) atm and the partial pressure of oxygen was \(0.0275\) atm. Calculate \(K_{\mathrm{p}}\) for each of these reactions.

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