Question

For the following reaction, \(K_{c}=115\) at a particular temperature: $$ \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g) $$ A container initially holds the following concentrations: \(0.050 \mathrm{M}\) \(\mathrm{H}_{2}, 0.050 \mathrm{M} \mathrm{F}_{2},\) and \(0.10 \mathrm{M} \mathrm{HF} .\) When equilibrium is reached, what is the concentration of HF?

Short answer

[HF] at equilibrium is approximately 0.1686 M.

Step-by-step solution

- Write the expression for the equilibrium constant

For the given reaction \ \( \text{H}_{2}(g) + \text{F}_{2}(g) \rightleftharpoons 2 \text{HF}(g) \), the expression for the equilibrium constant, \(K_{c}\), is written as: \ \[ K_{c} = \frac{[\text{HF}]^2}{[\text{H}_{2}][\text{F}_{2}]} \]

- Define the change in concentrations

Let the change in the concentration of \( \text{HF} \) when equilibrium is reached be \( +2x \). Since the reaction shows that for every 2 moles of \( \text{HF} \) produced, 1 mole each of \( \text{H}_{2} \) and \( \text{F}_{2} \) is consumed, the change in the concentrations of \( \text{H}_{2} \) and \( \text{F}_{2} \) will be \( -x \).

- Write the equilibrium concentrations in terms of x

Initial concentrations: \ \( [\text{H}_{2}]_{initial} = 0.050 \text{M} \) \ \( [\text{F}_{2}]_{initial} = 0.050 \text{M} \) \ \( [\text{HF}]_{initial} = 0.10 \text{M} \) \ Therefore, the equilibrium concentrations can be expressed as: \ \( [\text{H}_{2}]_{eq} = 0.050 - x \) \ \( [\text{F}_{2}]_{eq} = 0.050 - x \)\ \( [\text{HF}]_{eq} = 0.10 + 2x \)

- Substitute the equilibrium concentrations into the equilibrium constant expression

Substitute the expressions from Step 3 into the equilibrium constant equation: \ \[ 115 = \frac{(0.10 + 2x)^2}{(0.050 - x)(0.050 - x)} \]

- Simplify and solve for x

First, expand the equation: \ \[ 115 = \frac{(0.10 + 2x)^2}{(0.050 - x)^2} \] \ Next, take the square root of both sides: \ \[ \sqrt{115} = \frac{0.10 + 2x}{0.050 - x} \] \ Calculate \( \sqrt{115} \approx 10.724 \) to get: \ \[ 10.724 (0.050 - x) = 0.10 + 2x \] \ Distribute the \( 10.724 \): \ \[ 0.5362 - 10.724x = 0.10 + 2x \] \ Combine like terms: \ \[ 0.5362 - 0.10 = 10.724x + 2x \] \ \[ 0.4362 = 12.724x \] \ Finally, solve for \( x \): \ \[ x = \frac{0.4362}{12.724} \approx 0.0343 \]

- Calculate the equilibrium concentration of HF

Substitute the value of \( x \) back into the equilibrium concentration for \( [\text{HF}] \): \ \[ [\text{HF}]_{eq} = 0.10 + 2(0.0343) \approx 0.1686 \text{M} \]

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_c\), measures the ratio of the concentration of products to reactants at equilibrium for a given chemical reaction. For the reaction provided: $$ \text{H}_{2}(g) + \text{F}_{2}(g) \rightleftharpoons 2 \text{HF}(g) $$ The expression for \(K_c\) is written as: $$ K_c = \frac{[\text{HF}]^2}{[\text{H}_{2}][\text{F}_{2}]} $$ The concentration terms inside the brackets denote the molarity (moles per liter) of each substance at equilibrium. This expression tells you how much of each substance is present when the system has reached equilibrium. A higher \(K_c\) value, in this case, 115, indicates a greater concentration of products compared to the reactants when the reaction reaches equilibrium. Understanding the equilibrium constant helps in predicting the extent of the reaction and in calculating the concentrations of reactants and products.

Chemical Reaction Equilibrium

When a chemical reaction reaches equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of reactants and products remain constant over time. In the provided reaction: $$ \text{H}_{2}(g) + \text{F}_{2}(g) \rightleftharpoons 2 \text{HF}(g) $$ initially, you have 0.050 M of both \(\text{H}_{2}\) and \(\text{F}_{2}\) and 0.10 M of \(\text{HF}\). As the reaction proceeds towards equilibrium, \(\text{H}_{2}\) and \(\text{F}_{2}\) are consumed to produce more \(\text{HF}\). The key to solving this equilibrium problem is to set up an ICE table (Initial, Change, Equilibrium) and write the changes in concentrations in terms of a variable, \(x\). The changes are then used to express the equilibrium concentrations of all species involved. This process allows you to substitute these expressions into the equilibrium constant equation and solve for \(x\), thereby determining the equilibrium concentrations.

Concentration Changes

To determine the concentration changes at equilibrium, you start by identifying the initial concentrations and noting the changes that occur as the system reaches equilibrium. For our reaction: $$\text{H}_{2}(g) + \text{F}_{2}(g) \rightleftharpoons 2 \text{HF}(g)$$ Given initial concentrations:

• \([\text{H}_{2}]_{initial} = 0.050 \text{M}\)
• \([\text{F}_{2}]_{initial} = 0.050 \text{M}\)
• \([\text{HF}]_{initial} = 0.10 \text{M}\)

We denote the change in concentration of \(\text{HF}\) as \(+2x\) because 2 moles of \(\text{HF}\) are produced for every 1 mole of \(\text{H}_{2}\) and \(\text{F}_{2}\) consumed, which are reduced by \(x\). So, at equilibrium:

• \([\text{H}_{2}]_{eq} = 0.050 - x\)
• \([\text{F}_{2}]_{eq} = 0.050 - x\)
• \([\text{HF}]_{eq} = 0.10 + 2x\)

Finally, substituting these expressions into the \(K_c\) equation and solving for \(x\) allows calculation of the exact equilibrium concentrations. This approach ensures you account for all changes and correct equilibrium concentrations of the involved substances.