Question

At equilibrium, a gas mixture has a partial pressure of 0.7324 atm for \(\mathrm{HBr}\) and \(2.80 \times 10^{-3}\) atm for both hydrogen and bromine gases. What is \(K\) for the formation of two moles of HBr from \(\mathrm{H}_{2}\) and \(\mathrm{Br}_{2} ?\)

Short answer

Answer: The equilibrium constant, K, for the formation of two moles of HBr from H2 and Br2 gases is approximately 68,086.

Step-by-step solution

Write the expression for the equilibrium constant (K)

For the given balanced chemical equation, we can write the expression for the equilibrium constant (K) as: \(K = \frac{[\mathrm{HBr}]^2}{[\mathrm{H}_{2}][\mathrm{Br}_{2}]}\) Note that square brackets indicate partial pressures in atmospheres.

Find the partial pressures at equilibrium

We are given the equilibrium partial pressures for the reactants and products: - Partial pressure of HBr: 0.7324 atm - Partial pressure of H2 and Br2: \(2.80 \times 10^{-3}\) atm

Substitute the partial pressures into the K expression

Using the partial pressures from Step 2, substitute the values into the K expression from Step 1: \(K = \frac{(0.7324)^2}{(2.80 \times 10^{-3})(2.80 \times 10^{-3})}\)

Calculate K

Next, we'll calculate the equilibrium constant: \(K = \frac{(0.7324)^2}{(2.80 \times 10^{-3})(2.80 \times 10^{-3})} = 68,086\) The equilibrium constant for the formation of two moles of HBr from H2 and Br2 gases is approximately 68,086.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state of a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. Hence, the concentrations—or in the case of gases, the partial pressures—of the reactants and products remain constant over time, but are not necessarily equal.

For example, in the provided exercise, HBr, H₂, and Br₂ reach a state where their partial pressures do not change, indicating that they have attained chemical equilibrium. At this point, the amount of HBr being produced by the reaction of H₂ and Br₂ equals the amount being decomposed to form H₂ and Br₂.

To calculate the equilibrium constant, it's important to use the expression that considers the stoichiometry of the reaction, represented usually with a capitalized K. This constant is a ratio of the concentration (or pressure) of the products to the reactants, each raised to the power of their coefficients in the balanced equation. In this case, as the exercise demonstrates, the equilibrium constant could give students a quantitative understanding of the reaction's state at equilibrium.

Partial Pressure

The concept of partial pressure is vital when discussing reactions involving gases at equilibrium. It refers to the pressure one component of a mixture of gases would exert if it were alone occupying the entire volume of the mixture. Mathematically, it's a direct consequence of Dalton's Law of Partial Pressures.

In the context of the provided exercise, the partial pressures of HBr, H₂, and Br₂ gases are given, and these are used to calculate the equilibrium constant. When dealing with gases, the partial pressure can be used in the equilibrium expression instead of concentration, which aligns with the principles of gas behavior under different conditions of temperature and pressure. This introduces students to a practical and tangible aspect of chemical equilibria, where they must realize the role that physical states and conditions exert on the systems in equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle offers a qualitative explanation for how a system at equilibrium responds to changes in concentration, pressure, temperature, or volume. It states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust its equilibrium position to counteract the change in order to achieve a new equilibrium state.

For instance, if more H₂ is added to the reaction in the given problem, the principle suggests that the system would respond by forming more HBr to reduce the change—shifting the equilibrium towards the product side. Conversely, an increase in the quantity of HBr would shift the equilibrium towards the reactants. Students should understand that while this principle offers a prediction of the direction of change, it doesn't provide exact quantities or change the value of the equilibrium constant, which remains the same at a given temperature unless the reaction mechanism itself changes.