Question

A mixture of \(1.374 \mathrm{~g}\) of \(\mathrm{H}_{2}\) and \(70.31 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) is heated in a 2.00-L vessel at \(700 \mathrm{~K}\). These substances react according to $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ At equilibrium, the vessel is found to contain \(0.566 \mathrm{~g}\) of \(\mathrm{H}_{2}\). (a) Calculate the equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{Br}_{2},\) and \(\mathrm{HBr} .\) (b) Calculate \(K_{c}\)

Short answer

The equilibrium concentrations are [H₂] = 0.142 M, [Br₂] = 0.018 M, and [HBr] = 0.404 M. The equilibrium constant Kc is 639.

Step-by-step solution

Calculate the initial moles of reactants and moles of H₂ at equilibrium

First, we need to convert the given masses of H₂, Br₂, and the mass of H₂ at equilibrium into moles. We can use the molar masses of H₂ and Br₂ to do this: H₂ molar mass: 2g/mol Br₂ molar mass: 2 * 79.9g/mol = 159.8g/mol Initial moles of H₂: \(1.374 g \mathrm{H}_2 \times \frac{1 \mathrm{mol} \mathrm{H}_2}{2g} = 0.687 \mathrm{mol}\) Initial moles of Br₂: \(70.31 g \mathrm{Br}_2 \times \frac{1 \mathrm{mol} \mathrm{Br}_2}{159.8g} = 0.440 \mathrm{mol}\) Moles of H₂ at equilibrium: \(0.566 g \mathrm{H}_2 \times \frac{1 \mathrm{mol} \mathrm{H}_2}{2g} = 0.283 \mathrm{mol}\)

Calculate moles of reactants and products at equilibrium

Since we know the moles of H₂ decreased by (0.687 - 0.283) = 0.404 mol, we can deduce the loss of moles of Br₂ and the gain in moles of HBr: Moles of Br₂ at equilibrium = 0.440 - 0.404 = 0.036 mol (as 1 mol of Br₂ is used per mol of H₂ used) Moles of HBr at equilibrium = 2 * 0.404 = 0.808 mol (as 2 mol of HBr is formed per mol of H₂ used)

Calculate the equilibrium concentrations

We can now find the equilibrium concentrations of each substance by dividing the moles at equilibrium by the volume of the vessel: Concentration of H₂: \(\frac{0.283 \mathrm{mol}}{2.00 L} = 0.142 M\) Concentration of Br₂: \(\frac{0.036 \mathrm{mol}}{2.00 L} = 0.018 M\) Concentration of HBr: \(\frac{0.808 \mathrm{mol}}{2.00 L} = 0.404 M\) Thus, the equilibrium concentrations are: [H₂] = 0.142 M, [Br₂] = 0.018 M, and [HBr] = 0.404 M.

Calculate the equilibrium constant Kc

Using the balanced equation, we can write the expression for Kc: \(K_c = \frac{[\mathrm{HBr}]^2}{[\mathrm{H}_2][\mathrm{Br}_2]}\) Plugging in the equilibrium concentrations, we find Kc: \(K_c = \frac{(0.404 M)^2}{(0.142 M)(0.018 M)} = 639\) The equilibrium constant Kc for this reaction at 700 K is 639.

Key concepts

Chemical Equilibrium

Chemical equilibrium refers to the state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction.
At this point, the concentrations of reactants and products remain constant over time. It is important to understand that this doesn't mean the reactants and products are equal in concentration, just that their concentrations are stable.

For the reaction involving hydrogen gas (\(\text{H}_2\text{(g)}\)) and bromine gas (\(\text{Br}_2\text{(g)}\)) forming hydrobromic acid gas (\(\text{HBr}\text{(g)}\)), equilibrium is reached when the creation of \(\text{HBr}\) from \(\text{H}_2\) and \(\text{Br}_2\) balances with the decomposition of \(\text{HBr}\) back into \(\text{H}_2\) and \(\text{Br}_2\).

• This balance is dynamic, meaning that the reactions continue to occur, but there is no net change in the number of reactants and products.
• The concept of equilibrium is fundamental in chemical thermodynamics and kinetics, as it defines the point at which a system has achieved a minimum energy state and maximum disorder (or entropy), where no macroscopic changes are observed.

Concentration Calculations

Calculating the concentrations of substances at equilibrium is a critical step in analyzing chemical reactions.
Concentrations are typically expressed in moles per liter (Molarity, M). Understanding how to determine equilibrium concentrations helps in using this data to find variables like the equilibrium constant, \(K_c\).

• The initial concentrations are known from the given masses and the molar masses of the substances involved.
• For instance, in the given exercise, knowing the provided mass of \(\text{H}_2\) at the beginning and at equilibrium allows us to deduce how much \(\text{H}_2\) reacted.
• Using the stoichiometry of the reaction, the change in the moles of other reactants and products can be inferred.
• By dividing the moles of each substance by the total volume of the reaction mixture, one can find the equilibrium concentration of each component.

This allows us to systematically approach reactions and predict product concentrations based on initial conditions and equilibrium behavior.

Reaction Stoichiometry

Reaction stoichiometry describes the quantitative relationships between the amounts of reactants and products in a chemical reaction.
It is achieved using the balanced chemical equation for the reaction, which provides the molar ratios necessary for these calculations.

• In the example reaction \(\text{H}_2\text{(g)} + \text{Br}_2\text{(g)} \rightarrow 2\text{HBr}\text{(g)}\), stoichiometry dictates that 1 mole of \(\text{H}_2\) reacts with 1 mole of \(\text{Br}_2\) to produce 2 moles of \(\text{HBr}\).
• Understanding this ratio allows chemists to calculate how much product is expected from given amounts of reactants or, conversely, how much reactant is needed to produce a desired quantity of product.
• Essentially, stoichiometry enables calculations that adhere to the principle of conservation of mass, ensuring all atoms presented as reactants are accounted for as products.

The stoichiometric coefficients derived from the balanced equation are fundamental to understanding the transformation that a chemical reaction undergoes, facilitating practical applications such as yield predictions and equilibrium calculations.