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 as follows: $$ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) $$ 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 HBr. (b) Calculate \(K_{c}\).

Short answer

The equilibrium concentrations of H₂, Br₂, and HBr are 0.281 M, 0 M, and 0.802 M, respectively. Kc is undefined, indicating that the reaction proceeds entirely to the right, and the products (HBr) are favored over the reactants (H₂ and Br₂).

Step-by-step solution

Convert grams to moles

First, we need to convert the given masses of H₂, Br₂, and H₂ at equilibrium into moles. To do this, we can use the molar masses of each element: - H₂: \(1.008 \mathrm{~g/mol}\) - Br₂: \(159.8 \mathrm{~g/mol}\) - Initial moles of H₂: \(\frac{1.374 \mathrm{~g}}{1.008 \mathrm{~g/mol}} = 1.364 \mathrm{~mol}\) - Initial moles of Br₂: \(\frac{70.31 \mathrm{~g}}{159.8 \mathrm{~g/mol}} = 0.440 \mathrm{~mol}\) - Moles of H₂ at equilibrium: \(\frac{0.566 \mathrm{~g}}{1.008 \mathrm{~g/mol}} = 0.562 \mathrm{~mol}\)

Calculate the change in moles for the reaction

Now that we have the initial moles of H₂ and Br₂ and the moles of H₂ at equilibrium, we can determine the change in moles for the reaction: - Change in moles of H₂: \(1.364 \mathrm{~mol} - 0.562 \mathrm{~mol} = 0.802 \mathrm{~mol}\) - Change in moles of Br₂: Since the molar ratio of H₂:Br₂ is 1:1 in the balanced equation, the change in moles of Br₂ is also 0.802 mol. - Moles of HBr formed: Since the molar ratio of H₂:2HBr is 1:2 in the balanced equation, the moles of HBr formed is twice the change in moles of H₂: \(2 \times 0.802 \mathrm{~mol} = 1.604 \mathrm{~mol}\).

Calculate the equilibrium concentrations

Next, we can use the volume of the container to convert the moles at equilibrium to concentration (in mol/L): - Concentration of H₂ at equilibrium: \(\frac{0.562 \mathrm{~mol}}{2.00 \mathrm{~L}} = 0.281 \mathrm{~M}\) - Concentration of Br₂ at equilibrium: \(\frac{0.440 \mathrm{~mol} - 0.802 \mathrm{~mol}}{2.00 \mathrm{~L}} = -0.181 \mathrm{~M} = 0 \mathrm{~M}\) (Since Br₂ is completely consumed, its concentration at equilibrium is 0.) - Concentration of HBr at equilibrium: \(\frac{1.604 \mathrm{~mol}}{2.00 \mathrm{~L}} = 0.802 \mathrm{~M}\).

Calculate Kc

Finally, we can use the equilibrium concentrations to calculate Kc for the reaction: - Kc = \(\frac{[\mathrm{HBr}]^2}{[\mathrm{H}_{2}][\mathrm{Br}_{2}]}\) - Kc = \(\frac{(0.802 \mathrm{~M})^2}{(0.281 \mathrm{~M})(0 \mathrm{~M})}\) Since the denominator contains 0 M, the value of Kc is undefined. This indicates that the reaction proceeds entirely to the right, meaning that the products (HBr) are favored over the reactants (H₂ and Br₂).

Key concepts

Equilibrium Concentrations

When studying chemical reactions, equilibrium concentrations refer to the amounts of reactants and products present once a reversible chemical reaction reaches a state of balance. This means the rate at which the reactants are converting into products is equal to the rate at which the products are reverting back into reactants. In our problem, we have hydrogen gas (\(\mathrm{H}_2\)) and bromine gas (\(\mathrm{Br}_2\)) reacting to form hydrogen bromide (\(\mathrm{HBr}\)).
To find the equilibrium concentrations, we first need to convert the mass of each substance into moles because concentration is typically expressed in molarity (M), which is moles per liter. We calculated the initial amount of each reactant and product in moles. Then, by finding how much of \(\mathrm{H}_2\) remains at equilibrium, we were able to determine the change in moles for the reaction.

Once we know the moles at equilibrium, these values are divided by the volume of the reaction container (in liters) to convert them into concentrations. Using a balanced chemical equation allowed us to understand the stoichiometry, which tells us how the molecules interact and change during the reaction.

The final concentrations at equilibrium were: - The concentration of \(\mathrm{H}_2\) was 0.281 M. - The concentration of \(\mathrm{Br}_2\) was 0 M, indicating it was completely consumed.- The concentration of \(\mathrm{HBr}\) was 0.802 M.

Reaction Stoichiometry

Reaction stoichiometry involves using the balanced chemical equation to understand the proportional relationships between reactants and products in a chemical reaction. In the given exercise, the reaction is: \( \mathrm{H}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{HBr}(g) \).

In stoichiometry, the coefficients in the balanced equation reveal the relative amounts of each substance involved in the reaction. Here, the coefficients show a 1:1 ratio for \(\mathrm{H}_2\) and \(\mathrm{Br}_2\), and a 1:2 ratio between \(\mathrm{H}_2\) and \(\mathrm{HBr}\).

From the problem, we know the change in moles of \(\mathrm{H}_2\), so we can infer the change in \(\mathrm{Br}_2\) using the 1:1 stoichiometric ratio. Similarly, because the reaction produces 2 moles of \(\mathrm{HBr}\) for every mole of \(\mathrm{H}_2\) consumed,- The change in moles for \(\mathrm{H}_2\) and \(\mathrm{Br}_2\) was each 0.802 mol- The increase in moles for \(\mathrm{HBr}\) was double or 1.604 mol.

The stoichiometry helps in predicting the amount of products formed from given amounts of reactants, and vice versa, so it is crucial for solving equilibrium problems.

Equilibrium Constant (Kc)

The equilibrium constant, \(K_c\), is a number that expresses the ratio of the concentration of products to reactants at equilibrium for a reversible chemical reaction at a given temperature. It provides insight into whether a reaction favors products or reactants once equilibrium is achieved. In our example, the constant is defined as:\[ K_c = \frac{[\mathrm{HBr}]^2}{[\mathrm{H}_2][\mathrm{Br}_2]} \]

To calculate \(K_c\), we substitute the equilibrium concentrations of each compound into this expression. However, because the concentration of \(\mathrm{Br}_2\) is zero at equilibrium, the denominator in the expression \(\frac{[\mathrm{HBr}]^2}{[\mathrm{H}_2][\mathrm{Br}_2]}\) becomes zero. This results in an undefined \(K_c\).

An undefined \(K_c\) indicates that the reaction strongly favors the product, \(\mathrm{HBr}\), and proceeds almost entirely to completion under the given conditions, consuming all available reactants. In real-world applications, knowing \(K_c\) helps chemists understand and predict the outcomes of chemical reactions and assess how conditions affect the balance between reactants and products.