Problem 36 - Chemistry The Central Science

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 \(K_{c} .\)

Step-by-Step Solution

Answer

The equilibrium concentrations of H₂, Br₂, and HBr are 0.140 mol/L, 0.020 mol/L, and 0.400 mol/L, respectively. The equilibrium constant Kc for the given reaction is approximately 57.14.

Step 1: Calculate the initial moles of H2 and Br2

To find the initial moles of H2 and Br2, we need to use their masses and molar masses. The molar mass of H2 is 2.02 g/mol, and for Br2, it's 159.8 g/mol. Moles of H2 = mass / molar mass = 1.374 g / 2.02 g/mol ≈ 0.680 mol Moles of Br2 = mass / molar mass = 70.31 g / 159.8 g/mol ≈ 0.440 mol

Step 2: Calculate the moles of H2 and Br2 reacted, and the moles of HBr formed

To find the moles of H₂ and Br₂ reacted, we can subtract the moles of H₂ at equilibrium from the initial moles of H₂. Moles of H₂ reacted = Initial moles of H₂ - moles of H₂ at equilibrium Moles of H₂ reacted = 0.680 mol - (0.566 g / 2.02 g/mol) ≈ 0.680 mol - 0.280 mol = 0.400 mol Since the stoichiometry of the reaction shows that one mole of H₂ and one mole of Br₂ react to give two moles of HBr, the moles of Br₂ reacted and the moles of HBr formed can be determined. Moles of Br₂ reacted = 0.400 mol (same as moles of H₂ reacted) Moles of HBr formed = 2 * moles of H₂ reacted = 2 * 0.400 mol = 0.800 mol

Step 3: Calculate the equilibrium moles of H2, Br2, and HBr

Now, we can find the equilibrium moles of all species. Moles of H₂ at equilibrium = 0.680 mol - 0.400 mol = 0.280 mol Moles of Br₂ at equilibrium = 0.440 mol - 0.400 mol = 0.040 mol Moles of HBr at equilibrium = 0 + 0.800 mol = 0.800 mol

Step 4: Calculate the equilibrium concentrations of H2, Br2, and HBr

We can now find the equilibrium concentrations by dividing the moles by the volume of the vessel (2.00 L). [H₂] = 0.280 mol / 2.00 L = 0.140 mol/L [Br₂] = 0.040 mol / 2.00 L = 0.020 mol/L [HBr] = 0.800 mol / 2.00 L = 0.400 mol/L

Step 5: Calculate the equilibrium constant Kc

Using the equilibrium concentrations, we can now calculate the equilibrium constant Kc using the formula: Kc = ([HBr]²) / ([H₂] * [Br₂]) Kc = (0.400 mol/L)² / (0.140 mol/L * 0.020 mol/L) ≈ 57.14 The equilibrium constant Kc for the given reaction is approximately 57.14.

Key Concepts

Equilibrium Concentrations

Understanding equilibrium concentrations is vital in analyzing reactions. When a chemical reaction reaches equilibrium, the concentrations of its reactants and products become steady. This does not mean that the reactants and products cease reacting. Instead, the rate at which the products are formed is equal to the rate at which they convert back into reactants. In the given problem, we're calculating the equilibrium concentrations of

• Hydrogen ( \( \mathrm{H}_2 \) ),
• Bromine ( \( \mathrm{Br}_2 \) ),
• And Hydrogen Bromide ( \( \mathrm{HBr} \) ).

To find the concentrations, we use the established equilibrium moles and divide each by the volume of the reaction vessel, which is 2.00 L. For example, the number of moles of \( \mathrm{H}_2 \) at equilibrium is 0.280 moles. So, its concentration is \[ [\mathrm{H}_2] = \frac{0.280}{2.00} = 0.140 \text{ mol/L} \]This method is repeated for \( \mathrm{Br}_2 \) and \( \mathrm{HBr} \) to find their respective equilibrium concentrations.

Reaction Stoichiometry

Reaction stoichiometry is the relationship between quantities of reactants and products in a chemical reaction. Understanding stoichiometry allows us to determine how much of each reactant is needed to form a desired amount of product or vice versa. In our reaction of \( \mathrm{H}_2 + \mathrm{Br}_2 \leftrightarrows 2 \mathrm{HBr} \), stoichiometry tells us that 1 mole of \( \mathrm{H}_2 \) reacts with 1 mole of \( \mathrm{Br}_2 \) to form 2 moles of \( \mathrm{HBr} \). This 1:1:2 ratio is crucial.Since 0.400 moles of \( \mathrm{H}_2 \) reacted, we can conclude that:

• The same amount of \( \mathrm{Br}_2 \) reacted, 0.400 moles.
• Twice that amount of \( \mathrm{HBr} \) was produced, totaling 0.800 moles.

Understanding these ratios ensures you account for all reactants and products involved in a reaction, which is indispensable when making further calculations like equilibrium constants.

Equilibrium Constant (Kc)

The equilibrium constant (\( K_c \)) is a numerical value that indicates the ratio of the concentrations of products to reactants at equilibrium. It depends on temperature and is specific for a given reaction. In our chemical equation:\[ \mathrm{H}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{HBr}(g) \]The \( K_c \) expression is written as:\[ K_c = \frac{[\mathrm{HBr}]^2}{[\mathrm{H}_2 ][\mathrm{Br}_2]} \]Substituting in the equilibrium concentrations:

• \( [\mathrm{HBr}] = 0.400 \text{ mol/L} \)
• \( [\mathrm{H}_2] = 0.140 \text{ mol/L} \)
• \( [\mathrm{Br}_2] = 0.020 \text{ mol/L} \)

We calculate:\[ K_c = \frac{(0.400)^2}{0.140 \times 0.020} = 57.14 \]A high \( K_c \) value typically indicates that at equilibrium, the reaction favors the formation of products. This information is valuable in predicting how a reaction will behave under different conditions and is central to the field of chemical engineering.