Question

A mixture of \(\mathrm{H}_{2} \mathrm{~S}\), and \(\mathrm{H}_{2} \mathrm{~S}\) is held in a 1.0-L vessel at \(90^{\circ} \mathrm{C}\) until the following equilibrium is achieved: $$ \mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g) $$ At equilibrium the mixture contains \(0.46 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(0.40 \mathrm{~g} \mathrm{H}_{2}\). (a) Write the equilibrium- constant expression for this reaction. (b) What is the value of \(K_{c}\) for the reaction at this temperature? (c) Why can we ignore the amount of \(S\) when doing the calculation in part \((b)\) ?

Short answer

(a) The equilibrium constant expression is: \(K_c = [\mathrm{H}_2\mathrm{S}] / [\mathrm{H}_2]\) (b) The value of Kc at this temperature is: \(K_c = 0.0678\) (c) We can ignore the amount of sulfur solid because its concentration does not change during the reaction.

Step-by-step solution

Write the balanced chemical equation

We are given the balanced chemical equation for the reaction as: $$ \mathrm{H}_2(g) + \mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2}\mathrm{S}(g) $$

Write the equilibrium constant expression

The equilibrium constant expression for the reaction is given by: $$ K_c = \frac{[\mathrm{H}_2\mathrm{S}]}{[\mathrm{H}_2]} $$ Note that we don't include the concentration of sulfur solid in the equilibrium constant expression, as its concentration remains constant during the reaction.

Calculate the concentrations at equilibrium

We need to find the concentrations of \(\mathrm{H}_2\mathrm{S}\) and \(\mathrm{H}_2\) at equilibrium, given the masses. To do this, we will first convert the masses to moles and then divide by the volume of the vessel. Molar masses: $$ \mathrm{H}_2 = 2 \cdot 1.008 \, \text{g/mol} = 2.016\, \text{g/mol} $$ $$ \mathrm{H}_2\mathrm{S} = 2 \cdot 1.008 + 32.1 \, \text{g/mol} = 34.116\, \text{g/mol} $$ Moles: $$ \mathrm{n}_{\mathrm{H}_2} = \frac{0.40\, \text{g}}{2.016\, \text{g/mol}} = 0.1988\, \text{mol} $$ $$ \mathrm{n}_{\mathrm{H}_2\mathrm{S}} = \frac{0.46\, \text{g}}{34.116\, \text{g/mol}} = 0.01349\, \text{mol} $$ Concentration: $$ [\mathrm{H}_2] = \frac{0.1988\, \text{mol}}{1.0\, \text{L}} = 0.1988\, \text{M} $$ $$ [\mathrm{H}_2\mathrm{S}] = \frac{0.01349\, \text{mol}}{1.0\, \text{L}} = 0.01349\, \text{M} $$

Calculate the value of Kc

We can now calculate Kc using the equilibrium concentrations: $$ K_c = \frac{[\mathrm{H}_2\mathrm{S}]}{[\mathrm{H}_2]} = \frac{0.01349\, \text{M}}{0.1988\, \text{M}} = 0.0678 $$

Why we can ignore the amount of sulfur solid

We can ignore the amount of sulfur solid in the calculation of Kc because it is present in its solid state. Only gaseous and aqueous species are included in equilibrium constant expressions, as the concentrations of solids and liquids do not change during the reaction. So, the final answers are: (a) The equilibrium constant expression is: \(K_c = [\mathrm{H}_2\mathrm{S}] / [\mathrm{H}_2]\) (b) The value of Kc at this temperature is: \(K_c = 0.0678\) (c) We can ignore the amount of sulfur solid because its concentration does not change during the reaction.

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as \(K_c\), plays a crucial role in understanding chemical reactions at equilibrium. It is a dimensionless number that provides insights into the extent of a reaction at a given temperature. If you think of a balanced equation such as \(H_2(g) + S(s) ightleftharpoons H_2S(g)\), the equilibrium constant expression is formulated by taking the concentration of the products and dividing it by the concentration of the reactants. In this specific reaction, the expression becomes \(K_c = \frac{[\mathrm{H}_2\mathrm{S}]}{[\mathrm{H}_2]}\). These brackets represent molarity, the concentration of a substance in a large reaction mixture.

It's important to remember that solids like sulfur are not included in the equilibrium constant expression. Why? Because they do not impact the concentrations of reactants or products in the gas phase, as they remain constant throughout the reaction at equilibrium.

Equilibrium Concentration

Equilibrium concentration refers to the amount of each reactant and product present in a reaction mixture when the reaction has reached equilibrium. At this point, the rate of the forward reaction is equal to the rate of the reverse reaction. To find the equilibrium concentrations, we typically convert the mass of each substance to moles, and then use the volume of the reaction vessel to find the molarity, which is moles per liter.

For example, in the reaction \(H_2(g) + S(s) \rightleftharpoons H_2S(g)\), we know the equilibrium mixture contains \(0.46\, \text{g}\) of \(H_2S\) and \(0.40\, \text{g}\) of \(H_2\). First, these masses are converted to moles using their respective molar masses. Afterward, these moles are divided by the volume (1.0 L in this case) to obtain the equilibrium concentrations:

• \([H_2S] = 0.01349\, \text{M}\)
• \([H_2] = 0.1988\, \text{M}\)

This is pivotal for calculating the equilibrium constant and understanding the behavior of the chemical system at equilibrium.

Mole Calculation

Mole calculations are fundamental in chemistry for converting between the mass of a substance and the amount in moles, giving us a clearer view of the quantities involved in chemical reactions. A mole is a quantity that represents \(6.022 \times 10^{23}\) particles, usually atoms or molecules. In the context of determining equilibrium concentrations, you first need to find the number of moles of each substance from their given masses.

Consider this practical example: Given \(0.40\, \text{g}\) of \(H_2\) and \(0.46\, \text{g}\) of \(H_2S\) in a reaction mixture, you would:

• Use the molar mass of \(H_2\) (\(2.016\, \text{g/mol}\)) and \(H_2S\) (\(34.116\, \text{g/mol}\)) to find the moles.
• For \(H_2\): \(n = \frac{0.40}{2.016} = 0.1988\, \text{mol}\)
• For \(H_2S\): \(n = \frac{0.46}{34.116} = 0.01349\, \text{mol}\)

Such conversions are key to analyzing chemical reactions and finding equilibrium states. So, mole calculations bridge the gap between the physical mass of materials and their chemical behavior in reactions.