Problem 18 - General Chemistry: Principles and Modern Applications

Question

A mixture of \(1.00 \mathrm{g} \mathrm{H}_{2}\) and \(1.06 \mathrm{g} \mathrm{H}_{2} \mathrm{S}\) in a 0.500 Lflask comes to equilibrium at \(1670 \mathrm{K}: 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{S}_{2}(\mathrm{g}) \rightleftharpoons\) \(2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) .\) The equilibrium amount of \(\mathrm{S}_{2}(\mathrm{g})\) found is \(8.00 \times 10^{-6}\) mol. Determine the value of \(K_{p}\) at 1670 K.

Step-by-Step Solution

Answer

By converting mass of gases into moles, then determining their pressures using the ideal gas law, and finally including those pressures in the expression for the equilibrium constant Kp, the value of Kp at 1670 K is found.

Step 1 : Calculate moles of reactants

The first step is to calculate the moles of H2 and H2S, using their respective molar masses. For H2, it's 2 g/mol, so there are 1 g / 2 g/mol = 0.5 moles of H2. For H2S, it's 34.08 g/mol, so there are 1.06 g / 34.08 g/mol = 0.0311 moles of H2S.

Step 2 : Chemical Equations

Using the stoichiometry of the balanced chemical reaction, \(2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{S}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\)we can see that 2 moles of hydrogen reacts with 1 mole of sulfur to form 2 moles of hydrogen sulfide.However, we were given that the equilibrium amount of S2 is \(8.00 \times 10^{-6}\)mol ,which would have reacted with \(2*8.00 \times 10^{-6}\)mol= \(1.6 \times 10^{-5} \)mol of H2. Since the initial moles of H2 and H2S were 0.5 mol and 0.0311 mol respectively, at equilibrium, moles of H2 will be 0.5-\(1.6 \times 10^{-5} \)mol=0.499984 mol and moles of H2S will be 0.0311 +\(1.6 \times 10^{-5} \)=0.031116 mol.

Step 3 : Ideal Gas Law

Use the Ideal Gas Law PV=nRT to calculate pressures at equilibrium. Remember that the pressure for an individual gas is the partial pressure. Rearranging the formula for pressure we get P=nRT/V. For \( R=0.0821 L atm / K mol \), \( T=1670 K \), and \( V=0.500L \), calculate the pressure for each gas.

Step 4 : Equilibrium Constant Kp

Plug the pressures obtained in step 3 into the equilibrium constant Kp expression which from the balanced chemical reaction is : \(Kp=(P_{H_{2}S})^2/(P_{H_{2}})^2*P_{S_{2}}\) . This will give the equilibrium constant Kp at 1670 K.

Key Concepts

Equilibrium Constant

When a chemical reaction reaches a state where the concentrations of reactants and products remain constant over time, it is said to be in equilibrium. The equilibrium constant, represented as \(K_p\) for reactions involving gases, is a numerical value that reflects the ratio of the pressures of products to reactants at equilibrium. For the given chemical reaction, \(2 \mathrm{H}_{2} + \mathrm{S}_{2} \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{S}\), the \(K_p\) expression would be:

• \(K_p = \frac{(P_{\mathrm{H}_2\mathrm{S}})^2}{(P_{\mathrm{H}_2})^2 \cdot P_{\mathrm{S}_2}}\)

This equation tells us that the equilibrium constant is influenced by the pressure of hydrogen sulfide (\(\mathrm{H}_2\mathrm{S}\)) squared, divided by the pressure of hydrogen (\(\mathrm{H}_2\)) squared, multiplied by the pressure of sulfur gas (\(\mathrm{S}_2\)).
Understanding how to calculate \(K_p\) is crucial because it helps predict how changes in conditions will affect the system's equilibrium position.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that allows us to relate the quantities of a gas to its volume, temperature, and pressure. The formula is \(PV = nRT\), where:

• \(P\) is the pressure of the gas
• \(V\) is the volume
• \(n\) is the number of moles
• \(R\) is the ideal gas constant (0.0821 L atm / K mol)
• \(T\) is the temperature in Kelvin

To calculate the pressure of each gas in the equilibrium mixture, we rearrange the equation to solve for \(P\):

• \(P = \frac{nRT}{V}\)

This formula ensures that we can find the pressures needed to insert into the \(K_p\) expression. Knowing this relationship is vital for determining the partial pressures, which will further be used in calculating equilibrium constants for chemical reactions.

Stoichiometry

Stoichiometry is like the cookbook of chemistry, providing the 'recipes' necessary to predict how much reactant is needed and how much product will form in a chemical reaction. Every balanced chemical equation shows the proportion of moles of reactants and products. For the equilibrium reaction \(2 \mathrm{H}_{2} + \mathrm{S}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{S}\), the stoichiometry tells us:

• 2 moles of \(\mathrm{H}_2\) react with 1 mole of \(\mathrm{S}_2\) to form 2 moles of \(\mathrm{H}_2\mathrm{S}\).

Knowing this is crucial for calculating how much of each substance is present in equilibrium. In this exercise, based on the stoichiometry, we could determine how many moles of \(\mathrm{H}_2\) would react with a given amount of \(\mathrm{S}_2\), and thereby calculate the changes in the number of moles at equilibrium.

Partial Pressure

In a mixture of gases, the pressure exerted by each individual gas is known as its partial pressure. The total pressure of the gas mixture is the sum of the partial pressures of all the gases present. You can think of partial pressure as the individual 'share' of the total gas pressure each gas contributes. It's calculated using the Ideal Gas Law for each component in the mixture, by plugging in the number of moles, temperature, and volume for the specific gas.
In the context of chemical equilibrium, partial pressures are essential because they are used in the expression for \(K_p\). Each gaseous reactant and product in the balanced equation contributes to the \(K_p\) expression through its partial pressure.

• For example, in the equilibrium of \(\mathrm{H}_2\), \(\mathrm{S}_2\), and \(\mathrm{H}_2\mathrm{S}\), their partial pressures help determine the position of equilibrium at a given temperature and thus allow for the calculation of \(K_p\).

Understanding partial pressures is therefore key for solving equilibrium problems in gaseous systems.