Question

Refer to Example \(15-4 . \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) at \(747.6 \mathrm{mmHg}\) pressure and a \(1.85 \mathrm{g}\) sample of \(\mathrm{I}_{2}(\mathrm{s})\) are introduced into a \(725 \mathrm{mL}\) flask at \(60^{\circ} \mathrm{C} .\) What will be the total pressure in the flask at equilibrium? $$\begin{aligned} \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{s}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})+\mathrm{S}(\mathrm{s}) & \\ K_{\mathrm{p}}=& 1.34 \times 10^{-5} \mathrm{at} 60^{\circ} \mathrm{C} \end{aligned}$$

Short answer

The total pressure in the flask at equilibrium is approximately equal to 0.8556 atm.

Step-by-step solution

Calculating Initial Partial Pressures

Firstly, convert the pressure of \( \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) from mmHg to atmospheres, by dividing 747.6 mmHg by 760 mmHg. Hence, the initial pressure \( P_{\mathrm{H}_{2} \mathrm{S}} = 0.9836 \, \text{atm} \), and the pressure of \( \mathrm{I}_{2}(\mathrm{s})\) is not considered since it's a solid.

Calculating number of Moles

Let's find out the number of moles of \( \mathrm{I}_{2}(\mathrm{s})\) . The molecular weight of Iodine is about 253.8 g/mol. Therefore, number of moles \( n_{\mathrm{I}_{2}} = 1.85 \, \text{g} / 253.8 \, \text{g/mol} = 7.28 \times 10^{-3} \, \text{moles} \). Since the reaction dictates that one mole of \( \mathrm{H}_{2} \mathrm{S}\) reacts with one mole of \( \mathrm{I}_{2}\) to give products, we will have this amount of moles reacted for both. The remaining \( \mathrm{H}_{2} \mathrm{S}\) can be found using the Ideal Gas Law. \( P_{\mathrm{H}_{2} \mathrm{S}} V = n_{\mathrm{H}_{2} \mathrm{S}} RT \) This gives us \( n_{\mathrm{H}_{2} \mathrm{S}} = P_{\mathrm{H}_{2} \mathrm{S}} V / RT \). On substituting values, we find \( n_{\mathrm{H}_{2} \mathrm{S}} = 0.0403 \, \text{moles} \), and therefore, the moles of \( \mathrm{H}_{2} \mathrm{S}\) remaining = 0.0403 - 7.28 \times 10^{-3} = 0.03302 moles.

Calculating Equilibrium Pressures

The change in pressure at equilibrium for \( \mathrm{H}_{2} \mathrm{S}\) can be found by using the change in moles and the Ideal Gas Law: our 'change' is \( -0.03302 \, \text{moles} \), therefore \( \Delta P_{\mathrm{H}_{2} \mathrm{S}} = RT ( \Delta n_{\mathrm{H}_{2} \mathrm{S}} ) / V = -0.4278 \, \text{atm} \). However, we cannot have a negative pressure; this means that all of the \( \mathrm{H}_{2} \mathrm{S}\) has reacted. Therefore, the pressure of \( \mathrm{H}_{2} \mathrm{S}\) at equilibrium is 0. The pressure of \( \mathrm{HI} \) at equilibrium can be calculated since it is produced from 2 \( \mathrm{I}_{2}\) : \( P_{\mathrm{HI}} = 2 RT (n_{\mathrm{I}_{2}} ) / V = 2 \times 0.4278 \, \text{atm} = 0.8556 \, \text{atm} \).

Calculating Total Pressure

Now, we can calculate the total pressure in the flask at equilibrium. As we know, the total pressure equals the sum of partial pressures. So, we add the pressure of \( \mathrm{H}_{2} \mathrm{S}\) and the pressure of \( \mathrm{HI}\). Hence, \( P_{\text{total}} = P_{\mathrm{H}_{2} \mathrm{S}} + P_{\mathrm{HI}} = 0 + 0.8556 \, \text{atm} = 0.8556 \, \text{atm} \).

Key concepts

Calculating Equilibrium Pressures

Understanding how to calculate equilibrium pressures is essential for students studying chemical reactions involving gases. To determine the pressure of gases at equilibrium, we consider both the initial pressures and the changes that occur as the reaction proceeds towards equilibrium.

Let's break this down. In a closed system, the total pressure at equilibrium can be influenced by the initial concentrations or pressures of reactants and products, as well as the stoichiometry of the reaction. As the reaction moves forward or backward towards equilibrium, the moles of gases involved in the reaction change, which in turn affects the pressure.

In the given exercise, following the stoichiometry of the reaction, the moles of gases consumed or produced directly affect their partial pressures. Once we understand the mole to mole ratios dictating the reaction, we can work out the new amount of moles at equilibrium and hence calculate the pressures using the Ideal Gas Law. This involves recognizing that the pressure of solids and liquids are not included in the calculations for gases.

Ideal Gas Law Application

The Ideal Gas Law provides a clear mathematical relationship between the pressure (P), volume (V), number of moles (n), and temperature (T) for an ideal gas: \( PV = nRT \). Here, R is the gas constant. This law is a crucial tool when you're working with gases involved in a chemical reaction.

For example, in the textbook exercise, we use the Ideal Gas Law to calculate the moles of \( \mathrm{H}_{2} \mathrm{S} \) at the start. We also use this law to determine the change in pressure as the reaction moves towards equilibrium. The application requires us to be careful with units and conditions since R can have different values depending upon the units used for pressure and volume.

Chemical Reaction Stoichiometry

Stoichiometry is the term we use to describe the quantitative relationship between reactants and products in a chemical reaction. It's like a recipe that tells us how much of each substance we need to react together to create certain products.

In the context of the given problem, stoichiometry tells us that one mole of \( \mathrm{H}_{2} \mathrm{S} \) reacts with one mole of iodine (\( \mathrm{I}_{2} \) - which is solid and thus doesn't exert pressure) to produce two moles of hydrogen iodide gas (\( \mathrm{HI} \) ) and solid sulfur. This molar ratio is fundamental when calculating how much reactant is consumed and product is formed, which in turn affects the pressure changes in the reaction vessel.

Chemical Equilibrium Constant

The concept of the chemical equilibrium constant, denoted as \( K \) or \( K_{p} \) for gases when dealing with partial pressures, is the core of understanding the balance that exists in a reversible reaction. It is a ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients, at equilibrium.

In our exercise scenario, \( K_{p} \) is provided, signifying the extent to which the reaction will favor the formation of products at equilibrium under the given conditions. While we didn't use \( K_{p} \) in the direct calculation of pressures in this particular problem, it serves as an essential guide for predicting the direction of the reaction and for more complex equilibrium calculations. Knowing the equilibrium constant can also allow us to determine concentrations or pressures of constituents at equilibrium when not all the initial conditions are known.