Question

At a certain temperature, \(K=0.29\) for the decomposition of two moles of iodine trichloride, \(\mathrm{ICl}_{3}(s),\) to chlorine and iodine gases. The partial pressure of chlorine gas at equilibrium is three times that of iodine gas. What are the partial pressures of iodine and chlorine at equilibrium?

Short answer

Answer: At equilibrium, the partial pressures of iodine and chlorine gases are 0.3563 atm and 1.069 atm, respectively.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the decomposition of two moles of iodine trichloride to iodine and chlorine gases is: \[2\mathrm{ICl}_3(s) \rightleftharpoons \mathrm{I}_2 \uparrow(g) + 3\mathrm{Cl}_2 \uparrow(g)\]

Write the equilibrium expression

For this reaction, the equilibrium constant K is related to the partial pressures of iodine and chlorine gases at equilibrium: \[K = \frac{[\mathrm{I}_2][\mathrm{Cl}_2]^3}{[\mathrm{ICl}_3]^2}\] However, as the iodine trichloride is in solid form, its concentration does not appear in the equilibrium expression. Therefore, we can simplify the expression to: \[K = [\mathrm{I}_2] [\mathrm{Cl}_2]^3\]

Use the given relation between partial pressures

The problem states that the partial pressure of chlorine gas at equilibrium is three times that of iodine gas. We can represent this relationship algebraically: \[[\mathrm{Cl}_2] = 3[\mathrm{I}_2]\]

Solve the equations for partial pressures

Now we can substitute the relationship between the partial pressures of iodine and chlorine gases into the equilibrium expression: \[K = [\mathrm{I}_2](3[\mathrm{I}_2])^3\] Plug in the given value for K: \[0.29 = [\mathrm{I}_2](3[\mathrm{I}_2])^3\] Now, we will solve for the partial pressure of iodine, \([\mathrm{I}_2]\): \begin{align*} 0.29 &= [\mathrm{I}_2](27[\mathrm{I}_2]^3)\\ \frac{0.29}{27} &= [\mathrm{I}_2]^4\\ \left(\frac{0.29}{27}\right)^{1/4} &= [\mathrm{I}_2]\\ 0.3563&= [\mathrm{I}_2] \end{align*} Now that we have the partial pressure of iodine, we can find the partial pressure of chlorine using the given relationship: \[[\mathrm{Cl}_2] = 3[\mathrm{I}_2] \implies [\mathrm{Cl}_2] = 3(0.3563) = 1.069\] So, at equilibrium, the partial pressures of iodine and chlorine gases are 0.3563 and 1.069 atm, respectively.

Key concepts

Equilibrium Constant

Understanding the equilibrium constant is essential when studying chemical reactions that do not go to completion. The equilibrium constant (\(K\text{eq}\) or simply, K) is a numerical value that expresses the ratio of the concentration of the products to the concentration of the reactants when a chemical reaction is at equilibrium. Each reactant and product in the reaction is raised to the power of its coefficient in the balanced chemical equation.

Let's apply this concept to an exercise related to the chemical reaction where iodine trichloride decomposes into iodine and chlorine gases. When we say that the equilibrium constant, K, is 0.29, that number helps us to quantify the point at which the reactants and the products reach a state of balance. A low K value typically indicates that, at equilibrium, the concentration of reactants is higher than that of the products. In this case, it tells us that the decomposition of iodine trichloride favorably produces a small amount of iodine and chlorine gases at the given temperature.

Partial Pressure

Partial pressure is a crucial concept in understanding gases involved in chemical reactions, especially when they are in a mixture. It is the pressure exerted by an individual gas in a mixture of gases. The total pressure in the system is the sum of the partial pressures of each gas present. Dalton's Law describes how, in a mixture of non-reacting gases, each gas acts as if it were the only gas present and exerts pressure independently.

In order to determine the partial pressures of iodine and chlorine in our exercise, we utilize the given that chlorine's partial pressure is three times that of iodine. This is a critical piece of information because it allows us to set up a direct relationship between their partial pressures and solve for their individual values using the equilibrium constant. It beautifully demonstrates how partial pressures can be related algebraically and quantified to reflect the composition of a gas at equilibrium.

Balanced Chemical Equation

Writing a balanced chemical equation is the foundation of working with chemical reactions. It ensures that the Law of Conservation of Mass is satisfied, meaning the number of atoms for each element is the same on both the reactants and products side. In the given exercise, the balanced chemical equation was \[2\mathrm{ICl}_3(s) \rightleftharpoons \mathrm{I}_2(g) + 3\mathrm{Cl}_2(g)\].Notice every atom on the left has a corresponding atom on the right. It means that two moles of solid iodine trichloride decompose to yield one mole of iodine gas and three moles of chlorine gas.

This balance is not just a matter of bookkeeping; it determines the stoichiometry - or the proportional relationship between reactants and products. These coefficients directly influence the equilibrium expression that we use to calculate the equilibrium constant and the partial pressures. Without a balanced equation, all subsequent calculations in our exercise would be off, making the importance of correctly balancing equations quite clear.