Question

When heated at high temperatures, iodine vapor dissociates as follows: $$ \mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{I}(g) $$ In one experiment, a chemist finds that when 0.054 mole of \(\mathrm{I}_{2}\) was placed in a flask of volume \(0.48 \mathrm{~L}\) at \(587 \mathrm{~K},\) the degree of dissociation (i.e., the fraction of \(\mathrm{I}_{2}\) dissociated) was \(0.0252 .\) Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the reaction at this temperature.

Short answer

\(K_c = 2.93 \times 10^{-4}\), \(K_p = 0.0141\).

Step-by-step solution

Determine Moles of Iodine Atoms Formed

First, we need to calculate the moles of iodine atoms (I) formed due to the dissociation of iodine molecules (\(\mathrm{I}_2\)). The degree of dissociation \(\alpha\) is given as 0.0252. The initial moles of \(\mathrm{I}_2\) is 0.054. Therefore, the moles of \(\mathrm{I}_2\) dissociated is \(0.054 \times 0.0252 = 0.00136\). Since each mole of \(\mathrm{I}_2\) produces two moles of \(\mathrm{I}\), the moles of \(\mathrm{I}\) formed are \(2 \times 0.00136 = 0.00272\) moles.

Calculate Final Moles of I2

After dissociation, the moles of \(\mathrm{I}_2\) remaining is the initial moles minus the moles dissociated, which calculates to \(0.054 - 0.00136 = 0.05264\) moles.

Calculate Equilibrium Concentrations

The equilibrium concentration of \(\mathrm{I}_2\) is given by \(\frac{0.05264}{0.48} = 0.1097 \mathrm{~mol/L}\). The equilibrium concentration of \(\mathrm{I}\) is \(\frac{0.00272}{0.48} = 0.00567 \mathrm{~mol/L}\).

Calculate Kc

The expression for the equilibrium constant \(K_c\) is \(K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]}\). Substituting the equilibrium concentrations, we have: \(K_c = \frac{(0.00567)^2}{0.1097} = 2.93 \times 10^{-4}\).

Calculate Kp using the Ideal Gas Equation

We use the relationship between \(K_c\) and \(K_p\): \(K_p = K_c (RT)^{\Delta n}\). Here, \(\Delta n = 1\) because \(1\) mole of \(\mathrm{I}_2\) produces \(2\) moles of \(\mathrm{I}\). The temperature \(T\) is 587 K, and \(R\) (ideal gas constant) is 0.0821 L·atm/mol·K. Thus, \(K_p = 2.93 \times 10^{-4} \times (0.0821 \times 587) = 0.0141\).

Key concepts

Dissociation

Dissociation in chemistry refers to the process where molecules break down into smaller particles, such as atoms, ions, or radicals. In the context of iodine vapor, dissociation occurs when iodine molecules \(\text{I}_2\) split into individual iodine atoms, \(2\text{I}\). This transformation happens when iodine vapor is heated to high temperatures, such as at 587 K, as mentioned in the exercise. The degree of dissociation is a measure of the fraction of the initial molecules that have dissociated. For iodine vapor, \(\alpha = 0.0252\) means 2.52% of all iodine molecules dissociate at the given conditions. Understanding dissociation is crucial for calculating the equilibrium constant, which is a key indicator of chemical reaction dynamics.

Equilibrium Constant (Kc and Kp)

The equilibrium constant is a fundamental concept in chemical equilibrium, allowing us to understand the balance between reactants and products in a chemical reaction. There are two types of equilibrium constants -

• \(K_c\): This constant is used when concentrations are measured in molarity (mol/L). It captures the equilibrium state of a reaction in a closed system and is calculated using the formula: \(K_c = \frac{[\text{I}]^2}{[\text{I}_2]}\).
• \(K_p\): This constant applies to situations where pressures are relevant, particularly in gaseous reactions. The formula connecting \(K_p\) and \(K_c\) is: \(K_p = K_c (RT)^{\Delta n}\), where \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) refers to the change in the number of moles of gas.

For the iodine dissociation, we see \(\Delta n = 1\) because one mole of \(\text{I}_2\) becomes two moles of \(\text{I}\). Thus, by calculating \(K_c\) and \(K_p\), we determine the favored reaction state under those specific conditions.

Iodine Vapor

Iodine vapor is formed when solid iodine enters a gaseous state, typically at high temperatures. At 587 K, iodine exists as vapor, which is crucial for observing and studying chemical reactions like dissociation.

Properties of Iodine:

• Atomic number: 53
• Density: 4.933 g/cm³ (as solid)
• Vapor pressure: High enough under increased temperature for relevant experiments

Color and Odor:

Iodine vapor exhibits a violet color and distinct sharp odor, noticeable during experiments.

Applications:

Its properties make iodine vapor valuable in laboratory settings and industrial applications, including antiseptics and photographic materials.
Being aware of how iodine behaves in its vapor form assists in comprehending its role in reactions, such as the dissociation process described in the exercised scenario.