Problem 42 - Chemistry: The Central Science

Question

(a) At \(800 \mathrm{~K}\) the equilibrium constant for \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) is \(K_{c}=3.1 \times 10^{-5} .\) If an equilibrium mixture in a 10.0-L vessel contains \(2.67 \times 10^{-2} \mathrm{~g}\) of \(\mathrm{I}(\mathrm{g})\), how many grams of \(I_{2}\) are in the mixture? (b) For \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), \quad K_{p}=3.0 \times 10^{4} \mathrm{at}\) \(700 \mathrm{~K} .\) In a 2.00-L vessel the equilibrium mixture contains \(1.17 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(0.105 \mathrm{~g}\) of \(\mathrm{O}_{2}\). How many grams of \(\mathrm{SO}_{2}\) are in the vessel?

Step-by-Step Solution

Answer

In the given equilibrium mixture of the reaction \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\), there are \(3.58 × 10^{-2} \mathrm{~g}\) of \(I_2(g)\) present.

Step 1: Calculate the initial concentration of I(g)

The initial concentration of I(g) can be calculated using the formula: \[c = \frac{n}{V},\] where c is the concentration, n is the amount (moles) of the substance, and V is the volume of the vessel. First, we need to convert the mass of I(g) to moles using the molar mass of I(g) (\(126.9 \mathrm{~g/mol}\)): \(n_{\mathrm{I}} = \frac{2.67 × 10^{-2}\mathrm{~g}}{126.9\mathrm{~g/mol}} = 2.10 × 10^{-4}~ \text{moles}\) Now, calculate the concentration: \(c_{\mathrm{I}} = \frac{2.10 × 10^{-4} \mathrm{~mol}}{10.0 \mathrm{~L}} = 2.10 × 10^{-5} \mathrm{~M}\)

Step 2: Write the expression for the equilibrium constant

The equilibrium constant Kc expression for the reaction \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) is: \[K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I_{2}}]},\] where [I] and [I₂] are equilibrium concentrations of I(g) and I₂(g), respectively.

Step 3: Set up an equation to find the concentration of I2(g)

From the equilibrium constant expression, we can write: \([I]^2 = K_c × [I_2]\) Plug in the known values: \((2.10 × 10^{-5})^2 = (3.1 × 10^{-5}) × [I_2]\) Now, solve for [I₂]: \([I_2] = \frac{(2.10 × 10^{-5})^2}{3.1 × 10^{-5}} = 1.41 × 10^{-5}\)

Step 4: Convert the concentration of I2(g) to grams

Now that we have found the concentration of I₂(g), we need to find the mass in grams. First, find the moles of I2(g) and then convert it to grams using the molar mass of I2(g) (\(253.8 \mathrm{~g/mol}\)): n of I₂ = \(1.41 × 10^{-5} \mathrm{~M} × 10.0 \mathrm{~L} = 1.41 × 10^{-4} \mathrm{~mol}\) mass of I₂ = \(1.41 × 10^{-4} \mathrm{~mol} × 253.8 \mathrm{~g/mol} = 3.58 × 10^{-2} \mathrm{~g}\) So, there are \(3.58 × 10^{-2} \mathrm{~g}\) of \(I_2(g)\) in the mixture.

Key Concepts

Equilibrium Constant

In chemistry, the equilibrium constant (represented as Kc for concentrations and Kp for partial pressures) is a number that quantifies the ratio of concentrations of products to reactants at equilibrium. For the reaction \(I_2(g) \rightleftharpoons 2 I(g)\), the equilibrium constant is \(K_c=3.1 \times 10^{-5}\). This low value suggests that, at equilibrium, the reactant (\(I_2\)) is favored over the product (\(I\)) at 800 K.

Understanding the equilibrium constant is crucial because it provides insight into the direction of the reaction and the extent of the reaction at equilibrium. If Kc is much greater than 1, the products are greatly favored, and the reaction proceeds almost to completion. Conversely, if Kc is much less than 1, as in this case, the reactants are favored, and the reaction does not proceed far to the right before reaching equilibrium.

Equilibrium Concentration

The equilibrium concentration refers to the concentration of each reactant and product in a reaction mixture at the state of equilibrium. It's important to calculate these concentrations to understand the composition of an equilibrium mixture. For the provided exercise, the process involves first finding the moles of I(g) from its mass and then using the volume to calculate its molar concentration.

Knowing the equilibrium constant allows us to set up an equation relating the concentration of the iodine molecules (\(I_2\)) to that of the iodine atoms (\(I\)). For example, \[K_c = \frac{[I]^2}{[I_2]}\] becomes the starting point to find the unknown equilibrium concentration of \(I_2\) by substituting the known concentrations and solving for \(I_2\). Getting to grips with finding equilibrium concentrations is essential for predicting how a chemical system will respond to changes in conditions, such as pressure or temperature.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It's a bridging factor between the mass of a substance and the number of moles, which chemists use to count particles since moles reflect amounts on an atomic scale.

In the exercise, molar mass plays a pivotal role in converting mass to moles, which is necessary for calculating concentration. Iodine (I) has a molar mass of \(126.9 g/mol\), allowing the conversion of \(2.67 \times 10^{-2} g\) of I(g) to moles, which is then used to find its concentration in the mixture. Similarly, for iodine molecules \(I_2\), the molar mass is \(253.8 g/mol\), which is needed to convert the calculated moles of \(I_2\) back to grams.

Students should become comfortable using molar mass as it is fundamental to stoichiometric calculations, allowing for conversions between mass and moles, affecting both the understanding of and calculations pertaining to chemical reactions.