Question

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 I(g), how many grams of \(\mathrm{I}_{2}\) are in the mixture?

Short answer

In the equilibrium mixture of \(I_2 (g) \rightleftharpoons 2I (g)\) at 800K and 10.0 L volume, if there are \(2.67 \times 10^{-2}\,\text{g}\) of I(g) present, there are \(3.56 \times 10^{-2}\,\text{g}\) of \(I_2\) present.

Step-by-step solution

Balanced chemical equation

The balanced chemical equation is given as: \[I_2 (g) \rightleftharpoons 2I (g).\]

Convert mass of I into moles

We're given that the mass of I (g) in the equilibrium mixture is \(2.67 \times 10^{-2}\mathrm{g}\). To find the moles of I, we use the molar mass of Iodine, which is 127 g/mol. Moles of I = \(\frac{mass}{molar\,mass}\) Moles of I = \(\frac{2.67\times10^{-2}\,\text{g}}{127\,\text{g/mol}}\) Moles of I = \(2.1 \times 10^{-4}\,\text{mol}\)

Find moles of I2 at equilibrium using the equilibrium constant expression

We can write the expression for the equilibrium constant \(K_c\) for the given reaction as: \[K_c = \frac{[\text{I}]^2}{[\text{I}_2]}\] We have to find moles of I2 in the equilibrium mixture. Since the volume of the vessel is 10.0 L, we can use molar concentrations and write the expression for \(K_c\) as \[K_c = \frac{(\frac{moles\, of\, I}{volume})^2}{\frac{moles\, of\, I_2}{volume}}.\] We can rearrange this expression to isolate moles of I2: Moles of I2 = \(\frac{volume \times (moles\, of\, I)^2}{(K_c \times volume^2)}\) Moles of I2 = \(\frac{10.0\, \text{L} \times (2.1 \times 10^{-4}\,\text{mol})^2}{(3.1\times 10^{-5} \times (10.0\, \text{L})^2)}\) Moles of I2 = \(1.4 \times 10^{-4}\,\text{mol}\)

Convert moles of I2 into grams

Now, we need to convert moles of I2 into grams using the molar mass of I2: Mass of I2 = moles of I2 × molar mass of I2 = \(1.4 \times 10^{-4}\,\text{mol}\) × \(254\,\text{g/mol} = 3.56 \times 10^{-2}\,\text{g}\) So, there are \(3.56 \times 10^{-2}\,\text{g}\) of I2 in the equilibrium mixture.

Key concepts

Equilibrium Constant

Imagine a tug-of-war between two equally matched teams. No one side wins; they're in a deadlock, or 'equilibrium.' In chemistry, reactions can reach a similar state, known as chemical equilibrium, where the rates of the forward and reverse reactions are balanced. At this point, the concentrations of the reactants and products remain constant over time.

The numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium is called the equilibrium constant, denoted as \(K_c\). In our specific exercise, the reaction between iodine molecules \(I_2(g)\) and iodine atoms I(g) has an equilibrium constant \(K_c=3.1 \times 10^{-5}\). This small value suggests that, at equilibrium, the concentration of reactants (iodine molecules) is much higher than that of the products (iodine atoms), hinting that at 800 K, the reaction favors the formation of \(I_2(g)\) over I(g).

The equilibrium constant is crucial because it allows us to predict the direction in which a reaction will proceed and calculate the concentrations of reactants and products at equilibrium.

Molar Mass

In chemistry, molar mass acts like the 'recipe' that tells us how much a single 'ingredient' (atom or molecule) weighs. It's a bridge that helps us translate back and forth between the mass of a substance and the number of atoms or molecules it contains, which we measure in 'moles'.

For instance, in the given problem, the molar mass of an iodine atom (I) is crucial. It's provided as 127 g/mol. This means every mole of iodine atoms weighs 127 grams. We also look at the molar mass of an iodine molecule (\(I_2\)), which is double the molar mass of a single atom since \(I_2\) consists of two iodine atoms, making it 254 g/mol. Understanding the molar mass enables us to convert between grams and moles, an essential step in solving equilibrium problems.

Conversion of Mass to Moles

Alchemists of old sought a mystical 'Philosopher's Stone' to turn lead into gold, while modern chemists use the molar mass to turn mass into moles. This concept is fundamental when dealing with chemical reactions since reactions occur on the atomic or molecular level, not based on the mass of the substances involved.

To convert mass to moles, we divide the given mass by the molar mass of the substance, as shown in our exercise. Here's the magical formula: \( Moles = \frac{mass}{molar\,mass} \). This step is essential because the equilibrium constant expression is based on concentrations (moles per liter), not on mass. Applying this conversion to 2.67 x 10^-2 g of iodine atoms, we calculated the moles of I to be a very small number, 2.1 x 10^-4 moles, which is the starting point for finding out how much \(I_2\) is in the mixture.

Equilibrium Concentration Calculation

Now, onto the grand finale of our equilibrium performance: calculating the equilibrium concentrations. It's like solving a mystery using clues left behind at the scene (the equilibrium constant and moles of substances). We've done just that in step 3 of our solution, deciphering how many moles of \(I_2\) are present at equilibrium within the fixed volume of the container.

We rearrange the equilibrium constant expression to solve for the moles of \(I_2\), taking into account the volume of the reaction vessel, 10.0 L. Then with a bit of algebraic magic, we find the moles of \(I_2\), and finally, convert those moles back into grams using the molar mass. Through this calculation, we've determined the equilibrium mixture contains 3.56 x 10^-2 g of \(I_2\). Knowing how to calculate equilibrium concentrations is a fundamental skill for any chemist and is essential for predicting how chemical systems will behave in different conditions.