Problem 63 - Chemistry: Principles and Reactions

Question

At \(1800 \mathrm{~K}\), oxygen dissociates into gaseous atoms: $$\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}(g)$$ \(K\) for the system is \(1.7 \times 10^{-8} .\) If one mole of oxygen molecules is placed in a \(5.0\) -L flask and heated to \(1800 \mathrm{~K}\), what percentage by mass of the oxygen dissociates? How many \(\mathrm{O}\) atoms are in the flask?

Step-by-Step Solution

Answer

Answer: The percentage by mass of the oxygen that dissociates is 0.00491%, and the number of dissociated oxygen atoms is approximately \(1.89 \times 10^{21}\) atoms.

Step 1: Write the balanced chemical equation

The given balanced chemical equation is: $$\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}(g)$$

Step 2: Set up an ICE table

We will set up an ICE table to determine the initial, change, and equilibrium amounts in moles for each substance. O2 (g) -> 2O (g) Initial (I) 1 0 Change (C) -x +2x Equilibrium (E) 1-x 2x

Step 3: Use the given equilibrium constant value

We are given the equilibrium constant, K = \(1.7 \times 10^{-8}\). Thus, we can now set up an equilibrium expression from the ICE table: $$K = \frac{[O]^2}{[O_2]}$$ _(Note: Concentrations are expressed in moles/L)_ Substitute initial values from the ICE table and the value of K: $$ 1.7 \times 10^{-8} = \frac{(2x\frac{mol}{5L})^2}{(1-x)\frac{mol}{5L}}$$

Step 4: Solve for x

Now we can solve the equation for x, which represents the moles of O2 dissociated: $$ 1.7 \times 10^{-8} = \frac{4x^2} {(1-x)}$$ Solving this equation yields x ≈ 0.00157.

Step 5: Find the mass percentage dissociated

Knowing that 1 mole of O2 = 32 g, we can calculate the mass percentage of dissociated oxygen. Percentage dissociated = \(\frac{0.00157\times32}{32} \times 100 \%\) ≈ 0.00491 %

Step 6: Calculate the number of O atoms

Now we can determine the number of dissociated O atoms using Avogadro's number (6.022 x 10^23 atoms/mol) and the value of x: Number of O atoms = 2x × 6.022 x 10^23 ≈ 1.89 x 10^21 atoms So, the percentage of oxygen dissociates is 0.00491% and the number of dissociated O atoms in the flask is approximately \(1.89 \times 10^{21}\) atoms.

Key Concepts

Equilibrium Constant

In any chemical reaction that can occur in both directions, that is, a reaction that can proceed forward and backward, a point is reached where the rates of the forward and reverse reactions are equal. This is known as chemical equilibrium. At equilibrium, the concentrations of the reactants and products remain constant over time. To quantify this state, we use the equilibrium constant, denoted as K.

For a general reaction where \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant expression is given by \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \), where the concentrations of the products and reactants are raised to the power of their stoichiometric coefficients. A larger value of K indicates that the equilibrium position favors products and vice versa for a smaller value of K.

In the given exercise, the equilibrium constant \( K = 1.7 \times 10^{-8} \) suggests that very little of the oxygen dissociates at \(1800 \mathrm{K}\), favoring the reactants. This is confirmed by the calculated small percentage of dissociation.

ICE Table Method

The ICE table method is a systematic approach to solving equilibrium problems. 'ICE' stands for Initial, Change, and Equilibrium, and each row of the table captures the corresponding quantities of reactants and products in a reaction:

• Initial: The starting concentrations or moles before the reaction begins.
• Change: The amount by which the reactants and products change as the system moves towards equilibrium.
• Equilibrium: The concentrations or moles of reactants and products when the reaction has reached equilibrium.

By setting up an ICE table and writing an equilibrium expression that relates the equilibrium constant to the concentrations of products and reactants, one can solve for the unknown change, which is often represented as 'x'. This technique simplifies the process of determining how much of each substance is present at equilibrium, as demonstrated in solving the oxygen dissociation problem in the exercise.

Dissociation of Oxygen

Dissociation is a process where a molecule breaks down into smaller entities. Specifically, the dissociation of oxygen (\( \mathrm{O}_2 \)) into atomic oxygen (\( \mathrm{O} \)) can be represented by the equation \( \mathrm{O}_{2}(g) \rightleftharpoons 2\mathrm{O}(g) \). This reaction is endothermic, meaning it absorbs energy from the surroundings, and higher temperatures can drive the reaction further towards the dissociated atoms.

Atomic oxygen is highly reactive due to its unpaired electrons, and these unpaired electrons are what make oxygen atoms important in various chemical and biological processes. The exercise deals with the extent of oxygen dissociation at a high temperature, where most \( \mathrm{O}_2 \) molecules remain intact due to the low equilibrium constant, indicating that such conditions are not sufficient to dissociate a significant amount of molecular oxygen.

Percentage by Mass Calculation

The percentage by mass of a substance within a mixture or a reaction system is a measurement of the concentration of that substance. To calculate the percentage by mass of a dissociated substance, as in the exercise, we can use the mass of the dissociated portion divided by the total mass of the substance before dissociation, multiplied by 100%.

In the context of chemical equilibrium, this calculation helps to understand the extent of a reaction by showing what portion of a reactant has been converted to products. For the oxygen dissociation reaction, the small percentage by mass that dissociates (approximately 0.00491%) indicates that at \(1800 \mathrm{K}\), the reaction favors the reactants and very little product is formed.

Avogadro's Number

Avogadro's number is a fundamental constant in chemistry that reflects the number of particles, usually atoms or molecules, in one mole of a substance. Its value is approximately \(6.022 \times 10^{23} \) particles per mole. This number is crucial in converting between the number of moles of a substance and the actual number of particles it contains.

In practice, Avogadro's number allows us to link the macroscopic world that we can measure, like grams or liters, to the microscopic world of atoms and molecules. When the exercise asks us to calculate how many \( \mathrm{O} \) atoms are in the flask, we use Avogadro's number to convert from moles of atomic oxygen to the number of oxygen atoms, revealing the astonishingly large number of particles involved in chemical reactions, even those with small percentages of dissociation.