Problem 5 - Chemistry and Chemical Reactivity

Question

A mixture of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at \(1000 \mathrm{K}\) contains the gases at the following concentrations: \(\left|\mathrm{SO}_{2}\right|=\) $$5.0 \times 10^{-3} \mathrm{mol} / \mathrm{L},\left|\mathrm{O}_{2}\right|=1.9 \times 10^{-3} \mathrm{mol} / \mathrm{L}, \text { and } $$ \(\left[\mathrm{SO}_{3}\right]=6.9 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Is the reaction at equilibrium? If not, which way will the reaction proceed to reach equilibrium?$$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) K_{c}=279 $$

Step-by-Step Solution

Answer

The reaction is not at equilibrium and will proceed forward to form more \(\mathrm{SO}_3\).

Step 1: Write the Equilibrium Expression

The equilibrium expression for the given reaction \(2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\) is given by: \[ K_c = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2[\mathrm{O}_2]} \]

Step 2: Calculate the Reaction Quotient (Q)

The reaction quotient \(Q\) is calculated using the concentrations of the reactants and products at the given moment: \[ Q = \frac{[6.9 \times 10^{-3}]^2}{[5.0 \times 10^{-3}]^2 \times [1.9 \times 10^{-3}]} \] Calculate each component:- \([\mathrm{SO}_3]^2 = (6.9 \times 10^{-3})^2 = 4.761 \times 10^{-5}\)- \([\mathrm{SO}_2]^2 = (5.0 \times 10^{-3})^2 = 2.5 \times 10^{-5}\)- \([\mathrm{O}_2] = 1.9 \times 10^{-3}\) So, \(Q = \frac{4.761 \times 10^{-5}}{(2.5 \times 10^{-5}) \times (1.9 \times 10^{-3})} \approx 100.305\)

Step 3: Compare Q and Kc

Now compare the calculated value of \(Q\) with the given \(K_c\) which is 279: \(Q \approx 100.305\) and \(K_c = 279\).

Step 4: Determine the Direction of Reaction

Since \(Q < K_c\), the reaction is not at equilibrium. Therefore, the reaction will proceed in the forward direction to reach equilibrium, towards forming more \(\mathrm{SO}_3\).

Key Concepts

Reaction Quotient

The reaction quotient, often symbolized as \(Q\), is used to determine the current state of a chemical reaction in terms of concentrations of reactants and products. It serves as a snapshot of a reaction at any given moment. Unlike the equilibrium constant, which is only applicable at equilibrium, \(Q\) can be calculated at any set of conditions.

To calculate \(Q\), we apply the same formula used to find the equilibrium constant \(K_c\). For the reaction \(2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\), the expression is:

• \(Q = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2[\mathrm{O}_2]}\).

Once you have calculated \(Q\), you compare it with \(K_c\) to find out if the reaction is at equilibrium or which direction it needs to shift to reach equilibrium. If \(Q = K_c\), the system is at equilibrium. A \(Q < K_c\) indicates a shift towards products, while \(Q > K_c\) indicates a shift towards reactants.

Equilibrium Constant

The equilibrium constant, \(K_c\), is a valuable number in understanding chemical reactions. It communicates the ratio of concentrations of products to reactants at equilibrium. Unlike the reaction quotient, \(K_c\) is only calculated when a system has reached equilibrium.

Expressing \(K_c\) depends heavily upon the specific balanced chemical equation under consideration.

• In the equation \(2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\), the equilibrium expression is \(K_c = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2[\mathrm{O}_2]}\).

The value of \(K_c\) provides insight into the position of the equilibrium. A high \(K_c\) value, greater than one, implies that at equilibrium, the products are favored. Conversely, a low \(K_c\), less than one, means reactants are preferred. \(K_c\) is fixed for a given reaction at a certain temperature, making it a critical reference point when using the reaction quotient \(Q\) to determine the direction in which a reaction must proceed.

Le Chatelier's Principle

Le Chatelier's Principle offers a straightforward means of predicting how a change in conditions will affect the position of equilibrium for a chemical reaction. This principle asserts that if a system at equilibrium experiences a change in concentration, temperature, or pressure, the system will adjust to counteract that change.

In practical terms:

• Adding more reactants to a system will shift the equilibrium towards the products.
• Increasing the concentration of products will shift the equilibrium back towards the reactants.
• In endothermic reactions, raising the temperature shifts the equilibrium towards the products, whereas in exothermic reactions, it shifts towards the reactants.
• When it comes to pressure, if the number of moles of gas differs between reactants and products, a change in volume will favor the side with fewer moles of gas.

Understanding Le Chatelier's Principle helps predict how a system will respond when a reaction like \(2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\) needs to reach equilibrium after changes in concentration.