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 equilib- rium? If not, which way will the reaction proceed to reach equilibrium? $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_{3}(\mathrm{g}) K_{\mathrm{c}}=279 $$

Step-by-Step Solution

Answer

The reaction is not at equilibrium; it will shift toward the reactants.

Step 1: Understand the Reaction

The chemical reaction is \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\). This means that the concentration of reactants SO\(_2\) and O\(_2\) will convert into product SO\(_3\) during the reaction. The equilibrium constant, \(K_c\), is given as 279, which indicates the ratio of concentrations of products to reactants at equilibrium.

Step 2: Calculate the Reaction Quotient (Q)

The equation for the reaction quotient \(Q\) is similar to that of the equilibrium constant \(K_c\), but it uses the initial concentrations:\[ Q = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2[\mathrm{O}_2]} \]Substitute the given concentrations:\[ Q = \frac{(6.9 \times 10^{-3})^2}{(5.0 \times 10^{-3})^2 \times (1.9 \times 10^{-3})} \]Calculate \(Q\):\[ Q = \frac{4.761 \times 10^{-5}}{4.75 \times 10^{-8}} = 1.002 \times 10^{3} \]

Step 3: Compare Q to Kc

Compare the reaction quotient \(Q = 1.002 \times 10^{3}\) with \(K_c = 279\). Since \(Q > K_c\), this indicates that the concentration of products is higher than at equilibrium. Therefore, the reaction will shift towards the reactants to achieve equilibrium.

Key Concepts

Reaction Quotient

In chemistry, the reaction quotient \( Q \) is a crucial concept that helps determine the direction in which a chemical reaction will proceed. The reaction quotient is calculated in a similar way to the equilibrium constant \( K_c \), but it utilizes the concentrations of reactants and products at any point in time, not just at equilibrium.
Think of \( Q \) as a snapshot of a reaction's status at a specific moment. It provides a ratio:

• Numerator: Concentration of products, each raised to the power of their coefficients in the balanced equation.
• Denominator: Concentration of reactants, each raised to the power of their coefficients in the balanced equation.

The formula for the given reaction is:\[Q = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2[\mathrm{O}_2]}\]When comparing \( Q \) to \( K_c \), if

• \( Q < K_c \), the reaction will shift towards the products to reach equilibrium.
• \( Q > K_c \), the reaction will shift towards the reactants to achieve equilibrium.

In our exercise, since \( Q = 1.002 \times 10^3 \) and it is greater than \( K_c = 279 \), more reactants will form as the system seeks balance.

Equilibrium Constant

The equilibrium constant \( K_c \) is a central part of understanding chemical reactions that reach an equilibrium state. It is a measure of the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients at equilibrium conditions. For the reaction of \(2 \mathrm{SO}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{g})\), the equilibrium constant is represented as:\[K_c = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2[\mathrm{O}_2]}\]This value provides insights into the position of equilibrium:

• When \( K_c \) is large (greater than 1), the reaction mixture contains more products than reactants.
• When \( K_c \) is small (less than 1), it indicates more reactants than products.

At \(1000 \mathrm{K}\), the equilibrium constant for this reaction is \(279\), indicating that at equilibrium, products are favored over reactants, but not overwhelmingly so.By comparing \( Q \) to \( K_c \), we determine how the system needs to adjust. In our example, because \( Q > K_c \), the system initially contains too much product, prompting a shift to form more reactants to move towards equilibrium.

Chemical Reaction Dynamics

Chemical reaction dynamics explore how and why chemical reactions occur the way they do. The term encompasses both the speed or rate of a reaction and the direction in which it progresses. Reactions can naturally shift towards their reactants or products, governed largely by principles of thermodynamics and kinetics.
**Le Chatelier's Principle**
This principle helps explain the direction shifts when a system experiences a disturbance. If a reaction is disrupted or is not at equilibrium, the system adjusts to minimize the change, either producing more products or reactants as needed.
**Equilibrium and Reaction Dynamics**
In equilibrium, the forward and reverse reactions occur at the same rate. However, when a reaction mixture is not at equilibrium (as determined by comparing \( Q \) with \( K_c \)), the balance is disrupted. This imbalance causes the reaction dynamics to push the system towards a new equilibrium
by altering rates.
For instance, in our given chemical system, initially, \(Q\) is greater than \(K_c\), indicating there are more products than desired at equilibrium. Thus:

• The reaction dynamically shifts towards the reactants. Essentially, it speeds up the reverse rate to decrease the product concentration and increase reactants until equilibrium is restored.

Understanding these dynamics helps in predicting how reactions will respond to changes in conditions.