Problem 26 - Chemistry The Central Science

Question

Consider the following equilibrium, for which\(K_{p}=0.0752\) at \(480^{\circ} \mathrm{C} :\) $$2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g)$$ \begin{equation} \begin{array}{l}{\text { (a) What is the value of } K_{p} \text { for the reaction }} \\ {4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ?} \\ {\text { (b) } \mathrm{What} \text { is the value of } K_{p} \text { for the reaction }} \\\ {\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?} \\ {\text { (c) What is the value of } K_{c} \text { for the reaction in part (b)? }}\end{array} \end{equation}

Step-by-Step Solution

Answer

The values of the equilibrium constants for the reactions are as follows: (a) For the reverse reaction, \(K_p^{'} = 13.298\). (b) For the simplified reaction, \(K_p^{''} = 0.274\). (c) For the simplified reaction, \(K_c = 36.87\).

Step 1: Find Kp for the reverse reaction (4 HCl(g) + O2(g) ⇌ 2 Cl2(g) + 2 H2O(g))

To find Kp for the reverse reaction, we simply take the inverse of the given Kp value for the initial reaction. The inverse of Kp for the given reaction will be the Kp for the reverse reaction. $$K_p^{'} = \frac{1}{K_{p}} = \frac{1}{0.0752} = 13.298$$ So, the value of Kp for the reverse reaction is 13.298.

Step 2: Find Kp for the simplified reaction (Cl2(g) + H2O(g) ⇌ 2 HCl(g) + 1/2 O2(g))

To find Kp for the simplified reaction, first notice that it is just half of the initial reaction. When we divide a reaction by a factor, the equilibrium constant is raised to the power of that factor. In this case, we are dividing the initial reaction by 2, so we need to take the square root of the initial Kp value. $$K_p^{''} = \sqrt{K_{p}} = \sqrt{0.0752} = 0.274$$ So, the value of Kp for the simplified reaction is 0.274.

Step 3: Find Kc for the simplified reaction (Cl2(g) + H2O(g) ⇌ 2 HCl(g) + 1/2 O2(g))

To find the value of Kc for the simplified reaction, we first need to know the relationship between Kp and Kc. The two constants are connected by the following formula: $$K_{p} = K_{c}(RT)^{\Delta n}$$ Where R is the gas constant (0.08206 L*atm/mol*K), T is the temperature in Kelvin (480+273.15=753.15K), and Δn is the change in the number of moles of gas between reactants and products (in this case, Δn = 2 - 3 + 0.5 = -0.5). Using the Kp value for the simplified reaction, we can find Kc. $$K_{c} = \frac{K_{p}}{(RT)^{\Delta n}} = \frac{0.274}{(0.08206 \times 753.15)^{-0.5}} = 36.87 $$ So, the value of Kc for the simplified reaction is 36.87.

Key Concepts

Equilibrium Constant

The Equilibrium Constant, represented by either \( K_c \) or \( K_p \), provides vital information about the tendency of a chemical reaction to occur at equilibrium. When a system is at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, meaning that the concentrations of reactants and products remain constant over time.

• \( K_c \) is used when concentrations are in moles per liter (\( mol/L \)).
• \( K_p \) is used when dealing with gases, using partial pressures (usually in atmospheres).

For a given reaction, reversing the reaction inverts the equilibrium constant. For example, if a reaction has \( K_p = 0.0752 \), its reverse will have \( K_p = \frac{1}{0.0752} = 13.298 \). Dividing a reaction by a factor \( n \) requires the equilibrium constant to be taken to the \( n \)th root, such as taking the square root when halving a reaction. These constants provide insights into the composition of the reaction mixture at equilibrium. A large \( K \) value means that, at equilibrium, a significant amount of reactants have converted to products, indicating a product-favored system. Conversely, a small \( K \) value suggests that few reactants have turned into products, signifying a reactant-favored equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle is a useful concept related to equilibrium in chemistry. It predicts how an equilibrium system responds to external changes. According to this principle, if a system at equilibrium experiences a change in concentration, temperature, volume, or pressure, the system will adjust to (partially) counteract that change and a new equilibrium will be established.
When you change the concentration of one of the reactants or products, the equilibrium position shifts to restore balance.

• Adding a reactant causes the system to shift towards the products to consume the added reactant.
• Removing a reactant causes the equilibrium to shift towards the reactants to replace it.

Temperature changes can affect \( K \), depending on whether the reaction is endothermic (absorbs heat) or exothermic (releases heat). In particular:

• Increasing the temperature for an endothermic reaction increases \( K \), as the system wants to absorb more heat by shifting toward the products.
• For exothermic reactions, increasing the temperature typically decreases \( K \), as the system counters extra heat by shifting towards the reactants.

Le Chatelier's Principle helps predict the effects of alterations to the system and manage conditions under which the reaction reaches desired outputs.

Thermodynamics in Chemistry

Thermodynamics in Chemistry involves the study of energy changes during chemical reactions and physical transformations. It's crucial for understanding how reactions proceed under various conditions.
Key thermodynamic concepts include:

• Gibbs Free Energy (\( \Delta G \)): Determines spontaneity of a reaction. When \( \Delta G < 0 \), a process is spontaneous under constant temperature and pressure. Conversely, \( \Delta G > 0 \) means non-spontaneity.
• Enthalpy (\( \Delta H \)): Reflects the heat absorbed or released during a reaction. Exothermic reactions release heat (\( \Delta H < 0 \)), while endothermic reactions absorb heat (\( \Delta H > 0 \)).
• Entropy (\( \Delta S \)): Indicates the degree of disorder or randomness. Systems tend to evolve towards higher entropy.

Thermodynamics also helps explain the temperature dependence of equilibrium constants with the Van't Hoff equation, connecting changes in \( K \) to temperature and enthalpy:\[\ln \left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^0}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]Where \( R \) is the universal gas constant, \( T_1 \) and \( T_2 \) are the initial and final temperatures, and \( \Delta H^0 \) is the enthalpy change. The equation helps understand how shifts in temperature impact the position of equilibrium, crucial for controlling reaction conditions.