For each of the equilibria in Problem 15.57, write the equilibrium constant expression for Kp and give the equation that relates Kp and Kc for parts (a), (b), and (c).

Verified step by step guidance

Step 1: Identify the chemical equilibrium reactions given in Problem 15.57. Each reaction will have a specific form, such as aA(g) + bB(g) ⇌ cC(g) + dD(g).

Step 2: Write the equilibrium constant expression for Kp for each reaction. Kp is expressed in terms of the partial pressures of the gases involved: Kp = (P_C^c * P_D^d) / (P_A^a * P_B^b), where P represents the partial pressure of each gas.

Step 3: Recall the relationship between Kp and Kc. The equation that relates Kp and Kc is: Kp = Kc(RT)^Δn, where R is the ideal gas constant (0.0821 L·atm/mol·K), T is the temperature in Kelvin, and Δn is the change in moles of gas (moles of gaseous products - moles of gaseous reactants).

Step 4: For each reaction, calculate Δn by subtracting the sum of the coefficients of the gaseous reactants from the sum of the coefficients of the gaseous products.

Step 5: Use the relationship Kp = Kc(RT)^Δn to express Kp in terms of Kc for each reaction, substituting the appropriate values for R, T, and Δn.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium Constant (Kp and Kc)

The equilibrium constant (K) quantifies the ratio of the concentrations of products to reactants at equilibrium. Kp is used for gas-phase reactions and is expressed in terms of partial pressures, while Kc is used for reactions in solution and is expressed in terms of molar concentrations. The relationship between Kp and Kc is crucial for converting between these two forms, especially when dealing with gaseous reactions.

Relationship between Kp and Kc

The relationship between Kp and Kc is given by the equation Kp = Kc(RT)^(Δn), where R is the ideal gas constant, T is the temperature in Kelvin, and Δn is the change in the number of moles of gas (moles of products minus moles of reactants). This equation allows chemists to convert between the two constants depending on the conditions of the reaction, particularly when dealing with gaseous reactants and products.

Equilibrium Expressions

Equilibrium expressions are mathematical representations of the concentrations or partial pressures of reactants and products at equilibrium. For a general reaction aA + bB ⇌ cC + dD, the equilibrium expression is written as K = [C]^c[D]^d / [A]^a[B]^b for Kc, and similarly for Kp using partial pressures. Understanding how to construct these expressions is essential for calculating equilibrium constants and analyzing chemical equilibria.

Recommended video:

Guided course

03:20

Equilibrium Constant Expressions

Related Practice

Open Question

For each of the following equilibria, write the equilibrium constant expression for Kc. (a) CH4(g) + H2O(g) ⇌ CO(g) + 3 H2(g) (b) 3 F2(g) + Cl2(g) ⇌ 2 ClF3(g) (c) H2(g) + F2(g) ⇌ 2 HF(g)

Open Question

For each of the following equilibria, write the equilibrium constant expression for Kc. (a) 2 C2H4(g) + O2(g) ⇌ 2 CH3CHO(g) (b) CO(g) + H2O(g) ⇌ CO2(g) + H2(g) (c) 4 NH3(g) + 5 O2(g) ⇌ 4 NO(g) + 6 H2O(g)

Textbook Question

For each of the equilibria in Problem 15.56, write the equi-librium constant expression for Kp and give the equation that relates Kp and Kc. (a)

The reaction 2 AsH31g2 ∆ As21g2 + 3 H21g2 has Kp = 7.2 * 107 at 1073 K. At the same temperature, what is Kp for each of the following reactions? (a) As21g2 + 3 H21g2 ∆ 2 AsH31g2

340

views

Textbook Question

The vapor pressure of water at 25 °C is 0.0313 atm. Cal- culate the values of Kp and Kc at 25 °C for the equilibrium H2O1l2 ∆ H2O1g2.

1116

views

Open Question

Chlorine monoxide and dichlorine dioxide are involved in the catalytic destruction of stratospheric ozone. They are related by the equation 2 ClO(g) ⇌ Cl2O2(g) for which Kc is 4.96 * 10^11 at 253 K. For an equilibrium mixture in which [Cl2O2] is 6.00 * 10^-6 M, what is [ClO]?