Textbook Question
Consider the following equilibrium: 2 H2(g) + S2(g) ⇌ 2 H2S(g) Kc = 1.08 × 107 at 700°C (b) Does the equilibrium mixture contain mostly H2 and S2 or mostly H2S?
Ch.15 - Chemical Equilibrium
Chapter 15, Problem 25
At 1000 K, Kp = 1.85 for the reaction SO2(g) + 1/2 O2(g) ⇌ SO3(g). (a) What is the value of Kp for the reaction SO3(g) ⇌ SO2(g) + 1/2 O2(g)? (b) What is the value of Kp for the reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g)?
Verified step by step guidance1
To find the value of Kp for the reverse reaction SO3(g) ⇌ SO2(g) + 1/2 O2(g), we need to take the reciprocal of the given Kp for the forward reaction. This is because the equilibrium constant for a reverse reaction is the inverse of the equilibrium constant for the forward reaction.
For part (b), we need to determine the Kp for the reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g). Notice that this reaction is the original reaction multiplied by 2. When a reaction is multiplied by a coefficient, the equilibrium constant is raised to the power of that coefficient.
Therefore, for the reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g), the Kp is the square of the Kp for the original reaction SO2(g) + 1/2 O2(g) ⇌ SO3(g).
To summarize, for part (a), calculate the reciprocal of the given Kp to find the Kp for the reverse reaction. For part (b), square the given Kp to find the Kp for the reaction with doubled coefficients.
Ensure to express the final Kp values with appropriate units if necessary, and verify the calculations by considering the stoichiometry and the principles of equilibrium constants.
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Equilibrium Constant (Kp)
The equilibrium constant, Kp, is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. It is calculated using the partial pressures of gases involved in the reaction. A higher Kp value indicates a greater concentration of products at equilibrium, while a lower value suggests a higher concentration of reactants.
Recommended video:
Guided course
03:20
Equilibrium Constant Expressions
Reaction Direction and Kp
When a reaction is reversed, the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. This means if Kp for a reaction is known, Kp for the reverse reaction can be easily calculated as 1/Kp. Understanding this relationship is crucial for determining Kp values for related reactions.
Recommended video:
Guided course
02:10Calculate Kp Example
Stoichiometry and Kp
The stoichiometry of a reaction affects the equilibrium constant. When the coefficients of a balanced chemical equation are multiplied by a factor, the equilibrium constant is raised to the power of that factor. For example, if a reaction is doubled, the new Kp is Kp^2. This principle is essential for calculating Kp for reactions that are multiples of the original reaction.
Recommended video:
Guided course
02:10Calculate Kp Example
Related Practice
Textbook Question
Consider the following equilibrium: 2 H2(g) + S2(g) ⇌ 2 H2S(g) Kc = 1.08 × 107 at 700°C (c) Calculate the value of 𝐾𝑐 if you rewrote the equation H2(g) + 1/2 S2(g) ⇌ H2S(g)
Textbook Question
At 1000 K, 𝐾𝑝 = 1.85 for the reaction SO2(𝑔) + 12 O2(𝑔) ⇌ SO3(𝑔) (c) What is the value of 𝐾𝑐 for the reaction in part (b)?
Textbook Question
Consider the following equilibrium, for which 𝐾𝑝 = 0.0752 at 480°C: 2 Cl2(𝑔) + 2 H2O(𝑔) ⇌ 4 HCl(𝑔) + O2(𝑔) (a) What is the value of 𝐾𝑝 for the reaction 4 HCl(𝑔) + O2(𝑔) ⇌ 2 Cl2(𝑔) + 2 H2O(𝑔)?