Consider the following equilibrium, for which K p = 0.0752 at 480°C: 2 Cl 2 (g) + 2 H 2 O(g) ⇌ 4 HCl(g) + O 2 (g) (b) What is the value of K p for the reaction Cl 2 (g) + H 2 O(g) ⇌ 2 HCl(g) + 1/2 O2(g)?

Welcome back everyone. The reaction to so three plus N two in equilibrium with two, no three plus S two has KP of 4.5 multiplied by 10, the power of negative seventh at 1000. Kelvin. What is the KP of the reaction below at the same temperature? Let's analyze the original equilibrium. And it tells us that we have two moles of sulfur trioxide 03, they react with one mole of nitrogen and we are producing two moles of no three and one mole of s too, let's call the equilibrium constant of this reaction. KP one. Our target equation, would it be two, no three plus S two in equilibrium with CSO three plus and two. Let's call the equilibrium constant as KKP two. We want to understand the relationship between these equations. And what we notice is that the reactants of reaction number one become our products and the products of reaction number one become our reactants in reaction. Number two. So we're essentially flipping reaction number one, if we are flipping reaction number one to obtain reaction number two, this is equivalent to multiplying reaction number one by a negative one. This is how we are obtaining reaction number two. Simply speaking, we're taking the reciprocal value of the equilibrium constant to get the second equilibrium constant. So whenever we are reversing our equilibrium, the new equilibrium constant KP two is simply the original equilibrium constant KP one raised to the power of negative one or the reciprocal valley of KP one. So we simply want to take the reciprocal value of 4.5 multiplied by sense, the power of negative seventh, which gives us 2.2 multiplied by 10, the power of sixth. And that's the equilibrium constant of the reverse reaction. Thank you for watching.

Ch.15 - Chemical Equilibrium

All textbooks

Brown 14th Edition

Ch.15 - Chemical Equilibrium

Problem 26b

Chapter 15, Problem 26b

Consider the following equilibrium, for which K_p= 0.0752 at 480°C: 2 Cl₂(g) + 2 H₂O(g) ⇌ 4 HCl(g) + O₂(g) (b) What is the value of K_p for the reaction Cl₂(g) + H₂O(g) ⇌ 2 HCl(g) + 1/2 O2(g)?

Verified step by step guidance

Understand that the equilibrium constant Kp is dependent on the stoichiometry of the balanced chemical equation. The given reaction is: 2 Cl2(g) + 2 H2O(g) ⇌ 4 HCl(g) + O2(g) with Kp = 0.0752.

Recognize that the new reaction Cl2(g) + H2O(g) ⇌ 2 HCl(g) + 1/2 O2(g) is derived by halving the coefficients of the original reaction. This means the stoichiometry of the reaction has changed.

Recall that when the coefficients of a balanced equation are multiplied by a factor, the equilibrium constant is raised to the power of that factor. In this case, the coefficients are halved, which corresponds to raising Kp to the power of 1/2.

Express the mathematical relationship for the new Kp: Kp(new) = (Kp(original))^(1/2).

Substitute the given Kp value into the equation: Kp(new) = (0.0752)^(1/2). This will give you the equilibrium constant for the new reaction.

Verified Solution

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Was this helpful?

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 Kp value less than 1 indicates that reactants are favored at equilibrium, while a value greater than 1 suggests that products are favored.

Reaction Stoichiometry

Reaction stoichiometry refers to the quantitative relationship between the amounts of reactants and products in a chemical reaction, as represented by the balanced chemical equation. In this case, the stoichiometry of the reactions must be carefully considered to relate the Kp values of different but related reactions, as they can be derived from one another by manipulating the coefficients of the balanced equations.

Manipulating Equilibrium Expressions

When dealing with equilibrium constants, it is essential to understand how to manipulate equilibrium expressions. If a reaction is reversed, the new Kp is the reciprocal of the original. If the coefficients of a balanced equation are multiplied by a factor, the Kp is raised to the power of that factor. This principle allows for the calculation of Kp for related reactions based on known values.

Recommended video:

Guided course

03:20

Equilibrium Constant Expressions

Related Practice

Open Question

At 1000 K, 𝐾_𝑝= 1.85 for the reaction SO₂(𝑔) + 1/2 O₂(𝑔) ⇌ SO₃(𝑔) (a) What is the value of K_p for the reaction SO₃(𝑔) ⇌ SO₂(𝑔) + 1/2 O₂(𝑔)? (b) What is the value of K_p for the reaction 2 SO₂(𝑔) + O₂(𝑔) ⇌ 2 SO₃(𝑔)?

Textbook Question

At 1000 K, 𝐾_𝑝= 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)?

1087

views

Textbook Question

Consider the following equilibrium, for which 𝐾_𝑝= 0.0752 at 480°C: 2 Cl₂(𝑔) + 2 H₂O(𝑔) ⇌ 4 HCl(𝑔) + O₂(𝑔) (a) What is the value of 𝐾_𝑝 for the reaction 4 HCl(𝑔) + O₂(𝑔) ⇌ 2 Cl₂(𝑔) + 2 H₂O(𝑔)?

1045

views

Textbook Question

The following equilibria were attained at 823 K:

CoO(s) + H₂(g) → Co(s) + H₂O(g) K_c = 67

CoO(s) + CO(g) → Co(s) + CO₂(g) K_c = 490

Based on these equilibria, calculate the equilibrium constant for H₂(g) + CO₂(g) → CO(g) + H₂O(g) at 823 K.

127

views

Textbook Question

Consider the equilibrium N₂(𝑔) + O₂(𝑔) + Br₂(𝑔) ⇌ 2 NOBr(𝑔) Calculate the equilibrium constant 𝐾𝑝 for this reaction, given the following information at 298 K:

2 NO(𝑔) + Br₂(𝑔) ⇌ 2 NOBr(𝑔) 𝐾_𝑐= 2.02

NO(𝑔) ⇌ N₂(𝑔) + O₂(𝑔) 𝐾_𝑐= 2.1×10³⁰

989

views

Open Question

Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: 2 Hg2O(s) ⇌ 4 Hg(l) + O2(g). (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibrium-constant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

Consider the following equilibrium, for which Kp = 0.0752 at 480°C: 2 Cl2(g) + 2 H2O(g) ⇌ 4 HCl(g) + O2(g) (b) What is the value of Kp for the reaction Cl2(g) + H2O(g) ⇌ 2 HCl(g) + 1/2 O2(g)?

Recommended similar problem, with video answer:

Verified Solution

Key Concepts

Equilibrium Constant (Kp)

Reaction Stoichiometry

Manipulating Equilibrium Expressions

Consider the following equilibrium, for which K_p= 0.0752 at 480°C: 2 Cl₂(g) + 2 H₂O(g) ⇌ 4 HCl(g) + O₂(g) (b) What is the value of K_p for the reaction Cl₂(g) + H₂O(g) ⇌ 2 HCl(g) + 1/2 O2(g)?