Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 63

How do the following changes affect the value of the \(K_{P}\) for a gas-phase endothermic reaction: (a) increase in the total pressure by adding a noble gas, \((\mathbf{b})\) addition of a reactant, \((\mathbf{c})\) increase in the temperature (d) increase in the volume, \((\mathbf{e})\) decrease in the temperature?

Short Answer

Expert verified
The effect of the changes on the value of the \(K_P\) for a gas-phase endothermic reaction are: (a) Adding a noble gas does not affect the value of \(K_P\) as it doesn't interact with the reactants. (b) Addition of a reactant doesn't change the value of \(K_P\) as the equilibrium is re-established. (c) An increase in temperature causes the \(K_P\) value to increase as the reaction shifts towards the product side. (d) An increase in volume causes the \(K_P\) value to increase or decrease, depending on whether there are more moles of gas molecules on the product side or the reactant side. (e) A decrease in temperature causes the \(K_P\) value to decrease as the reaction shifts towards the reactant side.

Step by step solution

01

Use Le Chatelier's Principle to Analyze the Effect of Pressure Increase due to Noble Gas

For an endothermic reaction, if we increase the total pressure by adding a noble gas, this won't result in new collisions because noble gases don't interact with the reactants. As a result, the change in pressure won't affect the equilibrium constant \(K_P\). Therefore, there is no change in the value of \(K_P\) in this case.
02

Analyze the Effect of Reactant Addition on the \(K_P\) Value

The addition of a reactant will shift the reaction towards the product according to Le Chatelier's Principle. This will temporarily affect the concentration of reactants and products but at equilibrium, the reaction comes back and the \(K_P\) value remains the same. So, in this case, there is no change in the value of \(K_P\).
03

Determine the Effect of Temperature Increase on the \(K_P\) Value

Increasing the temperature for an endothermic reaction will lead to the reaction absorbing more heat. Le Chatelier's Principle states that the reaction will shift towards the side that can use that extra heat, which means it will shift towards the product side. As the reaction shifts in the direction of the products, the value of \(K_P\) will increase.
04

Explain the Effect of Volume Increase on the \(K_P\) Value

When the volume of the container for a gas-phase endothermic reaction is increased, the pressure decreases. According to Le Chatelier's Principle, the reaction would shift towards the side having larger moles of gaseous components to increase the pressure. Therefore, if the number of moles of gas molecules is larger on the product side, then \(K_P\) will increase. However, if the moles of gas molecules are larger on the reactant side, then \(K_P\) will decrease.
05

Discuss the Effect of Temperature Decrease on the \(K_P\) Value

Decreasing the temperature for an endothermic reaction will result in the reaction absorbing less heat. According to Le Chatelier's Principle, the reaction will shift towards the side that can produce more heat, which means it will shift towards the reactant side. As the reaction shifts in the direction of the reactants, the value of \(K_P\) will decrease.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

As shown in Table \(15.2,\) the equilibrium constant for the reaction $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\( is \)K_{p}=4.23 \times 10^{-7}\( at \)300^{\circ} \mathrm{C}\(. Pure \)\mathrm{NH}_{3}$ is placed in a 1.00-L flask and allowed to reach equilibrium at this temperature. There are $1.05 \mathrm{~g} \mathrm{NH}_{3}$ in the equilibrium mixture. (a) What are the masses of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) in the equilibrium mixture? (b) What was the initial mass of ammonia placed in the vessel? (c) What is the total pressure in the vessel?

\(\mathrm{NiO}\) is to be reduced to nickel metal in an industrial process by use of the reaction $$\mathrm{NiO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{CO}_{2}(g)$$ At \(1600 \mathrm{~K}\), the equilibrium constant for the reaction is $K_{p}=6.0 \times 10^{2} .\( If a CO pressure of \)20 \mathrm{kPa}$ is to be employed in the furnace and total pressure never exceeds \(101.3 \mathrm{kPa}\), will reduction occur?

The following equilibria were attained at \(298 \mathrm{~K}:\) $$\begin{array}{c} \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s) \quad K_{c}=5.6 \times 10^{9} \\ \mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q) \\ K_{c}=1.6 \times 10^{7}\end{array}$$ Based on these equilibria, calculate the equilibrium constant $\text { for } \mathrm{AgCl}(s)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}(a q)+\mathrm{Cl}^{-}(a q)$ at \(298 \mathrm{~K}\).

Which of the following statements are true and which are false? (a) For the reaction $2 \mathrm{~A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{A}_{2} \mathrm{~B}(g) K_{c}$ and \(K_{p}\) are numerically the same. (b) It is possible to distinguish \(K_{c}\) from \(K_{p}\) by comparing the units used to express the equilibrium constant. \((\mathbf{c})\) For the equilibrium in (a), the value of \(K_{c}\) increases with increasing pressure.

Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: $2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)$. (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 equilibriumconstant expression in terms of molarities for the reaction, using (solv) to indicate solvation.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free