/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 123 A student studies the equilibriu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A student studies the equilibrium $$ \mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}) $$ at a high temperature. She finds that the total pressure at equilibrium is \(40 . \%\) greater than it was originally, when only \(\mathrm{I}_{2}\) was present at a pressure of \(1.00 \mathrm{~atm}\) in the same sealed container. Calculate \(K_{\mathrm{p}}\).

Short Answer

Expert verified
\( K_p \approx 1.07 \)

Step by step solution

01

Initial setup of pressures

Initially, only \( \text{I}_2 \) is present at 1.00 atm. At equilibrium, the total pressure increases by 40%, so it becomes \( 1.00 \times 1.40 = 1.40 \) atm.
02

Define change in pressures

Let the change in pressure due to the dissociation of \( \text{I}_2 \) be \( x \). Thus, the pressures of \( \text{I}_2 \) and \( \text{I} \) at equilibrium are \( 1.00 - x \) atm and \( 2x \) atm, respectively, because \( 2 \) moles of \( \text{I} \) form for every mole of \( \text{I}_2 \) dissociated.
03

Expression for total equilibrium pressure

Total pressure at equilibrium will be the sum of \( \text{I}_2 \) and \( \text{I} \) pressures: \( (1.00 - x) + 2x = 1.40 \) atm.
04

Solve for x

Simplifying the total pressure expression gives \( 1.00 + x = 1.40 \). Thus, \( x = 0.40 \).
05

Calculate partial pressures at equilibrium

With \( x = 0.40 \), the partial pressures are: \( \text{I}_2 = 1.00 - 0.40 = 0.60 \) atm and \( \text{I} = 2 \times 0.40 = 0.80 \) atm.
06

Calculate the equilibrium constant Kp

The equilibrium constant \( K_p \) is given by \( K_p = \left(\frac{P_{\text{I}_2}}{P_{\text{I}}^2}\right) \). Substituting the equilibrium pressures gives \( K_p = \frac{(0.80)^2}{0.60} = 1.067 \approx 1.07 \).

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products no longer change over time. This does not mean that the reactants and products are in equal amounts, but rather that their rates of formation and decomposition are equal. In other words, the forward and reverse reactions occur at the same rate.

In our example of the iodine dissociation reaction \( \text{I}_2(g) \rightleftharpoons 2 \text{I}(g) \), chemical equilibrium is reached when the rate at which \( \text{I}_2 \) molecules break down into iodine atoms equals the rate at which iodine atoms recombine to form \( \text{I}_2 \). This balance results in stable amounts of both \( \text{I}_2 \) and \( \text{I} \) at the equilibrium state.
  • The system must be closed, meaning no substances enter or leave the container.
  • Equilibrium can be disturbed by changing conditions such as pressure, volume, or temperature.
  • Reaches a point where the Gibbs free energy is minimized.
Partial Pressure
Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. In a system with a mixture of gases, like the iodine dissociation, the total pressure is the sum of individual partial pressures of the gases.

In the dissociation reaction \( \text{I}_2(g) \rightleftharpoons 2 \text{I}(g) \), initially, only \( \text{I}_2 \) is present at 1.00 atm. When equilibrium is reached, the total pressure increased to 1.40 atm—40% greater than the original. To find individual partial pressures at equilibrium:
  • Let the pressure change due to the reaction be \( x \).
  • The partial pressure of \( \text{I}_2 \) is reduced to \( 1.00 - x \).
  • The partial pressure for \( \text{I} \), which forms from the dissociation of \( \text{I}_2 \), becomes \( 2x \).
With \( x \) calculated as 0.40 atm, the partial pressures at equilibrium are determined as follows:
- \( \text{I}_2 = 0.60 \) atm.
- \( \text{I} = 0.80 \) atm.
Dissociation Reaction
A dissociation reaction is a process where a compound separates into two or more components. These reactions are especially common with gases and liquids and often involve breaking of specific chemical bonds.

The dissociation of iodine gas, \( \text{I}_2(g) \), into individual iodine atoms, \( \text{I}(g) \), is one such reaction. In this case, each molecule of \( \text{I}_2 \) dissociates into two iodine atoms, a process that affects the equilibrium position and the total pressure inside the reaction vessel.
  • Dissociation reactions often require energy input to overcome the bond energies.
  • The extent of dissociation can significantly influence the pressure relationships in the system.
  • In this exercise, dissociation results in an increase in total pressure as new, additional molecules are formed.
Understanding how these reactions work helps to predict changes in pressure, composition, and even the equilibrium constant \( K_p \), which in this reaction was calculated to be 1.07.

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

Write equilibrium constant expressions, in terms of reactant and product concentrations, for each of these reactions. $$ \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \quad K_{\mathrm{c}}=1.0 \times 10^{-14} $$ \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})\) $$ \begin{array}{c} K_{\mathrm{c}}=1.8 \times 10^{-5} \\ \mathrm{~N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \end{array} $$ Assume that all gases and solutes have initial concentrations of \(1.0 \mathrm{~mol} / \mathrm{L}\). Then let the first reactant in each reaction change its concentration by \(-x\). (a) Using the reaction table (ICE table) approach, write equilibrium constant expressions in terms of the unknown variable \(x\) for each reaction. (b) Which of these expressions yield quadratic equations? (c) How would you go about solving the others for \(x ?\)

At \(2300 \mathrm{~K}\) the equilibrium constant for the formation of \(\mathrm{NO}(\mathrm{g})\) is \(1.7 \times 10^{-3}\) $$ \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) $$ (a) Analysis of the contents of a sealed flask at \(2300 \mathrm{~K}\) shows that the concentrations of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) are both \(0.25 \mathrm{M}\) and that of \(\mathrm{NO}\) is \(0.0042 \mathrm{M}\). Determine if the system is at equilibrium. (b) If the system is not at equilibrium, in which direction does the reaction proceed? (c) Calculate all three equilibrium concentrations.

At \(25^{\circ} \mathrm{C}\) the vapor pressure of water is \(0.03126 \mathrm{~atm}\). (a) Calculate \(K_{\mathrm{p}}\) and \(K_{\mathrm{c}}\) for $$ \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ (b) Calculate the value of \(K_{\mathrm{p}}\) for this same system at \(100 .{ }^{\circ} \mathrm{C}\) (c) Suggest a general rule for calculating \(K_{\mathrm{p}}\) for any liquid in equilibrium with its vapor at its normal boiling point.

Nitrosyl chloride, NOCl, decomposes to \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) at high temperatures. $$ 2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) $$ Suppose you place \(2.00 \mathrm{~mol} \mathrm{NOCl}\) in a \(1.00-\mathrm{L}\) flask, seal it, and raise the temperature to \(462^{\circ} \mathrm{C}\). When equilibrium has been established, \(0.66 \mathrm{~mol} \mathrm{NO}\) is present. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the decomposition reaction from these data.

Decomposition of ammonium dichromate is shown in the designated series of photos. In a closed container this process reaches an equilibrium state. Write a balanced chemical equation for the equilibrium reaction. How is the equilibrium affected if (a) more ammonium dichromate is added to the equilibrium system? (b) more water vapor is added? (c) more chromium(III) oxide is added?

See all solutions

Recommended explanations on Chemistry Textbooks

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.