/*! 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 41 (a) At \(1285^{\circ} \mathrm{C}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) At \(1285^{\circ} \mathrm{C}\) the equilibrium constant for the reaction \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) is \(K_{c}=1.04 \times 10^{-3} .\) A \(0.200-\mathrm{L}\) vessel containing an equilibrium mixture of thegases has \(0.245 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\mathrm{Br}(g)\) in the vessel? (b) For the reaction \(\mathrm{H}_{2}(g)+\mathrm{l}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g), K_{c}=55.3\) at \(700 \mathrm{~K}\). In a 2.00-L flask containing an equilibrium mixture of the three gases, there are \(0.056 \mathrm{~g} \mathrm{H}_{2}\) and \(4.36 \mathrm{~g} \mathrm{I}_{2}\). What is the mass of HI in the flask?

Short Answer

Expert verified
In summary, for part (a), we calculate the initial concentration of Brâ‚‚, set up the equilibrium expression using given \(K_c\), and solve for x to find the mass of Br which is approximately 0.0111 g. For part (b), we calculate the initial concentrations of Hâ‚‚ and Iâ‚‚, set up the equilibrium expression using given \(K_c\), and solve for x to find the mass of HI which is approximately 1.6 g.

Step by step solution

01

Identify the balanced equation and moles of the given substance

The balanced equation for this reaction is: \[\mathrm{Br}_{2}(g)\rightleftharpoons 2 \mathrm{Br}(g)\] We are given 0.245 g of Brâ‚‚ and we need to convert it into moles. To do this, we use the molar mass of Brâ‚‚ (which is approximately 160 g/mol): Moles of Brâ‚‚ = \[\frac{0.245 g}{160.0 g/mol}\]
02

Calculate initial and equilibrium concentrations of reactants and products

Now that we have the moles of Brâ‚‚, we can calculate its concentration in the given vessel: \[Initial\;concentration\;of\;Br_{2} = \frac{moles}{volume} = \frac{0.245g/160.0g/mol}{0.2 L}\] Since it's an equilibrium mixture, the concentration of Br decreases by x amount and the concentration of Brâ‚‚ will increase by 2x. At equilibrium, \[Br_{2} \to 2Br\Rightarrow Initial \; \; \; \ \; Decreased\,by\,x \; \ \ \ Increased\,by\,2x\]
03

Use the equilibrium constant to set up an equation

Due to equilibrium, we have the following expression for equilibrium constant \(K_c\): \(K_c = \frac{[Br]^2}{[Br_{2}]}\) We are given the value of equilibrium constant as \(K_c= 1.04 \times 10^{-3}\). So, substituting given values in the equation: \[\frac{(2x)^2}{(Initial\,concentration - x)} = 1.04 \times 10^{-3}\]
04

Solve for x and find mass of Br

Now, we have to solve the above equation for x (which represents moles of Br). After obtaining x, we can convert moles of Br to mass. The mass of Br can be calculated using the following equation: \[Mass = Moles \times Molar\,Mass \] #Part b:#
05

Identify the balanced equation and moles of the given substance

The balanced equation for this reaction is: \[\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)\rightleftharpoons 2 \mathrm{HI}(g)\] We are given 0.056 g Hâ‚‚ and 4.36 g Iâ‚‚ and we need to convert them into moles by using the molar masses (Hâ‚‚: 2 g/mol, Iâ‚‚: 254 g/mol): Moles of Hâ‚‚ = \[\frac{0.056 g}{2 g/mol}\] Moles of Iâ‚‚ = \[\frac{4.36 g}{254 g/mol}\]
06

Calculate initial and equilibrium concentrations of reactants and products

Now that we have the moles of Hâ‚‚ and Iâ‚‚, we can calculate their concentrations in the given vessel: \[Initial\;concentration\;of\;H_{2} = \frac{moles}{volume} = \frac{0.056g/2g/mol}{2 L}\] \[Initial\;concentration\;of\;I_{2} = \frac{moles}{volume} = \frac{4.36g/254g/mol}{2 L}\] Since it's an equilibrium mixture, the concentration of Hâ‚‚ and Iâ‚‚ will decrease by x amount while the concentration of HI will increase by 2x. At equilibrium, \[H_{2} + I_{2} \longleftrightarrow 2HI \Rightarrow \;Initial\; \;\,Few\,decreased\,by\,x\; \,Increased\,by\,2x\]
07

Use the equilibrium constant to set up an equation

Due to equilibrium, we have the following expression for equilibrium constant \(K_c\): \[K_c = \frac{[HI]^2}{[H_{2}][I_{2}]}\] We are given the value of equilibrium constant as \(K_c=55.3\). So, substituting given values in the equation: \[\frac{(2x)^2}{(Initial\,concentration\,of\, H_{2} - x)(Initial\,concentration\,of\, I_{2}-x)} = 55.3\]
08

Solve for x and find mass of HI

Now, we have to solve the above equation for x (which represents moles of HI). After obtaining x, we can convert moles of HI to mass. The mass of HI can be calculated using the following equation: \[Mass = Moles \times Molar\,Mass\]

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.

Equilibrium Constant
In the world of chemistry, the concept of an equilibrium constant, denoted as \( K_c \), is a crucial element in understanding how chemical reactions occur and balance out under specific conditions. The equilibrium constant provides a quantitative measure of the position of equilibrium in a chemical reaction. It is derived from the concentrations of the reactants and products involved when the reaction is at equilibrium. The constant essentially tells us the ratio of products to reactants. If \( K_c \) is large, it indicates that, at equilibrium, the products are favored. Conversely, a small \( K_c \) suggests that the reactants are predominant. Keep in mind that \( K_c \) is a constant at a given temperature, hence it is dependent on the reaction temperature.
Molar Concentration
Molar concentration, often called molarity, is a way of expressing the amount of a substance present in a solution. It is calculated as the number of moles of a substance divided by the volume of the solution in liters.When working through chemical equilibrium problems, molar concentration plays a critical role as it directly influences the calculation of the equilibrium constant. The formula for molarity (M) is as follows:\[ M = \frac{n}{V} \]where
  • \( n \) represents the number of moles of solute,
  • \( V \) denotes the volume of solution in liters.
In equilibrium problems, knowing initial molar concentrations helps set up the equilibrium expression needed to solve for unknown variables, ultimately aiding in the determination of unknown masses, as seen in the example exercises.
Chemical Reactions
Chemical reactions involve the breaking and forming of chemical bonds, resulting in the transformation of substances. These reactions are typically represented by balanced equations showing reactants turning into products. In equilibrium reactions, unlike simple transformations that go to completion, the products can revert back into reactants. This situation is known as a dynamic equilibrium. It means that, although the system has reached a stable state, reactions continue to occur in both directions at equal rates. During equilibrium states:
  • The concentrations of reactants and products remain constant over time,
  • The rate of the forward reaction equals the rate of the reverse reaction.
Understanding the behavior of molecules in these reactions helps us manipulate conditions to reach desired outcomes, like increasing product yield.
Stoichiometry
Stoichiometry is the calculation process in chemistry that deals with the relative quantities of reactants and products in a chemical reaction. It's essential for predicting how much material you'll need or obtain from a reaction.In chemical equations, stoichiometric coefficients are used to represent the number of moles of each substance involved in the reaction. The balanced equation governs how these quantities relate to one another. For example, the equation \( A + 2B \rightarrow C \) implies that 1 mole of \( A \) reacts with 2 moles of \( B \) to produce 1 mole of \( C \).Stoichiometric calculations often involve converting grams to moles so that ratios can be compared directly. The exercises demonstrate using stoichiometry by finding initial moles of substances and tracking their changes to compute unknowns at equilibrium. This allows us to determine how much of each substance is present at equilibrium, ultimately relating back to concepts like equilibrium constants and molar concentration.

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

\(\mathrm{NiO}\) 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 150 torr is to be employed in the furnace and total pressure never exceeds 760 torr, will reduction occur?

Write the expressions for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\) (b) \(\mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{TiCl}_{4}(l)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g)\) (d) \(\mathrm{C}(s)+2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})\) (e) \(4 \mathrm{HCl}(a q)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}_{2}(g)\)

Consider the equilibrium \(\mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{SO}_{2}(g) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{SO}_{3}(\mathrm{~s}) .\) (a) Write the equilibrium- constant expression for this reaction in terms of partial pressures. (b) Why doesn't the concentration of \(\mathrm{Na}_{2} \mathrm{O}\) appear in the equilibrium-constant expression?

Consider the following equilibrium, for which \(K_{p}=0.0752\) at \(480^{\circ} \mathrm{C}\) $$ 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) $$ (a) What is the value of \(K_{p}\) for the reaction \(4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ?\) (b) What is the value of \(K_{p}\) for the reaction \(\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?\) (c) What is the value of \(K_{c}\) for the reaction in part (b)?

For the equilibrium $$ 2 \operatorname{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) $$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C}\). If \(0.025 \mathrm{~atm}\) of \(\mathrm{IBr}\) is placed in a 2.0-L container, what is the partial pressure of this substance after equilibrium is reached?

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.