/*! 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 15 Write the expression for \(K_{c}... [FREE SOLUTION] | 91Ó°ÊÓ

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Write the expression for \(K_{c}\) for the following reactions. \(\operatorname{In}\) each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) (c) \(\mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g)\) (d) \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) (e) \(\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(I) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q)\) (f) \(\mathrm{Fe}^{2+}(a q)+\mathrm{Zn}(s) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Fe}(s)\) (g) \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q)\)

Short Answer

Expert verified
(a) \[K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]}\] (Homogeneous) (b) \[K_c = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2[\mathrm{O}_2]}\] (Homogeneous) (c) \[K_c = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]}\] (Heterogeneous) (d) \[K_c = \frac{[\mathrm{CO}_2][\mathrm{H}_2]}{[\mathrm{H}_2\mathrm{O}][\mathrm{CO}]}\] (Homogeneous) (e) \[K_c = \frac{[\mathrm{HCO}_3^-][\mathrm{H}^+]}{[\mathrm{CO}_2][\mathrm{H}_2\mathrm{O}]}\] (Heterogeneous) (f) \[K_c = \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Fe}^{2+}]}\] (Heterogeneous) (g) \[K_c = [\mathrm{Ca}^{2+}][\mathrm{CO}_3^{2-}]\] (Heterogeneous)

Step by step solution

01

(a) I2(g) 2 I(g)

For this reaction, the equilibrium constant Kc can be written as: \[K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]}\] Since all the species involved in this reaction are in the gas phase, the reaction is homogeneous.
02

(b) 2 SO2(g) + O2(g) 2 SO3(g)

The equilibrium constant Kc for this reaction is: \[K_c = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2[\mathrm{O}_2]}\] All species in this reaction are in the gas phase, so this is a homogeneous reaction.
03

(c) CO2(g) + C(s) 2 CO(g)

The equilibrium constant Kc can be expressed as: \[K_c = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]}\] This reaction involves both gaseous and solid species, hence it is a heterogeneous reaction.
04

(d) H2O(g) + CO(g) CO2(g) + H2(g)

For this reaction, the equilibrium constant Kc is: \[K_c = \frac{[\mathrm{CO}_2][\mathrm{H}_2]}{[\mathrm{H}_2\mathrm{O}][\mathrm{CO}]}\] All species are in the gas phase, making this a homogeneous reaction.
05

(e) CO2(aq) + H2O(l) HCO3^-(aq) + H^(aq)

The equilibrium constant Kc for this reaction is: \[K_c = \frac{[\mathrm{HCO}_3^-][\mathrm{H}^+]}{[\mathrm{CO}_2][\mathrm{H}_2\mathrm{O}]}\] In this reaction, we have aqueous and liquid species involved, so it is a heterogeneous reaction.
06

(f) Fe^2+(aq) + Zn(s) Zn^2+(aq) + Fe(s)

The equilibrium constant Kc for this reaction can be expressed as: \[K_c = \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Fe}^{2+}]}\] This reaction involves both aqueous and solid species, so it is heterogeneous.
07

(g) CaCO3(s) Ca^2+(aq) + CO3^2-(aq)

The equilibrium constant Kc for this reaction is: \[K_c = [\mathrm{Ca}^{2+}][\mathrm{CO}_3^{2-}]\] Since this reaction involves solid and aqueous species, it is a heterogeneous reaction.

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Key Concepts

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

Equilibrium Constant (Kc)
The equilibrium constant, denoted as \(K_c\), is a crucial concept in chemistry that helps determine the extent of a reaction as it reaches equilibrium. This constant is derived from the concentrations of products and reactants at equilibrium. The general expression for \(K_c\) is given by:
\[K_c = \frac{[Products]^ ext{coefficients}}{[Reactants]^ ext{coefficients}}\]
Here, the concentrations are represented in terms of molarity (moles per litre) and the exponents in the formula represent the stoichiometric coefficients from the balanced chemical equation.
  • If \(K_c > 1\), the reaction favors the formation of products.
  • If \(K_c < 1\), the reaction favors the formation of reactants.
Understanding the value of \(K_c\) and interpreting its meaning allows chemists to predict whether a reaction will proceed to completion or remain balanced at equilibrium. It's important to note that \(K_c\) values are specific to a given temperature and do not change unless the temperature does.
Homogeneous and Heterogeneous Reactions
Reactions can either be homogeneous or heterogeneous based on the physical states of the reactants and products involved.
**Homogeneous Reactions** occur when all reactants and products are in the same phase. A common example is a reaction where all components are gases. In Step 2, the reaction \(2 \, \mathrm{SO}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2 \, \mathrm{SO}_3(g)\) is homogeneous since all substances are in the gas phase.
**Heterogeneous Reactions** involve reactants and products in different phases, such as a solid reacting with a gas. In Step 3, the equation \(\mathrm{CO}_2(g) + \mathrm{C}(s) \rightleftharpoons 2 \, \mathrm{CO}(g)\) is heterogeneous due to the involvement of solid carbon along with gaseous components.
  • Homogeneous reactions are easier to describe using \(K_c\) because concentration terms for all reactants and products are involved.
  • In heterogeneous reactions, pure solids and liquids are usually omitted from the \(K_c\) expression as their concentrations do not change substantially.
Recognizing the type of reaction helps in setting up the correct equilibrium expression, leading to accurate calculations of \(K_c\).
Gas Phase Reactions
Gas phase reactions are prominent in chemistry due to the ease with which gases can mix and react. These reactions occur entirely in the gaseous state, allowing for more straightforward reaction mechanisms and equilibrium calculations.
One classic example from this context is the reaction \(\mathrm{I}_2(g) \rightleftharpoons 2 \, \mathrm{I}(g)\) from Step 1, where both reactants and products are in the gas phase. The equilibrium constant for gas-phase reactions can often be calculated using partial pressures, denoted as \(K_p\), but they can also be expressed in terms of concentrations with \(K_c\).
For gas reactions, the equilibrium constant expression relies on the number of moles and can be adjusted using the relation:
\[K_p = K_c(RT)^{\Delta n}\]
where \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas during the reaction.
  • Gas phase reactions allow for dynamic changes as gases expand to fill their containers, leading to unique conditions at equilibrium.
  • Understanding the interplay between \(K_p\) and \(K_c\) is fundamental when evaluating gas-phase equilibria and predicting the response of equilibrium to changes in conditions such as pressure or temperature.
Emphasizing gas-phase reactions helps in visualizing the fluidity and reactivity of substances in a gaseous state, making it easier to conceptualize equilibrium in these systems.

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Most popular questions from this chapter

At \(800 \mathrm{~K},\) the equilibrium constant for the reaction \(\mathrm{A}_{2}(g) \rightleftharpoons 2 \mathrm{~A}(g)\) is \(K_{c}=3.1 \times 10^{-4}\). (a) Assuming both forward and reverse reactions are elementary reactions, which rate constant do you expect to be larger, \(k_{f}\) or \(k_{r} ?\) (b) If the value of \(k_{f}=0.27 \mathrm{~s}^{-1}\), what is the value of \(k_{r}\) at \(800 \mathrm{~K} ?\) (c) Based on the nature of the reaction, do you expect the forward reaction to be endothermic or exothermic? (d) If the temperature is raised to \(1000 \mathrm{~K}\), will the reverse rate constant \(k_{r}\) increase or decrease? Will the change in \(k_{r}\) be larger or smaller than the change in \(k_{f}\) ?

The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is established at \(550 \mathrm{~K}\). An equilibrium mixture of the three gases has partial pressures of \(10.13 \mathrm{kPa}, 20.27 \mathrm{kPa}\), and \(35.46 \mathrm{kPa}\) for \(\mathrm{NO}, \mathrm{Cl}_{2},\) and \(\mathrm{NOCl}\), respectively.(a) Calculate \(K_{p}\) for this reaction at \(500.0 \mathrm{~K}\). (b) If the vessel has a volume of \(5.00 \mathrm{~L},\) calculate \(K_{c}\) at this temperature.

Which of the following statements are true and which are false? (a) The equilibrium constant can never be a negative number. (b) In reactions that we draw with a single-headed arrow, the equilibrium constant has a value that is very close to zero. (c) As the value of the equilibrium constant increases, the speed at which a reaction reaches equilibrium increases.

Consider the reaction \(\mathrm{IO}_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}(a q) ; K_{c}=3.5 \times 10^{-2} .\) If you start with \(25.0 \mathrm{~mL}\) of a \(0.905 \mathrm{M}\) solution of \(\mathrm{NaIO}_{4}\), and then dilute it with water to \(500.0 \mathrm{~mL},\) what is the concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}\) at equilibrium?

Consider the following exothermic equilibrium (Boudouard reaction) $$2 \mathrm{CO}(g) \rightleftharpoons \mathrm{C}(s)+\mathrm{CO}_{2}(g)$$ How will each of the following changes affect an equilibrium mixture of the three gases: (a) a catalyst is added to the mixture; \((\mathbf{b}) \mathrm{CO}_{2}(g)\) is added to the system; \((\mathbf{c}) \mathrm{CO}(g)\) is added from the system; \((\mathbf{d})\) the reaction mixture is heated; (e) the volume of the reaction vessel is doubled; \((\mathbf{f})\) the total pressure of the system is increased by adding a noble gas?

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