/*! 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 16 Write the expressions for \(K_{c... [FREE SOLUTION] | 91Ó°ÊÓ

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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) \(\mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{TiCl}_{4}(l)\) (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) \(\mathrm{C}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)\) (e) \(4 \mathrm{HCl}(a q)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}_{2}(g)\) (f) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g)\) (g) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(l)\)

Short Answer

Expert verified
(a) Homogeneous reaction. Kc expression: \[K_c = \frac{[O_2]^3}{[O_3]^2}\] (b) Heterogeneous reaction. Kc expression: \[K_c = \frac{[TiCl_4]}{[Cl_2]^2}\] (c) Homogeneous reaction. Kc expression: \[K_c = \frac{[C_2H_6]^2 [O_2]}{[C_2H_4]^2 [H_2O]^2}\] (d) Heterogeneous reaction. Kc expression: \[K_c = \frac{[CH_4]}{[H_2]^2}\] (e) Homogeneous reaction. Kc expression: \[K_c = \frac{[H_2O]^2 [Cl_2]^2}{[HCl]^4 [O_2]}\] (f) Heterogeneous reaction. Kc expression: \[K_c = \frac{[CO_2]^{16} [H_2O]^{18}}{[C_8H_{18}]^2 [O_2]^{25}}\] (g) Heterogeneous reaction. Kc expression: \[K_c = \frac{[CO_2]^{16} [H_2O]^{18}}{[C_8H_{18}]^2 [O_2]^{25}}\]

Step by step solution

01

(a) Reaction:

\(2 O_3(g) \rightleftharpoons 3 O_2(g) \) This reaction is homogeneous since all species are in the gas phase.
02

(a) Kc expression:

\[K_c = \frac{[O_2]^3}{[O_3]^2} \]
03

(b) Reaction:

\( Ti(s) + 2 Cl_2(g) \rightleftharpoons TiCl_4(l) \) This reaction is heterogeneous since species are present in different phases (solid, liquid, and gas).
04

(b) Kc expression:

\[K_c = \frac{[TiCl_4]}{[Cl_2]^2} \]
05

(c) Reaction:

\( 2 C_2H_4(g) + 2 H_2O(g) \rightleftharpoons 2 C_2H_6(g) + O_2(g) \) This reaction is homogeneous since all species are in the gas phase.
06

(c) Kc expression:

\[K_c = \frac{[C_2H_6]^2 [O_2]}{[C_2H_4]^2 [H_2O]^2} \]
07

(d) Reaction:

\( C(s) + 2 H_2(g) \rightleftharpoons CH_4(g) \) This reaction is heterogeneous since species are present in different phases (solid and gas).
08

(d) Kc expression:

\[K_c = \frac{[CH_4]}{[H_2]^2} \]
09

(e) Reaction:

\( 4 HCl(aq) + O_2(g) \rightleftharpoons 2 H_2O(l) + 2 Cl_2(g) \) This reaction is homogeneous since all species are either in the gas or aqueous phase.
10

(e) Kc expression:

\[K_c = \frac{[H_2O]^2 [Cl_2]^2}{[HCl]^4 [O_2]} \]
11

(f) Reaction:

\( 2 C_8H_{18}(l) + 25 O_2(g) \rightleftharpoons 16 CO_2(g) + 18 H_2O(g) \) This reaction is heterogeneous since species are present in different phases (liquid and gas).
12

(f) Kc expression:

\[K_c = \frac{[CO_2]^{16} [H_2O]^{18}}{[C_8H_{18}]^2 [O_2]^{25}} \]
13

(g) Reaction:

\( 2 C_8H_{18}(l) + 25 O_2(g) \rightleftharpoons 16 CO_2(g) + 18 H_2O(l) \) This reaction is heterogeneous since species are present in different phases (liquid and gas).
14

(g) Kc expression:

\[K_c = \frac{[CO_2]^{16} [H_2O]^{18}}{[C_8H_{18}]^2 [O_2]^{25}} \]

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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 chemical equilibrium. It quantitatively describes the balance between reactants and products in a reversible chemical reaction. The value of \(K_c\) is determined by the concentration of the substances at equilibrium. This relationship is expressed by the law of mass action.

The equilibrium constant is given by the formula:
  • \(K_c = \frac{[Products]^{coefficients}}{[Reactants]^{coefficients}}\)
In this expression, the concentrations of products appear in the numerator, each raised to the power of its stoichiometric coefficient, and the concentrations of reactants are in the denominator, also raised to their respective coefficients.

Understanding \(K_c\) helps predict the direction of the reaction. A large \(K_c\) indicates that at equilibrium, products are favored, while a small \(K_c\) implies reactants predominate. By calculating \(K_c\), chemists can make informed decisions about how a reaction will proceed and adjust conditions to favor the desired outcome.
Homogeneous Reactions
Homogeneous reactions occur when all reactants and products are in the same phase, typically gaseous or aqueous. This congruence simplifies the study of these reactions since concentrations refer to a single phase, making calculations consistent across all reactants and products.

Examples from our exercise, such as \(2 \text{O}_3(g) \rightleftharpoons 3 \text{O}_2(g)\), are homogeneous because all substances involved are gases. Here, the equilibrium constant expression is straightforward:
  • \(K_c = \frac{[O_2]^3}{[O_3]^2}\)
To further understand homogeneous reactions, consider all components in a uniform environment. This uniform environment means that spatial differences, like solubility or phase interactions, do not complicate the reaction dynamics.

Homogeneous reactions are found in many real-world applications, including gaseous combustion processes and solution-based chemical industries. They are easier to control and study due to the uniform phase nature.
Heterogeneous Reactions
Unlike homogeneous reactions, heterogeneous reactions involve reactants and products in different phases, such as liquid, solid, and gas. The interaction between these phases adds complexity to the reaction dynamics and equilibrium constant calculations.

For example, in the reaction \(\text{Ti}(s) + 2 \text{Cl}_2(g) \rightleftharpoons \text{TiCl}_4(l)\), the equilibrium constant expression does not include the solid \(\text{Ti}\) because its activity is considered constant:
  • \(K_c = \frac{[TiCl_4]}{[Cl_2]^2}\)
In heterogeneous equilibria, the phases interact at interfaces, which can affect reaction rates and equilibria. Solid catalysts in reactions often exploit these interactions to improve efficiency or selectivity.

Such reactions are common in processes like industrial synthesis, where reactions often occur between gases and solids in reactors. Understanding phase behavior in heterogeneous reactions enables engineers and chemists to design more efficient industrial processes.

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Most popular questions from this chapter

Gaseous hydrogen iodide is placed in a closed container at \(425^{\circ} \mathrm{C},\) where it partially decomposes to hydrogen and iodine: \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) .\) At equilibrium it is found that \([\mathrm{HI}]=3.53 \times 10^{-3} \mathrm{M},\left[\mathrm{H}_{2}\right]=4.79 \times 10^{-4} \mathrm{M}\) and \(\left[\mathrm{I}_{2}\right]=4.79 \times 10^{-4} \mathrm{M} .\) What is the value of \(K_{c}\) at this temperature?

The equilibrium constant for the reaction $$\begin{array}{r} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) \\ \text { is } K_{c}=1.3 \times 10^{-2} \text {at } 1000 \mathrm{~K} \text { . (a) At this tempera } \end{array}$$ (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), or does it favor NOBr? (b) Calcu- $$\begin{array}{l} \text { late } K_{c} \text { for } 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \text { . } \\ K_{c} \text { for } \mathrm{NOBr}(g) \rightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) \end{array}$$ (c) Calculate

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

How do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction: (a) removal of a reactant (b) removal of a product, (c) decrease in the volume, (d) decrease in the temperature, (e) addition of a catalyst?

Two different proteins \(\mathrm{X}\) and \(\mathrm{Y}\) are dissolved in aqueous solution at \(37^{\circ} \mathrm{C}\). The proteins bind in a 1: 1 ratio to form XY. A solution that is initially \(1.00 \mathrm{~m} M\) in each protein is allowed to reach equilibrium. At equilibrium, \(0.20 \mathrm{~m} M\) of free \(\mathrm{X}\) and \(0.20 \mathrm{~m} M\) of free \(\mathrm{Y}\) remain. What is \(K_{c}\) for the reaction?

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