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Chapter 18: Chemical Equilibrium

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1 Chapter 18: Chemical Equilibrium
A State of Dynamic Balance

2 What is Equilibrium? A reversible reaction is a reaction that can take place in both the forward and reverse directions. When a reaction is reversible a double-headed arrow will be used.  Chemical equilibrium occurs when the forward and reverse reactions occur at the same speed. At equilibrium, the concentrations of reactants and products remain the same.

3 Law of Chemical Equilibrium
This law states that at a given temperature, a chemical system may achieve a state in which a ratio of reactant and product concentrations has a constant value. The general equation for an equilibrium reaction is aA + bB  cC + dD

4 Equilibrium Constant The equilibrium constant expression for an equilibrium system is given as Keq = [C]c[D]d [A]a[B]b The brackets refer to molar concentrations.

5 Homogeneous vs Heterogeneous Equilibria
In homogeneous equilibria, all of the compounds are in the same state. When more than one state is present, pure solids and liquids are not included in the equilibrium expression.

6 H2S (g) + 2H2O (g)  SO2 (g) + 3H2 (g)
Homogeneous Example Write the equilibrium constant expression for the reaction of hydrogen sulfide and water vapor to form sulfur dioxide and hydrogen. H2S (g) + 2H2O (g)  SO2 (g) + 3H2 (g)

7 Heterogeneous Example
Write the equilibrium constant expression for the high-temperature reaction of carbon dioxide and solid carbon to form carbon monoxide. CO2 (g) + C (s)  2 CO (g)

8 Practice Problems Write equilibrium constant expressions for the following equilibria. a. C2H4O (g)  CH4 (g) + CO (g) b. 3O2 (g)  2O3 (g) c. 2N2O (g) + O2 (g)  4NO (g) d. 4NH3 (g) + 3O2  2N2 (g) + 6H2O (g)

9 Practice Problem Write equilibrium constant expressions for the following equilibria. a. C4H10 (l)  C4H10 (g) b. NH4HS (s)  NH3 (g) + H2S (g) c. CO (g) + Fe3O4 (s)  CO2 (g) + 3 FeO (s) d. (NH4)2CO3 (s)  2NH3 (g) + CO2 (g) + H2O (g)

10 Calculating Keq The value of Keq is a constant for a given reaction at a given temperature. If Keq is greater than 1, then products are favored at equilibrium. If Keq is less than 1, reactants are favored. If you know the concentrations you can find the Keq.

11 2NO (g) + Br2 (g)  2NOBr (g)
Example Nitrogen monoxide and bromine react to form nitrosyl bromide according to the following equation. 2NO (g) + Br2 (g)  2NOBr (g) Calculate Keq for this equilibrium using the data [NOBr] = M, [NO] = M and [Br2] = M.

12 HCONH2 (g)  NH3 (g) + CO (g)
Practice Problems The following is the chemical equation for the decomposition of formamide. HCONH2 (g)  NH3 (g) + CO (g) Calculate Keq using the equilibrium data [HCONH2] = M, [NH3] = M and [CO] = M.

13 4H2 (g) + CS2 (g)  CH4 (g) + 2H2S (g)
Practice Problems Hydrogen and carbon disulfide react to form methane and hydrogen sulfide according to this equation. 4H2 (g) + CS2 (g)  CH4 (g) + 2H2S (g) Calculate Keq if the equilibrium concentrations are [H2] = M, [CS2] = M, [CH4] = M and [H2S] = M.

14 Le Châtelier’s Principle
This principle states that if a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress. Situations: Adding or removing components Changes in volume Changes in temperature

15 2NO (g) + Br2 (g)  2NOBr (g)
Example Use Le Chatelier’s principle to predict how each of the following changes would affect this equilibrium. 2NO (g) + Br2 (g)  2NOBr (g) Adding more NOBr Adding more NO Removing Br2 Decreasing the volume

16 Changes in Temperature
A change in temperature changes both the equilibrium position and the value of Keq. For example, consider H2 (g) + Cl2 (g)  2HCl (g) ΔH°= -185 kJ The forward reaction releases heat, so heat is a product as seen as H2 (g) + Cl2 (g)  2HCl (g) + heat

17 Temperature Changes (cont’d)
Raising the temperature causes the value of Keq to decrease, since heat is removed. If the temperature is lowered, the equilibrium relieves the stress by shifting to the right, increasing the concentration of product and the Keq.

18 PCl5 (g)  PCl3 (g) + Cl2 (g) + heat
Practice Problem Phosphorus pentachloride decomposes exothermically to form phosphorus trichloride and chlorine. PCl5 (g)  PCl3 (g) + Cl2 (g) + heat How would you regulate the temperature of this equilibrium in order to do the following? Increase the concentration of PCl5 Decrease the concentration of PCl3 Increase the amount of Cl2 in the system Decrease Keq

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