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Chemical Equilibrium Chapter 17. 17.1 A State of Dynamic Balance All chemical reactions are reversible. All chemical reactions are reversible. When both.

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Presentation on theme: "Chemical Equilibrium Chapter 17. 17.1 A State of Dynamic Balance All chemical reactions are reversible. All chemical reactions are reversible. When both."— Presentation transcript:

1 Chemical Equilibrium Chapter 17

2 17.1 A State of Dynamic Balance All chemical reactions are reversible. All chemical reactions are reversible. When both the forward and reverse reactions continue to occur at equal rates but no net change is observed. When both the forward and reverse reactions continue to occur at equal rates but no net change is observed. Use a double arrow to show this situation. Use a double arrow to show this situation. Outside forces can affect the equilibrium! Outside forces can affect the equilibrium!

3 The Reaction Quotient and Equilibrium Constant For the reaction 3H 2 + 2N 2  2NH 3 the relationships among the substances has been experimentally established For the reaction 3H 2 + 2N 2  2NH 3 the relationships among the substances has been experimentally established The nature of an equilibrium is dynamic; state changes are involved! The nature of an equilibrium is dynamic; state changes are involved! Keq = [C] c [D] d [A] a [B] b

4 The Equilibrium Constant The equilibrium constant is the numerical value that illustrates an equilibrium. The equilibrium constant is the numerical value that illustrates an equilibrium. Product concentrations always appear in the numerator. Product concentrations always appear in the numerator. Reactant concentrations always appear in the denominator. Reactant concentrations always appear in the denominator. Each concentration is always raised to the power of its stoichiometric coefficient in the balanced equation. Each concentration is always raised to the power of its stoichiometric coefficient in the balanced equation. When the reaction has reached equilibrium, the value of the constant K depends on the particular reaction and on the temperature. Units are never given for K! When the reaction has reached equilibrium, the value of the constant K depends on the particular reaction and on the temperature. Units are never given for K!

5 K eq If K eq > 1 then the reaction is product favored at equilibrium. If K eq > 1 then the reaction is product favored at equilibrium. If K eq < 1 then the reaction is reactant favored at equilibrium. If K eq < 1 then the reaction is reactant favored at equilibrium. In an homogeneous equilibrium, all reactants and products are in the same physical state. In an homogeneous equilibrium, all reactants and products are in the same physical state.

6 Practice Problems What would be the K eq for the reaction between hydrogen gas and iodine gas? The product is hydrogen iodide gas. What would be the K eq for the reaction between hydrogen gas and iodine gas? The product is hydrogen iodide gas. N 2 O 4 (g)  2NO 2 (g) N 2 O 4 (g)  2NO 2 (g) CO(g) + 3H 2 (g)  CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g)  CH 4 (g) + H 2 O(g) 2H 2 S(g)  2H 2 (g) + S 2 (g) 2H 2 S(g)  2H 2 (g) + S 2 (g)

7 K eq Heterogeneous equilibria contain reactants and products in different physical states. Heterogeneous equilibria contain reactants and products in different physical states. Pure substances are left out of the expression because their concentrations are always constant at a given temperature! Pure substances are left out of the expression because their concentrations are always constant at a given temperature!

8 Writing Equilibrium Constant Expressions Reactions involving solids, water, and pure liquids. Reactions involving solids, water, and pure liquids. Concentrations of any solid reactants and products are omitted from the equilibrium constant expression. Concentrations of any solid reactants and products are omitted from the equilibrium constant expression. The molar concentration of water (or of any liquid reactant or product) is omitted from the equilibrium constant expression. The molar concentration of water (or of any liquid reactant or product) is omitted from the equilibrium constant expression.

9 Practice Problem Write the equilibrium constant expression for each of the following reactions in terms of concentrations: Write the equilibrium constant expression for each of the following reactions in terms of concentrations: PCl 5 (g)  PCl 3 (g) + Cl 2 (g) PCl 5 (g)  PCl 3 (g) + Cl 2 (g) Cu(OH) 2 (s)  Cu 2+ (aq) + 2OH - (aq) Cu(OH) 2 (s)  Cu 2+ (aq) + 2OH - (aq) Cu(NH 3 ) 4 2+ (aq)  Cu 2+ (aq) + 4NH 3 (aq) Cu(NH 3 ) 4 2+ (aq)  Cu 2+ (aq) + 4NH 3 (aq) CH 3 CO 2 H(aq) + H 2 O(l)  CH 3 CO 2 - (aq) + H 3 O + (aq) CH 3 CO 2 H(aq) + H 2 O(l)  CH 3 CO 2 - (aq) + H 3 O + (aq)

10 17.2 Factors Affecting Equilibrium Le Chatelier’s Le Chatelier’s If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress. If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress. changes in concentration, volume (pressure), and temperature changes in concentration, volume (pressure), and temperature

11 Concentration Increasing the amount of reactants will push the equilibrium right (toward products) Increasing the amount of reactants will push the equilibrium right (toward products) Decreasing the amount of reactants will push the equilibrium left (toward reactants) Decreasing the amount of reactants will push the equilibrium left (toward reactants) Increasing the amount of products will push the equilibrium left (toward reactants) Increasing the amount of products will push the equilibrium left (toward reactants) Decreasing the amount of products will push the equilibrium right (toward products) Decreasing the amount of products will push the equilibrium right (toward products)

12 Volume changing volume (pressure) on a system will cause the system to move to the side with fewer gas molecules changing volume (pressure) on a system will cause the system to move to the side with fewer gas molecules same numbers on both sides = no net shift in reaction same numbers on both sides = no net shift in reaction

13 Temperature Changing temperature alters equilibrium position AND K eq ! Changing temperature alters equilibrium position AND K eq ! The shift will depend on whether the reaction is endothermic or exothermic. The shift will depend on whether the reaction is endothermic or exothermic.


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