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Chapter 18 Equilibrium A + B  AB We may think that all reactions change all reactants to products, or the reaction has gone to completion But in reality,

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Presentation on theme: "Chapter 18 Equilibrium A + B  AB We may think that all reactions change all reactants to products, or the reaction has gone to completion But in reality,"— Presentation transcript:

1 Chapter 18 Equilibrium A + B  AB We may think that all reactions change all reactants to products, or the reaction has gone to completion But in reality, products may start to change back into reactants; the reaction is reversible. A + B ↔ AB

2 CO 2 + H 2 O  C 6 H 12 O 6 + O 2 O 2 + C 6 H 12 O 6  CO 2 + H 2 O

3 Equilibrium Equilibrium is the state where the forward and reverse reactions balance because they are occurring at equal rates. NO OVERALL CHANGE!

4 The Law of Chemical Equilibrium –the point in the reaction where a ratio of reactant and product concentration has a constant value, K eq Keq < 1Reactants Favored Keq = 1Neither is favored Keq > 1Products favored

5 Homogeneous equilibrium –States of all compounds are the ____________ H 2 (g) + I 2 (g) ↔ 2HI (g) Heterogeneous equilibrium –States of compounds are ________________ CaCO 3 (g) ↔ CaO (s) + CO 2 (g) Remember we only pay attention to the GASES when we calculate K eq !

6 What is the correct the K eq ? N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g)

7 What is the correct the K eq ? 2 NbCl 4 (g) ↔ NbCl 3 (g) + NbCl 5 (g)

8 What is the correct Keq? H 2 O (l) ↔ H 2 O (g)

9 If Keq = 2.4, which is more favored? A. products B. reactants C. neither is favored D. not enough information

10 Chemical Equilibrium Problems I Write the equilibrium constant expression (K eq ) for these homogeneous equations. 1. N 2 O 4 (g) ↔ 2 NO 2 (g)

11 2. CO (g) + 3H 2(g) ↔ CH 4 (g) + H 2 O (g) 3. 2 H 2 S (g) ↔ 2 H 2 (g) + S 2 (g)

12 Write the equilibrium constant expression (K eq ) for these heterogeneous equations. 1.C 10 H 8 (s) ↔ C 10 H 8 (g) 2.CaCO 3 (s) ↔ CaO (s) + CO 2 (g)

13 3. C (s) + H 2 O (g) ↔ CO (g) + H 2 (g) 4. FeO (s) + CO (g) ↔ Fe (s) + CO 2(g)

14 1. Calculate the Keq for the following equation using the data: [N 2 O 4 ] = 0.0613 mol/L [NO 2 ] = 0.0627 mol/L N 2 O 4 (g) ↔ 2NO 2 (g) A. 0.98 B. 15.59 C. 0.064 D. 1.02 E. 2.05

15 2. [CO] = 0.0613 mol/L [H 2 ] = 0.1839 mol/L [CH 4 ]=0.0387 mol/L [H 2 O]=0.0387 mol/L CO (g) + H 2 (g) ↔ CH 4 (g) + H 2 O (g) A. 0.25 B. 0.72 C. 0.04 D. 3.93 E. 0.02

16 3. [H 2 ]=1.5 mol/L [N 2 ]=2.0 mol/L [NH 3 ]=1.8 mol/L 3H 2 (g) + N 2(g) ↔ 2NH 3 (g) A. 0.48 B. 0.24 C. 2.5 D. 0.40

17 3. If the Keq = 0.48, we know that the equilibrium favors… a. reactants b. products c. neither

18 4. [Mg]=2 mol/L[HCl]=3 mol/L [MgCl]=6 mol/L [H 2 ]=3 mol/L 2 Mg (s) + 2 HCl (g) ↔ 2 MgCl (g) + H 2 (g) A. 3 B. 12

19 4. Since Keq = 3, we know that the equilibrium favors a. reactants b. products c. neither

20 5. [H 2 ]=0.52 mol/L [I 2 ]=0.23 mol/L [HI]=1.7 mol/L H 2 (g) + I 2 (g) ↔ 2HI (g) A. 202 B. 0.04 C. 24.16

21 5. Since Keq= 24.16, we know that the equilibrium favors a. reactants b. products c. neither


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