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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When do you use this? You use Keq to find the concentration of a reversible reaction at equilibrium. You use Q to find the concentration of a reversible.
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Presentation transcript:

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

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

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

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

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 !

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

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

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

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

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)

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)

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)

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

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

2. [CO] = mol/L [H 2 ] = mol/L [CH 4 ]= mol/L [H 2 O]= mol/L CO (g) + H 2 (g) ↔ CH 4 (g) + H 2 O (g) A B C D E. 0.02

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 B C. 2.5 D. 0.40

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

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

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

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 C

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