1. Chemical Equilibrium Chapter 18 Modern Chemistry
Sections 1 & 2
The Nature of Chemical Equilibrium
Shifting Equilibrium
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2. The Nature of Chemical Equilibrium
Section 18.1
The Nature of Chemical Equilibrium
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3. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597
Vocabulary
Reversible Reaction
Chemical Equilibrium
Equilibrium Expression
Equilibrium Constant
LeChatelier’s Principle
Insert Holt Disc 2
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4. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597
Reversible Reactions
Insert Holt Disc 2
Insert Glencoe Disc 1
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5. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597
Reversible Reactions
Products can react to re-form the reactants.
Must occur in a “closed” system
2HgO(s)  2Hg(l) + O2(g)
2Hg(l) + O2(g) 2HgO(s)
Both of these reactions occur simultaneously
2HgO(s)  2Hg(l) + O2(g)
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6. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597
Rate of its forward reaction equals the rate of its reverse reaction ….
and the concentrations of its products and reactants remain unchanged
Eventually all reversible reactions will reach eq. if the system is closed and conditions don’t change.
Eq. is dynamic – always in motion.
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7. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597
Reaction Rate vs time
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8. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597
Equilibrium Demonstration
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9. Reaction Rate vs. Time Rateforward = Ratereverse p. 591
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10. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597
What is “favored” at Eq?
At equilibrium
equal rates!
reactants are favored
neither is favored
products are favored R P
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11. Equilibrium Expression
n A + m B  x C + y D
Dependant on temperature
Independent of initial concentrations
[ ] = concentration in mol/L
[C]x [D]y
products
Keq = =
[A]n [B]m
reactants
x, y, n, m = coefficients
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12. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597
Equilibrium Constant
If Keq is large (>1) then products are favored at eq.
If Keq is small (<1) then reactants are favored at eq.
PRODUCTS
Keq =
REACTANTS
PRODUCTS
Keq =
REACTANTS
Pure liquids and solid are omitted.
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13. Equilibrium Constants Table
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Sample Problem p. 594
An equilibrium mixture of N2, O2 , and NO gases at 1500 K is determined to consist of 6.4x10–3 mol/L of N2, 1.7x10–3 mol/L of O2, and 1.1x10–5 mol/L of NO. What is the equilibrium constant for the system at this temperature?
Keq = 1.1 x 10−5
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Practice Problems p. 595 #1
At equilibrium a mixture of N2, H2, and NH3 gas at 500°C is determined to consist of mol/L of N2, mol/L of H2, and mol/L of NH3.What is the equilibrium constant for the reaction N2(g) + 3H2(g)  2NH3(g) at this temperature?
0.286
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Practice Problems p. 595 #2
The reaction AB2C(g)  B2(g) + AC(g) reached equilibrium at 900 K in a 5.00 L vessel. At equilibrium mol of AB2C, mol of B2, and mol of AC were detected. What is the equilibrium constant at this temperature for this system? (Don’t forget to convert amounts to concentrations.)
4.9 x 10−3
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17. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597
Practice Problems p. 595 #3
A reaction between gaseous sulfur dioxide and oxygen gas to produce gaseous sulfur trioxide takes place at 600°C.At that temperature, the concentration of SO2 is found to be 1.50 mol/L, the concentration of O2 is 1.25 mol/L, and the concentration of SO3 is 3.50 mol/L. Using the balanced chemical equation, calculate the equilibrium constant for this system.
4.36
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18. Equilibrium Concentrations and Keq values
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19. Eq. Concentration Problem
For the decomposition reaction of ammonia, 0.75 M of ammonia are added to an empty 1 L flask. When the reversible reaction has achieved equilibrium the concentration of nitrogen in the flask is 0.15 M. Find the equilibrium concentrations of hydrogen and ammonia. Also find the Keq.
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20. Determining Eq. Concentrations
N2 (g)+3 H2 (g)  2NH3(g) I
initial C
change
E equil.
1/3 = 0.15/x
x = 0.45M
0 M
0 M
0.75 M
+0.15 M
+0.45 M
0.15 M
0.45 M
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21. Determining Eq. Concentrations
N2 (g)+3 H2 (g)  2NH3(g) I
initial C
change
E equil.
1/2 = 0.15/x
x = 0.30M
0 M
0 M
0.75 M
+0.15 M
+0.45 M
-0.30 M
0.15 M
0.45 M
0.45 M
If this side is + then the other side is -.
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22. Eq. Concentration Problem
For the synthesis reaction of hydrogen and iodine, 0.20M of hydrogen and 0.30M of iodine are added to an empty 1 L flask. When the reversible reaction has achieved equilibrium the concentration of hydrogen in the flask is 0.10 M. Find the equilibrium concentrations of iodine and hydrogen iodide. Also find the Keq.
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23. Determining Eq. Concentrations
H2 (g) + I2 (g)  2HI(g) I
initial C
change
E equil.
1/1 = 0.10/x
x = 0.10M
0.20 M
0.30 M
0 M
-0.10 M
-0.10 M
0.10 M
0.20 M
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24. Determining Eq. Concentrations
H2 (g) + I2 (g)  2HI(g) I
initial C
change
E equil.
1/2 = 0.10/x
x = 0.20M
0.20 M
0.30 M
0 M
-0.10 M
-0.10 M
+0.20 M
0.10 M
0.20 M
0.20 M
If this side is - then the other side is +.
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25. Eq Concentration Problem
A 0.20 M solution of HC2H3O2 is 5.0% ionized. Find the equilibrium concentrations of H+, C2H3O21- and HC2H3O2, also find the Keq.
answer
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Eq Conc. & Keq Problems
At a specific temperature and pressure, 1.2 moles of hydrogen, 0.40 moles of nitrogen and 1.3 moles of ammonia are put into a closed one liter flask. When allowed to reach equilibrium the amount of ammonia is 1.6 moles. Find the Keq for this system.
N2 (g) + 3H2 (g)  2NH3 (g)
answer
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Eq Conc. & Keq Problems
When .56 moles of SO3 is placed in a liter container, some of it decomposes. The equilibrium concentrations of SO2 is 0.42 moles / liter. Calculate the equilibrium concentration of O2 and SO3 and the Keq.
O2 = 0.21 M SO3 = 0.14 M Keq = .53
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Ch 18 Sec 1 Homework
Page 595 # 1-9
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29. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597
Ch 18 Sec 1 Homework
Equilibrium Concentrations and Keq Worksheet
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