chemical equilibrium chapter 18 modern chemistry
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Chemical Equilibrium Chapter 18 Modern Chemistry. Sections 1 & 2 The Nature of Chemical Equilibrium Shifting Equilibrium. Section 18.1. The Nature of Chemical Equilibrium. Vocabulary. Reversible Reaction Chemical Equilibrium Equilibrium Expression Equilibrium Constant - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597
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ChemicalEquilibriumChapter 18
Modern ChemistrySections 1 & 2
The Nature of Chemical EquilibriumShifting Equilibrium
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Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597
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The Nature of Chemical Equilibrium
Section 18.1
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Reversible Reaction Chemical Equilibrium Equilibrium Expression Equilibrium Constant LeChatelier’s Principle
Vocabulary
Insert Holt Disc 2
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Reversible Reactions
Insert Holt Disc 2
Insert Glencoe Disc 1
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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)
Reversible Reactions
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Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597
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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.
Chemical Equilibrium
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Reacti
on
Rate
vs t
ime
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Equilibrium Demonstration
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Reaction Rate vs. Time
Rateforward = Ratereverse
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What is “favored” at Eq? At equilibrium
reactants are favored
neither is favored
products are favored
equal rates!
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n A + m B x C + y D
Dependant on temperature Independent of initial concentrations
Equilibrium Expression
[C]x [D]y
[A]n [B]mKeq =
[ ] = concentration in
mol/L
x, y, n, m = coefficients
=productsreactants
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If Keq is large (>1) then products are favored at eq.
If Keq is small (<1) then reactants are favored at eq.
Equilibrium Constant
PRODUCTSREACTANTS
Keq =
PRODUCTS
REACTANTS
Keq =
Pure liquids and solid are omitted.
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Equilibrium Constants Table
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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
Sample Problem p. 594
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At equilibrium a mixture of N2, H2, and NH3 gas at 500°C is determined to consist of 0.602 mol/L of N2, 0.420 mol/L of H2, and 0.113 mol/L of NH3.What is the equilibrium constant for the reaction N2(g) + 3H2(g) 2NH3(g) at this temperature?
0.286
Practice Problems p. 595 #1
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The reaction AB2C(g) B2(g) + AC(g) reached equilibrium at 900 K in a 5.00 L vessel. At equilibrium 0.084 mol of AB2C, 0.035 mol of B2, and 0.059 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
Practice Problems p. 595 #2
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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
Practice Problems p. 595 #3
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Equilibrium Concentrationsand Keq values
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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.
Eq. Concentration Problem
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N2 (g)+3 H2 (g) 2NH3(g)
Determining Eq. Concentrations
Iinitia
l
Cchang
e
Eequil
.
0 M 0 M 0.75 M
0.15 M
+0.15 M
1/3 = 0.15/x
x = 0.45M+0.45 M
0.45 M
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N2 (g)+3 H2 (g) 2NH3(g)
Determining Eq. Concentrations
Iinitia
l
Cchang
e
Eequil
.
0 M 0 M 0.75 M
0.15 M
+0.15 M
1/2 = 0.15/x
x = 0.30M+0.45 M
0.45 M
-0.30 M
0.45 M
If this side is + then the other side is -.
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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.
Eq. Concentration Problem
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H2 (g) + I2 (g) 2HI(g)
Determining Eq. Concentrations
Iinitia
l
Cchang
e
Eequil
.
0.20 M 0.30 M 0 M
0.10 M
-0.10 M
1/1 = 0.10/x
x = 0.10M-0.10 M
0.20 M
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H2 (g) + I2 (g) 2HI(g)
Determining Eq. Concentrations
Iinitia
l
Cchang
e
Eequil
.
0.20 M 0.30 M 0 M
0.10 M
-0.10 M
1/2 = 0.10/x
x = 0.20M-0.10 M
0.20 M
+0.20 M
0.20 M
If this side is - then the other side is +.
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A 0.20 M solution of HC2H3O2 is 5.0% ionized. Find the equilibrium concentrations of H+, C2H3O2
1- and HC2H3O2, also find the Keq.
answer
Eq Concentration Problem
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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
Eq Conc. & Keq Problems
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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
Eq Conc. & Keq Problems
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Ch 18 Sec 1 Homework
Page 595 # 1-9
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Ch 18 Sec 1 Homework
Equilibrium Concentrations and Keq Worksheet