Unit 5 EQ1 - Compatibility ModeUnit 5 EQ1 Outline: ‐equilibrium ‐calculating Keq ‐Le...
Transcript of Unit 5 EQ1 - Compatibility ModeUnit 5 EQ1 Outline: ‐equilibrium ‐calculating Keq ‐Le...
Unit 5
Unit 5 EQ1: Consider the characteristics and applications of equilibrium systems
in chemical reactions
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Unit 5 EQ1
Outline:
‐equilibrium ‐calculating Keq‐Le Châtelier's Principle ‐equilibrium shift: press. conc. temp.
‐favoured reactions ‐factors affecting Keq‐the equilibrium constant Keq ‐ICE method
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Equilibrium
An equilibrium is a dynamic condition in which the forward rate of a process equals the reverse rate in a closed system.
Imagine two closed containers of gas are connected and the valve separating them is opened. Each gas would flow into the other until the rate of each gas entering and leaving becomes equal. This is equilibrium. When you put the lid on a bottle of water, the particles evaporating and condensing will reach equilibrium.
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liquid + heat ⇄ vapour
Equilibrium is shown by a double headed arrow which means that the process occurs in both directions.
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H2(g) + I2(g)⇄ 2HI(g)
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Favoured Reactions
Recall that chemical equilibrium involves a reversible reaction in which the forward rate equals the reverse rate, and therefore the concentrations of reactants and products remain constant. So far, we have assumed that neither forward nor reverse reactions were favoured, and the concentrations of reactants and products are roughly equal at equilibrium.
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If however, the forward reaction is favoured, the forward reaction
is nearly complete before the reverse reaction establishes
equilibrium. In this case, there is a higher concentration of
products than reactants at equilibrium.
HBr(aq) + H2O(l) ‐‐‐‐‐⇄ H3O
+(aq) + Br−(aq) forward favoured
If the reverse reaction is favoured, the forward reaction is just
starting when the reverse reaction establishes equilibrium. Here,
there is a higher concentration of reactants than products at
equilibrium.
H2CO3(aq) + H2O(l) ⇆----- H3O+(aq) + HCO3
−(aq)
reverse favoured
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Graphing Equilibria
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Le Châtelier's Principle
Because equilibria are dynamic, any change to the reactants or products will change the equilibrium. Le Châtelier's principle states that when an equilibrium is stressed, it will shift to relieve the stress. For example:
liquid + heat ⇄ vapour
if this system is stressed by adding more heat, the equilibrium will shift to the right, which means that it will produce more vapour until it reaches a new equilibrium.
Stresses include changes in concentration, pressure, temperature, volume, etc.
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Equilibrium Shift
Pressure: Increasing the pressure will shift the eq. toward the side which has fewer moles of gas particles. If they are
equal, or there are no gases, there is no change. (Haber process)
N2(g) + 3H2(g) ⇄ 2NH3(g)
increase pressure
shift to right
Note: adding a non‐reacting gas to increase the pressure does not affect the eq.
Recall that Le Chatelier’s principle states that an equilibrium will
shift to relieve a stress.
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Concentration: Increasing the concentration of one reactant or product will shift the eq. toward the other side.
HBr(aq) + H2O(l) ⇄ H3O+(aq) + Br−(aq)
decrease [HBr]
shift to left
increase [H3O+]
shift to left
increase [H2O]
no effect on equilibrium
(adding solids or liquids will not shift the equilibrium)
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Temperature: Increasing the temperature shifts the eq. away from the energy term.
N2(g) + 3H2(g) ⇄ 2NH3(g) + 92 kJ
decrease temperature
shift to right
CaCO3(s) + 556 kJ ⇄ CaO(s) + CO2(g)
decrease temp
shift to left
a
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Because of their good taste in music, catalysts do not favour one direction
catalyst
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Consider the following equilibrium and use the graph to identify the stress applied:
CO(g) + Cl2(g) ⇄ COCl2(g) + heat
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Consider the following equilibrium and use the graph to identify the stress applied:
CO(g) + Cl2(g) ⇄ COCl2(g) + heat
A = adding CO (shift right)
B = increase volume / decrease pressure (shift left)
C = increase heat (shift left)
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The Equilibrium Constant
The equilibrium constant (K or Keq) describes the extent to which the reaction has been carried out by the time it reaches equilibrium. In a reaction:
aA + bB ⇄ cC +dD
Keq = [C]c [D]d Where: [A]a = the conc. of A
[A]a [B]b to the power of its coefficient
Keq = eq. const. (units vary)
a
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In general
Keq = [products]
[reactants]
If Keq = >1 products favoured (more products around at eq.)
Keq = 1 equal conc. of reactants and products
Keq = < 1 reactants favoured (more reactants around at eq.)
The value of Keq is found experimentally and is dependent on
temperature. If gases are used, pressure takes the place of
concentration, however your text uses concentration. The conc. of
pure solids and liquids essentially remain constant and are left out
of equilibrium expressions.
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NOTE: Changing the pressure, concentration, or adding a catalyst
does not change the Keq. For example, more product may be
formed (as in a shift to the right), but the ratio of their
concentrations remains the same.
Changing the temperature does change the Keq. For an
exothermic reaction, increasing the temp decreases the Keq; For
an endothermic reaction, increasing the temp increases the Keq.
Reactions run to completion when a product in not available for a
reverse reaction. Examples of this include producing a gas that
escapes, a precipitate that is insoluble, or a product that does not
ionize.
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But how can shifting the equilibrium not change the Keq?
This is the wrong way to look at it, the equilibrium shifts to keep the Keq constant.
consider:
A + B C + D Keq = [C][D]
[A][B]
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scenario 1:
A + B C + D Keq = [C][D] = (2)(2) = 4
1 1 2 2 [A][B] (1)(1)
scenario 2: lower C to 1
shifts right A + B C + D Keq = [C][D] = (1)(y) = 4
x x 1 y [A][B] (x)(x)
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scenario 1:
A + B C + D Keq = [C][D] = (2)(2) = 4
1 1 2 2 [A][B] (1)(1)
scenario 2: lower C to 1
shifts right A + B C + D Keq = [C][D] = (1)(2.25) = 4
x x 1 y [A][B] (0.75)(0.75)
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Ex 1) Calculate the equilibrium constant for the reaction
N2 (g) + O2 (g) ⇄ NO (g) at constant temperature if the concentrations at
equilibrium are:
[N2] = 6.4 × 10−3 mol/L
[O2] = 1.7 × 10−3 mol/L
[NO] = 1.1 × 10−5 mol/L
N2(g) + O2(g) ⇄ 2NO(g)
Keq = [NO]2 =
[N2][O2]
Keq = [NO]2 = (1.1 × 10−5 M)2
[N2][O2] (6.4 × 10−3 M)(1.7 × 10−3 M)
= 1.1 × 10−5
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Ex 2) Given: H2(g) + Br2(g) ⇄ 2HBr(g), Keq = 3.5 × 104, and
constant temp.
a) Calculate the concentration of H2 if [HBr] = 9.8 × 10−2 M, and [Br2] =
4.3 × 10−3 M.
Keq = [HBr]2
[H2][Br2]
[H2] = [HBr]2
Keq [Br2]
[H2] = (9.8 × 10−2 M)2
(3.5 × 104)(4.3 × 10−3 M)
= 0.000063813...
= 6.4 × 10−5 M
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Ex 2) Given: H2(g) + Br2(g) ⇄ 2HBr(g), Keq = 3.5 × 104, and
constant temp.
b) Calculate the concentration of HBr if [H2]= 0.0356 M and
[Br2] = 0.0298 M
Keq = [HBr]2
[H2][Br2]
[HBr]2 = Keq [H2][Br2]
= (3.5 × 104)(0.0356 M)(0.0298 M)
[HBr]2 = 37.1308 M2
[HBr] =√ 37.1308 M2 = 6.0935... = 6.1 M
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The ICE method
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The ICE method
When reactants are converted into products, their concentrations decrease and the concentrations of the products increase according to molar ratios. When calculating Keq, we must use equilibrium concentrations, not initial concentrations. To find equilibrium concentrations, it is sometimes useful to use the ICE method.
Ex 1) Given a 1.0 M HCl solution produces [H3O+] of 0.9 M at
equilibrium, what is the equilibrium concentration of HCl?
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ICE
Initial
Change
Equilibrium
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ICE
HCl(aq) + H2O(l) ‐‐‐‐‐⇄ H3O
+(aq) + Cl−(aq)
Initial 1.0 M
Change
Equilibrium
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ICE
HCl(aq) + H2O(l) ‐‐‐‐‐⇄ H3O
+(aq) + Cl−(aq)
Initial 1.0 M
Change
Equilibrium 0.9 M
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ICE
HCl(aq) + H2O(l) ‐‐‐‐‐⇄ H3O
+(aq) + Cl−(aq)
Initial 1.0 M 0
Change
Equilibrium ? 0.9 M
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ICE
HCl(aq) + H2O(l) ‐‐‐‐‐⇄ H3O
+(aq) + Cl−(aq)
Initial 1.0 M 0
Change ‐X +X
Equilibrium ? 0.9 M
X = 0.9 M
[HCl]eq = 1.0 –X = 1.0 – 0.9 = 0.1 M
[Cl‐]eq = 0.9 M
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On the board:
Given: 2Al3+(aq) + 3S2‐(aq) Al2S3(aq)
If the concentration of S2‐ changes from 0.45 M to 0.30 M at equilibrium, what is the equilibrium concentration of Al2S3?
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2Al3+(aq) + 3S2‐(aq) Al2S3(aq) 3x = 0.15 so x = 0.050
I ? 0.45 M 0
C ‐2X ‐3X +X
E 0.70 M 0.30 M ?
0 + 0.05 = 0.050 M
2x = 0.10 so 0.70 M + 0.10 = 0.80 M
Keq = [Al2S3] = (0.050) = 3.7792… = 3.8
[Al3+]2 [S2‐]3 (0.70)2(0.30)3
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ICE assignment
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More practice:
Consider the following equilibrium
4NH3(l) + 5O2 (g) 4NO(g) + 6H2O(l)
If the reaction proceeds from a mixture and the concentration of NO increases from 1.02 M to 2.29 M at equilibrium, find the initial concentration of oxygen if its equilibrium concentration is 2.49 M.
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More practice:
Consider the following equilibrium
4NH3(l) + 5O2 (g) 4NO(g) + 6H2O(l)
If the reaction proceeds from a mixture and the concentration of NO increases from 1.02 M to 2.29 M at equilibrium, find the initial concentration of oxygen if its equilibrium concentration is 2.49 M.
If the initial concentrations of ammonia and water are both 1.00 M, calculate the Keq.
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4NH3(l) + 5O2(g) 4NO(g) + 6H2O(l)
I 1.00 M ? 1.02 M 1.00 M
C ‐4X* ‐5X +4X +6X*
E 2.49 M 2.29 M
2.29 M – 1.02 M = 1.27 M = 4x so x = 0.3175
2.49 M + 0.3175 = 2.8075 M = 2.81 M
The concentrations of pure substances remain constant and are left out of Keq so …
Keq = [NO]4 = (2.29)4 = 0.2873… = 0.287
[O2]5 (2.49)5
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