equilibrium calculations. how can we describe an equilibrium system mathematically? reactants...
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Equilibrium Calculations
How can we describe an equilibrium system mathematically?
reactants products⇌
reactants
The Keq is the equilibrium constant- a number that does not change.
Providing the temperature is kept constant.
products
= 3.0Keq =
Equilibrium Calculations An equilibrium system, at any given temperature, can be described by an equilibrium expression and equilibrium constant.
aA + bB ⇌ cC + dD
(aq) and (g) are included in the expression! (l) (pure liquids, meaning it is the only one in the equation) and (s) are not because they have constant concentration!
[A]a[B]b
[C]c[D]d
Keq =
Equilibrium Constant- a number Expression- mathematical equation
Keq =Products
Reactants
Only changes to (aq) and (g) reactants or products cause the equilibrium to shift.
Solids (s) and pure liquids (l) have constant concentrations. Solids & Liquids have fixed densities, cannot be compressed, so
their molar concentrations are constant (s) And (l) concentrations are already included in Keq value so we
don’t include them in the equation.
CaCO3(s) + 2H+(aq) + 2Cl-
(aq) ⇌ Ca2+(aq) + Cl2 (g)
+ CO2(g) + H2O(l)
(s) and (l) do not!
no shift right left left leftright no shift
SO3(g) + H2O(g) ⇌ H2SO4(l)
1. at 25oC, [SO3] = 0.200 M. [H2O] = 0.480 M, and [H2SO4] = 24 M.
Calculate the Keq.
The Keq has no units but concentration units that go in the expression must be M!
= 10.4
(0.200)(0.480)
1
[SO3] [H2O]
1don’t count (l)! Use 1Keq =
=
At equilibrium
No ICE
2. 0.500 mole PCl5, 0.40 mole H2O, 0.200 mole HCl, and 0.400 mole POCl3 are found in
a 2.0 L container at 125 oC. Calculate the Keq.
PCl5(s) + H2O(g) ⇌ 2HCl(g) + POCl3(g)
[HCl] = 0.200 moles = 0.10 M2.0 L
[POCl3] = 0.400 moles = 0.20 M2.0 L
= 0.20 M0.40 moles
2.0 L
[H2O] =
Keq = [HCl]2[POCl3]
[H2O]
Keq = [0.10]2[0.20]
[0.20]
Keq = 0.010
No ICE
at equilibrium
3. If 0.600 mole of SO3 and 0.0200 mole of SO2 are found in a 2.00 L container at
equilibrium at 25 oC. Calculate the [O2].
[SO3] = 0.600 mole/2.00 L = 0.300 M
[SO2] = 0.0200 mole/2.00 L = 0.0100 M
Keq = [SO 3]2
[SO2]2[O2]
798 = (0.300)2
(0.0100)2[O2]
(0.3)2 = 798(0.01)2[O2]
[O2] = (0.3)2
798(0.01)2
= 1.14 M
2SO2(g) + O2(g) ⇌ 2SO3(g)Keq = 798
1
Size of Keq & Effect of Temperature on Keq
products
reactants
Keq =
Big Keq
Keq = 10
reactants
products
Keq =
Little Keq
Keq = 0.1
Note that the keq cannot be a negative number!
Keq = 1
Keq =
products
reactants
Effect of Temp on Keq
Keq Only a temperature change can affect the value of Keq.
Changes in concentrations, pressure or surface area have NO effect on Keq.
– These changes correspond to increase in number of reacting molecules per liter.
– Increased once and then equilibrium is re-established.
– So ratios of products to reactants do not change.
PBr3(g) + Br2(g) ⇋ PBr5(g) + energy
@ 100 oC Keq = 0.17 @ 200 oC Keq = 0.091
How can you tell if Keq gets bigger or smaller?
Temperature increased
Shifted left, Keq decreased
Keq= [products] ------------- [reactants]
More questions…
If the value of Keq increases when the temperature decreases , is the reaction exothermic or endothermic?
More questions…
What will happen to the value of Keq in the following reaction if we added more [B]?
A + B C + 100 KJ.⇌
Hebden Practice
Page 60: Exercises 31-35
Page 62: Exercises 36-45