Download 2 Chemical equilibrium occurs when a reaction and its reverse

Chemical equilibrium occurs when a reaction
and its reverse proceed at the same rate.
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At equilibrium, the forward and reverse
reactions proceed at the same rate.
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At equilibrium the amount of each reactant
and product is constant.
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Depicting Equilibrium
N2O4 (g)
2 NO2 (g)
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The Equilibrium Constant
Forward reaction:
N2O4 (g) → 2 NO2 (g)
Rate law:
Rate = kf [N2O4]
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The Equilibrium Constant
Reverse reaction:
2 NO2 (g) → N2O4 (g)
Rate law:
Rate = kr [NO2]2
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The Equilibrium Constant
Therefore, at equilibrium
Ratef = Rater
kf [N2O4] = kr [NO2]2
kf
[NO2]2
=
kr
[N2O4]
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The Equilibrium Constant
The ratio of the rate constants is a
constant at that temperature, and the
expression becomes:
Keq =
kf
kr
=
[NO2]2
[N2O4]
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The Equilibrium Constant
In General:
aA + bB
cC + dD
The equilibrium expression for this
reaction would be:
Kc =
[C]c[D]d
[A]a[B]b
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Questions: What are the equilibrium
expressions for these equilibria?
1. SnO2 (s) + 2CO (g) ↔ Sn (s) + 2CO2 (g)
2. CaCO3 (s) ↔ CaO (s) + CO2 (g)
3. Zn (s) + Cu2+ (aq) ↔ Cu (s) + Zn2+ (aq)
4. 2O3 (g) ↔ 3O2 (g)
5. 2NO (g) + Cl2 (g) ↔ 2NOCl (g)
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The Equilibrium Constant
NOTE: pressure is proportional to
concentration for gases in a closed
system:
Kp =
(PC)c (PD)d
(PA)a (PB)b
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Relationship between Kc and Kp
From the ideal gas law we know that
PV = nRT
Rearranging it, we get
P=
n
RT
V
L*atm /
Where R is 0.0821 L*atm
mol*K
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Relationship between Kc and Kp
Plugging this into the expression for Kp
for each substance, the relationship
between Kc and Kp becomes:
Kp = Kc (RT)∆n
Where
∆n = (moles of gaseous product) −
(moles of gaseous reactant)
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Questions
6. If, N2 + 3H2 ↔ 2NH3
Has a Kc = 9.60 at 300˚C, what is Kp?
7. If 2SO3 ↔ 2SO2 + O2
Has a Kc = 4.08*10-3 at 1000 K, what
is Kp?
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Equilibrium Can Be Reached
from Either Direction
The ratio of [NO2]2 to [N2O4] remains
constant (within error) at this temperature no
matter what the initial concentrations of NO2
and N2O4 are.
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Data from the last two trials from the
table on the previous slide.
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What does the magnitude of the
Value of K Mean?
If K >> 1, the reaction is product-favored;
product predominates at equilibrium.
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Magnitude of the Value of K Mean?
If K << 1, the reaction is reactant;
reactant predominates at equilibrium.
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The equilibrium constant of a reaction in the
reverse reaction is the reciprocal of the
equilibrium constant of the forward reaction.
N2O4 (g)
2 NO2 (g)
[NO ]2
2 NO2 (g) Kc = [N O2 ] = 0.212 at 100°C
2 4
[N2O4]
N2O4 (g) Kc = [NO
2
2]
=
1
0.212
= 4.72 at 100°C
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Questions
8. N2 + O2 ↔ 2NO has a Kc of 1*10-30
at 25˚C, is this a good reaction if
you intend to collect NO?
9. What is the value of Kc for the
reverse reaction and what is the
equilibrium expression?
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If a reaction doubled, you square the constant.
In general you raise the constant to the power
of the coefficient:
[NO ]2
N2O4 (g)
2 NO2 (g) Kc = [N O2 ] = 0.212 at 100°C
2 4
2 N2O4 (g)
[NO ]
4 NO2 (g)Kc = [N O2 ]2 = (0.212)2 at 100°C
2 4
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The equilibrium constant for a net reaction is
the product of the equilibrium constants for
the individual steps.
Step 1: NO2 + NO2 → NO3 + NO
Step 2: NO3 + CO → NO2 + CO2
Kc1=
[NO2][CO2]
[NO3][NO]
Kc2=
[NO3][CO]
[NO2]2
[NO][CO2]
Kc=
[NO2][CO]
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Questions
10. HF(aq) ↔ H+(aq) + F- (aq) Kc = 6.8*10-4
H2C2O4 (aq) ↔ 2H+ (aq) + C2O42-(aq) Kc = 3.8*10-6
Find Kc for 2HF(aq) + C2O42-(aq) ↔ 2F-(aq) + H2C2O4 (aq
(aq))
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Summary
1. At eq. ∆[conc] stops but reactions don’t
2. At eq. the ratio of [prod] and [react] are
equal to a coefficient (Kc)
3. Kc and Kp have no units
4. Kp = Kc(RT)∆n
5. Krev = (Kfor)-1
6. Kc of a reaction that has been multiplied is
raised to that power.
7. Kc of combined reactions is the product of
those reactions.
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