Equilibrium I

Report
Equilibrium
Chapter 15
At room temperature colorless N2O4 decomposes to brown NO2.
N2O4(g)  2NO2(g)
(colorless)
(brown)
• We expect the forward reaction rate to slow
over time and the rate of the reverse
reaction rate to increase over time.
• At some time, the color stops changing
which indicates and we have an equilibrium
mixture of N2O4 and NO2.
• Chemical equilibrium is the point at which
the concentrations (partial pressures of
gases) of all species are constant.
• A dynamic equilibrium exists when the
rates of the forward and reverse reactions
are equal.
– The reaction does NOT stop, but there is no
NET change in the reactant or product
concentrations.
• Use a double arrow () to imply that the
process is an equilibrium.
Haber process:
• Industrial preparation of ammonia from nitrogen and
hydrogen:
N2(g) + 3H2(g)  2NH3(g)
• Process is carried out at high temperature
(500oC) and pressure (200 atm).
• Much of the NH3 produced industrially is used
as a fertilizer.
N2(g) + 3H2(g)  2NH3(g)
• If we start with a mixture of nitrogen and hydrogen (in
any proportions), the reaction will reach equilibrium
with constant concentrations of nitrogen, hydrogen and
ammonia.
• If we start with just ammonia, the reaction will reach
equilibrium the same equilibrium with constant
concentrations of nitrogen, hydrogen and ammonia.
• No matter what the starting composition of reactants and
products is, the equilibrium mixture contains the same
relative concentrations of reactants and products.
– Equilibrium can be reached from either direction.
• Consider a simple reaction.
A(g)  B(g)
• Assume that it is an elementary process.
We can write rate expressions for each reaction.
Forward reaction: A  B
Rate = kf[A] kf = rate constant (forward reaction)
Reverse reaction:
B  A
Rate = kr[B]
kr = rate constant (reverse reaction)
In equilibrium the forward and reverse rates are equal
Rearranging:
kf[A]= kr[B]
[B]
[ A]

kf
kr
 Kc
This is called the
Equilibrium constant!
“c” because it’s based on concentrations
• For gaseous substances we can use the ideal gas equation to
convert between concentration and pressure:
• PV = nRT so M = (n/V) = (P/RT)
• For substances A and B:
•
[A] = (PA/RT) and [B] = (PB/RT)
• Ratefwd = kf PA/RT and Raterev = krPB/RT
• At equilibrium:
kf PA/RT = krPB/RT
• Rearranging, we get:
PB
PA
RT 
RT
PB
PA

kf
kr
 Kp
This is called the
Equilibrium constant!
“p” because it’s based on pressure
15.2 The Equilibrium Constant
• We can write an expression for the relationship between the
concentration of the reactants and products at equilibrium.
• For a general reaction: aA + bB  cC + dD
• The equilibrium expression (law of mass action)
• is given by:
Kc 
c
[C ] [ D ]
d
a
b
[ A] [ B ]
c
Kp 
d
PC PD
a
* P in atm
b
PA PB
*equilibrium concentration/pressure of (g) and (aq) only!
• Where Kc and Kp are equilibrium constants (Keq).
Converting between Kp and Kc
Kp=Kc(RT)Δn
Δn = (sum of coefficients of gaseous product) –
(sum of coefficients of gaseous reactant)
R is the gas constant (0.0821)
T must be Kelvin
When Δn is zero  Kp=Kc
• Keq depends on stoichiometry.
• Keq varies with temperature.
• Keq does NOT depend on initial
concentrations (pressures)
• Keq does NOT depend on the reaction
mechanism.
• We generally omit the units of Keq
The Magnitude of Equilibrium
Constants
• The equilibrium constant, Keq, is the ratio of
products to reactants.
– the larger Keq the more products are present at
equilibrium.
– the smaller Keq the more reactants are present at
equilibrium.
• If Keq >> 1, then products dominate at
equilibrium and “equilibrium lies to the right”
• If Keq << 1, then reactants dominate at
equilibrium and “equilibrium lies to the left”
The Direction of the Chemical
Equation and K
•
•
•
•
An equilibrium can be approached from any direction.
Consider the reaction: N2O4(g)  2NO2(g)
P
o
The equilibrium constant (at 100 C) is: K  P  6.49
However, when we write the equilibrium expression for
the reverse reaction: 2NO2(g)  N2O4(g)
P
o
K 
 0 . 154
• The equilibrium constant (at 100 C) is:
P
• The equilibrium constant for a reaction in one direction is
the reciprocal of the equilibrium constant of the reaction in
the opposite direction.
A + B  2C K
2 C  A + B 1/K
2
N O2
eq
N 2O4
N2O4
eq
2
NO 2
• The equilibrium constant of a reaction that has
been multiplied by a number is the equilibrium
constant raised to a power equal to that number.
A + B  2C K
2 A + 2 B  4 C K2
• The equilibrium constant for a net reaction made
up of two or more steps is the product of the
equilibrium constants for the individual steps.
A + B  2C
K1
+ C + A  D
K2
2 A + B  C + D K1K2
• Homogeneous equilibria - Equilibria in
which all reactants and products are present
in the same phase
• Heterogeneous equilibria - Equilibria in
which one or more reactants or products are
present in a different phase
Heterogeneous equilibria
• Consider the decomposition of calcium carbonate:
CaCO3(s)  CaO(s) + CO2(g)
• Experimentally, the amount of CO2 does not depend
on the amounts of CaO and CaCO3. Why?
• The concentration of a pure solid or pure liquid equals
its density divided by its molar mass.




=


• Neither density nor molar mass is a variable.
• Since the concentrations of solids and liquids are
constant, they are not included in the equilibrium
expression.
• Since the concentration of a pure solid or pure
liquid is not included in the equilibrium constant
expression. We anticipate that the amount of CO2
formed will not depend on the amounts of CaO and
CaCO3 present.
*As long as CaO and
CaCO3 are both present
At equilibrium!
• Note: Although the concentrations of these
species are not included in the equilibrium
expression, they do participate in the
reaction and must be present for an
equilibrium to be established!
#1 Suppose that the gas phase reactions A → B and B → A
are both elementary processes with rate constants of
3.8  10-2 s-1 and 3.1  10-1 s-1 respectively.
(a) What is the value of the equilibrium constant for the
equilibrium A  B?
(b) Which is greater at equilibrium, the partial
pressure of A or the partial pressure of B?
(a)
Kc =
kf
kr
=
3.8 x 10
-2
3.1 x 10
-1
s
-1
s
-1
= 0.12
(b) [A] Kc is less than 1 so reactants are favored
#2
Write the expression for Kc for the following reactions. In
each case indicate whether the reaction is homogeneous or
heterogeneous.
(a) 3 NO(g)  N2O(g) + NO2(g)
(b) Ni(CO)4(g)  Ni(s) + 4 CO(g)
(c) HF(aq)  H+1(aq) + F-1(aq)
(d) 2 Ag(s) + Zn+2(aq)  2 Ag+1(aq) + Zn(s)
(a)
Kc =
[N 2 O ][N O 2 ]
[N O ]
+
(c)
Kc =
(b )
3
Kc =

[H ][F ]
[C O ]
[N i(C O ) 4 ]
+
(d )
[H F ]
(a) and (c) are homogeneous
Kc =
4
[A g ]
[Z n
2
2
]
#3 When the following reactions come to
equilibrium, does the equilibrium mixture
contain mostly reactants or mostly products?
(a) N2 + O2  2 NO Kc = 1.5  10-10
(b) 2 SO2 + O2  2 SO3 Kp = 2.5  109
(a) mostly reactants (Kc is much less than 1)
(b) mostly products (Kp is much greater than 1)
#4
If Kc = 0.042 for
PCl3(g) + Cl2(g)  PCl5(g)
at 500 K, what is the value of Kp for this
reaction at this temperature?
 n = 1 m o l p ro d u cts - 2 m o l rea cta n ts = -1
K p = K c (R T )
n
=
0.042 [(0.0821)(500)]
-1
=
1 .0 x 1 0
-3
#5
At 1000 K, Kp = 1.85 for the reaction:
SO2(g) + ½ O2(g)  SO3(g)
(a) What is the value of Kp for the reaction
SO3(g)  SO2(g) + ½ O2(g)?
(b) What is the value of Kp for the reaction
2 SO2(g) + O2(g)  2 SO3(g)?
(a) 1 / 1.85 = 0.541
(b) (1.85)2 = 3.42
#6 Mercury (I) oxide decomposes into mercury and
oxygen: 2 Hg2O(s)  4 Hg(l) + O2(g)
(a) Write the equilibrium constant expression for
this reaction in terms of partial pressures
(b) Explain why we normally exclude pure solids
and liquids from equilibrium constant expressions.
(a)
K p = PO
2
(b) the concentration/pressure is constant for pure
solids and liquids
#7 Gaseous hydrogen iodide is placed in a closed
container at 425oC, where it partially decomposes
to hydrogen and iodine:
2 HI(g)  H2(g) + I2(g)
At equilibrium, it is found that
[HI] is 3.53  10-3 M, [H2] is 4.79  10-4 M, and
[I2] is 4.79  10-4 M.
What is the value of Kc at this temperature?
Kc =
[H 2 ][I 2 ]
[H I ]
2
-4
=
-4
(4.79 x 10 )( 4 . 79 x 10 )
-3
(3.53 x 10 )
2
= 0.0184
#8
The equilibrium: 2 NO(g) + Cl2(g)  2 NOCl(g) is
established at 500 K.
An equilibrium mixture of the three gases has partial pressures
of 0.095 atm, 0.171 atm, and 0.28 atm for NO, Cl2, and NOCl
respectively.
Calculate Kp for this reaction at 500 K.
Kp =
P
(P
2
NO
2
NOCl
=
)( PC l )
2
(0.28)
2
2
( 0 . 095 ) ( 0 . 171)
=
51
Calculating Equilibrium Constants
• Tabulate initial and equilibrium concentrations (or
partial pressures) for all species in the equilibrium.
• If an initial and an equilibrium concentration is given
for a species, calculate the change in concentration.
• Use the coefficients in the balanced chemical equation
to calculate the changes in concentration of all species.
• Deduce the equilibrium concentrations of all species.
• Use these equilibrium concentrations to calculate the
value of the equilibrium constant.
#9
A mixture of 0.10 mol of NO and 0.050 mol of H2, and 0.10 mol of H2O is
placed in a 1.0 L vessel at 300 K.
The following equilibrium is established:
2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g)
At equilibrium, [NO] = 0.062 M.
(a) Calculate the equilibrium concentrations of H2, N2, and H2O
(b) Calculate Kc
T h is is called
(a)
an " IC E " ch art
calculate the change in [NO], 0.10 – 0.062 = 0.04
2 NO +
initial 0.10
change - 0.04
equil 0.06
2 H2 
0.050
- 0.04
0.01
N2 +
0
+ 0.02
0.02
2 H2O
0.10
+ 0.04
0.14
(b)
Kc =
[N 2 ][H 2 O ]
2
[N O ] [H 2 ]
2
2
=
(0.02)(0.14)
2
2
(0.06) (0.01)
2
=
1000
#10 A mixture of 0.2000 mol CO2, 0.1000 mol H2, and 0.1600 mol H2O is placed in a 2.000 L
vessel. The following equilibrium is established at 500 K:
CO2(g) + H2(g)  CO(g) + H2O(g)
(a) Calculate the initial partial pressures of CO2, H2, and H2O
(b) At equilibrium, PH2O = 3.51 atm. Calculate the equilibrium partial pressures of CO2,
H2, and CO
(c) Calculate Kp for the reaction
(a)
P = n R T /V
PC O = (0.2000)(0.0821)(500) / 2.000 = 4.10 atm
2
P H = (0.1000)(0.0821)(500) / 2.000 = 2.05 atm
2
PH
(b)
2O
= (0.1600)(0.0821)(500) / 2.000 = 3.28 atm
the change in pressure of H2O = 3.51 – 3.28 = 0.23 atm
initial
change
equil
(c)
CO2
4.10
- 0.23
3.87
Kp =
+
(PC O )(PH 2 O )
(PC O 2 )(PH 2 )
H2 
2.05
- 0.23
1.82
=
CO +
0
+ 0.23
0.23
(0.23)(3.51)
(3.87)(1.82)
= 0.11
H2O
3.28
+ 0.23
3.51

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