Transcript CHEMICAL EQUILIBRIUM - RATES OF REACTION k Reactants products

CHEMICAL EQUILIBRIUM - RATES OF REACTION
kF
Reactants

products
kB
Chemical reactions are a dynamic process, that is,
reactions involve both forward and reverse processes.
Chemical Equilibrium is reached by as reaction mixture
when the rates of forward and reverse reactions becomes
equal.
kF = kB
NO net change appears obvious although the system is
still in constant motion.
LAW OF MASS ACTION
k  equilibrium constant
aA + bB  cC + dD
k = Products
Reactants
k = [C]c [D]d = Q reaction quotient
[B]b [A]a
Write the equilibrium equation for:
a. HC2H3O2  H+ + C2H3O2b. H2O + H2O  H3O+ + OHc. 4NH3(g) + 302(g)  2N2(g) + 6H2O(g)
d. N2(g) + 3H2(g)  2NH3(g)
Predicting the direction of reaction:
Q >K
forms more reactants

Q = K
equilibrium
Q < K
forms more products

Note: 1. kf = 1
kr
2. k = kn
If the balanced equation is multiplied
by a factor then the K(& Q) is
multiplied by the exponent.
K as either Kc or Kp
Kc = the equilibrium constant using concentrations.
Kp = the equilibrium constant using pressure
P = n  Pxn
RT v ••
v
Kp = Kc(RT) ngas
Write Kp the Kc for:
1. N2(g) + 3H2(g)  2NH3(g)
2. N2O4(g)  2NO2(g)
3. Calculate kp if kc = 0.105 for #1
4. 2SO3(g)  2SO2(g) + O2(g)
if Kc = 4.07 x 10-3, what is kp?
N2 + O2  2NO
a. Since Q is very small Q<<<1, very little, NO will form @
25°C. The equilibrium lies to the left favoring reactants.
N2 + O2 = 2NO
b. Kp = (PNO)2
PN2PO2
n = 2 - 2 = 0
kp = Kc
Kc = [NO]2
[N2][O2]
c. Kp = PN2 PO2
(PNO)2
d. n = 2 - 2 = 0 kp = kc
kc = [N2][O2]
[NO]2
DIRECTION OF REACTIONS AND Keg
1. The following reaction is a means of “fixing” nitrogen:
N2(g) + O2(g)  2 NO(g)
A. If the value for Q at 25°C is 1 x 10-30, describe
the feasibility of this reaction for Nitrogen
fixation.
B. Write the equilibrium expression, Kc
C. Write the equilibrium expression for
2NO(g)  N2(g) + O2(g)
D. Determine the Kc for “C”
HOMOGENEOUS EQUILIBRIA
H2(g) + I2(g)  2HI(g)
Kp = (PHI)2
(PI2)(PH2)
Kc = ?
2O3(g)  3O2(g)
Kp = (PO2)3
(PO3)2
Kc = ?
HETEROGENEOUS EQUILIBRIA
2H2O(l) 
H3O+(aq)_ + OH-(aq)
Kc = [H3O+][OH-]
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)
Kc = ?
HETEROGENEOUS EQUILIBRIA
Substance in more then 1 phase
1.
CaCO3(s)  CaO(s) + CO2(g)
Kc = [CaO][CO2]
[CaCO3]
How do the [ ] of solid express?
A. D
= g/cm3 = mol
MW
g/mol
cm3
Pure solids & liquid have constant [ ]
Kc = constant [CO2]
constant
Kc = Kc con = [CO2]
con
Each of the mixtures listed below was placed in a closed
container and allowed to stand. Which of these mixtures
is capable of attaining the equil, expressed by 1
a)
pure CaCO3
b)
CaO & PCO2 > Kp
c)
solid CaCO3 & PCO2 > Kp
d)
CaCO3 & CaO
CALCULATING THE Keq
1.
In one experiment, Haber introduced a mixture of H2
& N2 into a reaction vessel and allowed the system
to attain chemical equilibrium at 472°C. The
equilibrium mixture of gases were analyzed and
found to contain 0.1207M H 2, 0.0402M N2, and
0.00272M NH3. Calculate Keq.
2.
Nitryl Chloride, NO2Cl, is in equilibrium in a
closed container with NO2 and Cl2.
2 NO2Cl(g)  2 NO2(g) + Cl2(g)
Calculate Keq if [NO2Cl] = 0.00106M
[NO2] = 0.0108M & [Cl2] = 0.00538M
3.
For the Haber process
N2(g) + 3H2(g)  2NH3(g)
Kp = 1.45 x 10-5 at 500°C
If an equilibrium mixture of the three gas
started with partial pressures of 0.928 atm for
H2 and 0.432 atm for N2, what is the partial
pressure of NH3?
4.
A 1.00 L flask is filled with 1.00 mol of H2 and
2.00 mol I2 at 448°C is 50.5.
What are the equilibrium concentrations of H2, I2
& HI?
CALCULATING Keq
1. A mixture of 5.0 x 10-3 mol of H2 and 1.0 x 10-2 mol of I2 is
placed in a 5.0L container at 448°C and allowed to come to
equilibrium. Analysis of this equilibrium mixture shows that the
[HI] is 1.87 x 10-3 M.
Calculate Kc:
H2(g) + I2(g)  2HI(g)
2. At 448°C the equilibrium constant Kc for the
reaction below is 50.5.
H2(g) + I2(g)  2HI(g)
Predict how the reaction will proceed to reach
equilibrium if the initial amount of HI is 2.0 x 10-2 mol,
H2 is 1.0 x 10-2 mol, and I2 is 3.0 x 10-2 mol in a 2.00 L container.
APPLICATION OF Keq
1. Predicting the direction of reaction
Q = reaction quotient
at equil Q = K
(no net Rx)
Q > K  species on Rt (prod)
react to form left
K = [Equil] Q = [Non Equil]
Q < K  forms more products
Goal: Calculate Q to determine state of Rx, equil, more product
or more reaction
1. At 448°C the equilibrium constant Kc for the
reaction is 50.5.
H2(g) + I2(g)  2HI(g)
Predict how the Rx will proceed to reach equil at 448°C if the
initial amount of HI is 2.0 x 10-2 mol, 1.0 x 10-2 mol, H2 and 3.0
x 10-2 mol I2 in at 2.0 L container.
[HI]° = 2 x 10-2 mol/2.0L = 1.0 x 10-2M
[H2]° = 1.0 x 10-2mol/2.0L = 5.0 x 10-3M
[I2]° = 3.0 x 10-2mol/2.0L = 1.5 x 10-2M
Q = Prod = [HI]2 =
(1.0 x 10-2)2
= 1.3
React [H2][I2]
(5 x 10-3)(1.5 x 10-2)
since K = 50.5, Q = 1.3, Q < K [HI] will need to increase
and [H2][I2] will decrease to reach equilibrium.
2.
At 1000K the value of Kc for the reaction
2S03(g)  2SO2(g) + O2(g)
is 4.07 x 10-3. Calculate the value for Q
and predict the direction in which the reaction
will proceed towards equil if the initial
concentration of reactants are:
[SO3] = 2 x 10-3 M
[SO2] = 5 x 10-3 M
[O2] = 3 x 10-2 M
Q = 0.2 reaction will proceed from Rt to
left forming SO3.
CALCULATING Keq
1.
Sulfur Trioxide decomposes at High temperature in a
sealed container.
2 SO3(g)  2 SO2(g) + O2 (g)
Initially the vessel is filled at 1000K with SO3(g) at
a concentration of 6.09 x 10-3M. At equilibrium, the
[SO3] is 2.44 x 10-3M. Calculate Kc.
2.
Calculate the value for Q and predict the direction in
which the reaction will proceed towards equilibrium if
the initial concentrations are:
[SO3] = 2.0 x 10-3M
[SO2] = 5.0 x 10-3M
[O2] = 3.0 x 10-2M
LE CHATELIER’S PRINICIPLE
“If a system at equilibrium is disturbed by a change in
temperature, pressure, or concentration of one of its
components, that system will shift it’s equilibrium position
as to ‘counteract’ the effect of the disturbance.”
Equilibrium can be disturbed by:
-
adding or removing components
a change in pressure
a change in volume
a change in temperature
PREDICTING THE DIRECTION OF THE SHIFT
I.
CATALYST
A catalyst increases the rate at which equilibrium
is achieved but not the composition of the
equilibrium mixture.
II.
THE REACTION QUOTIENT
Q < K: the reaction shifts to the products
Q > K: the reaction shifts to the reactants
III.
CHANGES IN VOLUME
Reducing the volume of a gas at equilibrium causes
the system to shift in the direction that reduces
the number of moles of gas.
IV.
CHANGES IN TEMPERATURE
When heat is added to a system, the equilibrium
shifts in the direction that absorbs heat. Cooling
has the opposite effect and shifts the equilibrium
towards the side which produces heat.
Endothermic Reactions:
heat + Reactants  Products
An increase in temperature leads to a shift
towards the products, a decrease leads to
a shift towards the reactants. (Keq increases)
Exothermic Reactions:
Reactants  Products + heat
An increase in temperature leads to a shift
towards reactants. (Keq decreases)
Example:
N2O4(g)  2 NO2(g)
H = 58 kJ
In which direction will the equilibrium shift when each
of the following changes are made to a system at
equilibrium?
A) add N2O4
B) remove NO2
C) increase the total pressure by adding N2
D) increase the volume of the container
E) decrease the temperature
Example:
PCl5(g)  PCl3(g) + Cl2(g)
H = 88 kJ
In which direction will the equilibrium shift when each of the
following changes are made to a system at equilibrium?
A) add Cl2
B) temperature is increased
C) the volume of the reaction system is decreased
D) PCl5 is added
E) a catalyst is added
N2O4  2NO2
A) The system will adjust so to decrease [N2O4] shifts
to  products
B) Shifts to  reactants (NO2 removed)
C) N2 will increase total pressure but since it is not
involved in Rx the partial pressures of N2O4 and
NO2 are unchanged, no shift.
D)  volume shifts  to more moles of gas
E)  Temp - Rx is endo heat + N2O4  2 NO2
Temp  shifts so more heat produced  K is affected
1.
The equilibrium constant for the Haber process
at 472°C is Kc = 0.105. A 2.00 L flask is filled
with 0.500 mol of ammonia and is then allowed to
reach equilibrium at 472°C. What are the equilibrium
concentrations?
N2(g) + 3 H2(g)  2 NH3(g)
2.
For the reaction
PCl5(g)  PCl3(g) + Cl2(g)
at a certain temperature Kc equals 450. What will
happen when 0.10 mol of PCl5, 1.0 mol of PCl3, and
l.5 mol of Cl2 are added to a 2.0-L container and
the system is brought to the temperature at which
Kc=450. What are the equilibrium concentrations?
EFFECT OF VARIOUS DISTURBANCES ON AN EQUILIBRIUM SYSTEM
DISTURBANCE
NET DIRECTION OF REACTION
EFFECT ON
VALUE OF K
Concentration
Increase (reactant)
Decrease (reactant)
Toward formation of product
Toward formation of reactant
None
None
Pressure (volume)
Increase P
Decrease P
Temperature
Increase T
Decrease T
Catalyst added
Toward formation of lower
amount (mol) of gas1
Toward formation of higher
amount (mol) of gas
Toward absorption of heat
Toward release of heat
None; rates of forward and
reverse reactions increase equally
None
None
Increases H°rxn>0
Decreases if H°rxn<0
Increases H°rxn<0
Decreases H°rxn>0
None
VAN’T HOFF EQUATION
Changes in K due to T
In K2 = - H°rxn ( 1 - 1)
K1
R
T1 T2
R = 8.314 J/mol K
T = Kelvin
The formation of methanol is an important industrial
reaction in the processing of new fuels. At 298K,
Kp = 2.25 x 104 for the reaction
CO(g) + 2 H2(g)  CH3OH(l)
If H°rxn = -128 kJ/mol CH3OH, calculate
kp at 0°C.
CH4(g) + CO2(g)  2CO(g) + 2 H2(g)
A.
What is the percent yield of H2 when equimolar
mixture of CH4 and CO2 with a total pressure
of 20.0 atm reaches equilibrium at 1200K at
which Kp = 3.548 x 106?
B.
What is the percent yield of H2 for this system at
1300K, at which Kp = 2.626 x 107?
C.
Use the Van’t Hoff equation to find H°rxn.