Transcript N 2 (g)

WHEN A REACTION IS AT
EQUILIBRIUM, BOTH THE
FORWARD AND REVERSE
REACTIONS ARE
OCCURING AT THE SAME
RATE.
CONCENTRATION REMAINS
CONSTANT AT EQUILIBRIUM
…..BECAUSE
REACTANTS ARE BEING USED UP
TO PRODUCE PRODUCTS
AT THE SAME RATE THAT
PRODUCTS ARE BEING USED UP
TO PRODUCE THE REACTANTS.
Kc or Keq or KP
are
EQUILIBRIUM CONSTANTS.
THEY SHOW A RELATIONSHIP BETWEEN REACTANTS
AND PRODUCTS IN A REACTION.
CONCENTRATIONS ARE GIVEN IN MOLARITY [ ].
FOR THE REACTION:
aA + bB D cC + dD
Keq= [C]c [D]d
[A]a [B]b
NOTE, PRODUCTS OVER
REACTANTS.
N2O4 (g) D 2 NO2 (g)
When the reactant, N2O4 is put in an evacuated
container at 100 oC, it decomposes to NO2. In
the beginning only NO2 is formed, but as soon
as it forms, it begins going back to forming
N2O4. Eventually, the rate of the forward and
reverse reactions are equal. Thus, the reaction
has reached equilibrium.
The following data and diagram depict this reaction at equilibrium.
Time (s)
0
Conc. N2O4 (M)
0.100
Conc. NO2 (M)
0.000
20
40
60
0.070
0.050
0.040 0.040 0.040
0.060
0.100
0.120 0.120 0.120
0.120
80
100
NO2
0.100
Conc(M)
0.040
0.000
N2O4
0 20
Time (s)
60
•Regardless of quantities of reactants or products started
with
•Regardless of pressure
•Regardless of volume
THE RATIO OF PRODUCTS TO REACTANTS WILL
BE A CONSTANT AT EQUILIBRIUM AT ANY GIVEN
TEMPERATURE.
EX. [NO2 ]2
[N2O4 ]
Exp 1.
[NO2 ]2
[N2O4 ] = (0.120)2 / 0.040 = 0.36
Exp 2.
[NO2 ]2
[N2O4 ] = (0.072)2 / 0.014 = 0.37
Exp. 3.
[NO2 ]2
[N2O4 ] = (0.160)2 / 0.070 = 0.36
But if temp. is increased to 150 oC the Keq = 3.2
AN EQUILIBRIUM EXPRESSION IS
ASSOCIATED WITH A REACTION
N2O4 (g) D 2 NO2 (g)
Keq = [NO2]2 / [N2O4]
= 0.36
½ N2O4 (g) D NO2 (g)
Keq = [NO2] / [N2O4]1/2
= (0.36)1/2 = 0.60
2 NO2 (g) D N2O4 (g)
Keq = [N2O4] / [NO2]2
= 1 / 0.36 = 2.8
PURE LIQUIDS AND SOLIDS
ARE NOT INCLUDED IN THE
EQUILIBRIUM EXPRESSION
• CO2(g) + H2(g) D CO(g) + H2O( l )
Keq = [CO]
[H2 ][CO2]
• I2(s) D 2 I(g)
Keq = [I ]2
CuO(s) + H2(g) D Cu(s) + H2O(g)
Keq = [H2O]
[H2]
NOTICE THAT THE SOLIDS ARE NOT PRESENT
IN THE EQUILIBRIUM EXPRESSION.
CALCULATE Keq FOR THE FOLLOWING REACTION:
NH4Cl (s) D
NH3(g) + HCl(g)
2 moles of NH3 and 2 moles of HCl and 1 mole of NH4Cl are in
5.0 L at equilibrium.
Keq = [NH3][HCl]
[HCl] = 2MOLES
5.0 L = 0.4M
[NH3] = 2MOLES 5.0 L = 0.4M
Keq = 0.4 x 0.4 = 0.16
FOR THE FOLLOWING REACTION:
2HI(g) D H2(g)
+ I2(g)
STARTING WITH 0.100M HI, AND THEN AT EQUILIBRIUM,
[H2] = 0.010 M. CALCULATE [I2], [HI], AND Keq.
SOLUTION:
AT THE START, H2 AND I2 ARE ZERO AND ARE ALSO A 1:1
RATIO. THEY WILL HAVE THE SAME MOLAR CONC.
2 [HI] = [H2] THAT IS, 2 [HI]ARE REQUIRED TO MAKE ONE
[H2]. SO WHATEVER THE [H2] IS, THE [HI] IS DECREASING
BY DOUBLE THAT AMOUNT.
THEREFORE, 0.010 [H2] x 2 = 0.020M DECREASE IN [HI],
THEREFORE, 0.100 [HI] – 0.020 = 0.080M [HI] AT
EQUILIBRIUM.
CONTINUED…….
Keq= [I2][H2] [HI]2
=
(0.010) (0.010) (0.080)2
Keq= 0.016
Why do we care about the equilibrium constant?
• IT SHOWS TO WHAT EXTENT A REACTION WILL
PROCEED.
• IT SHOWS THE DOMINATING DIRECTION IN WHICH
THE REACTION WILL GO TO REACH EQUILIBRIUM.
• IT SHOWS THE CONCENTRATIONS OF SPECIES
PRESENT AT EQUILIBRIUM.
FOR
N2O4 (g) D
2NO2(g) Keq = 0.36
CALCULATE THE QUOTIENT OF PRODUCTS TO
REACTANTS (Q) AND DETERMINE THE DIRECTION
THE REACTION WILL SHIFT TO REACH
EQUILIBRIUM FOR THE FOLLOWING CONDITIONS:
a) 0.20 MOLE / 4.0 L N2O4 = 0.05 M
b) 0.20 MOLE / 4.0 L N2O4 AND 0.20 MOLE / 4.0 L NO2
Solution for part a:
Q =
02 = 0,
0.05
Q < K THEREFORE THE REACTION
WILL SHIFT TO THE RIGHT.
b) 0.20 MOLE / 4.0 L N2O4 = 0.05 M
0.20 MOLE / 4.0 L NO2 = 0.05 M
Q = 0.052 = 0.05 < 0.36 therefore the reaction
0.05
proceeds to the right
WE HAVE USED Keq, THE EQUILIBRIUM
CONSTANT, WHICH SHOWS THE RATIO OF THE
PRODUCTS TO THE REACTANTS AT
EQUILIBRIUM. WE KNOW THAT Kc REPRESENTS
MOLAR CONCENTRATIONS OF SPECIES AT
EQUILIBRIUM AND Kp REPRESENTS THE RATIO
OF PARTIAL PRESSURES OF GASES AT
EQUILIBRIUM.
THE REACTION QUOTIENT, Q, IS LIKE K BUT IT
IS NOT NECESSARILY AT EQUILIBRIUM.
COMPARING IT TO K IS A WAY IN WHICH WE
MAY DETERMINE THE DIRECTION IN WHICH
THE REACTION WILL PROCEED IN ORDER TO
RE-ESTABLISH OR REACH EQUILIBRIUM.
RELATIONSHIPS OF K, Q,
AND DGo
• Q < K…………………….
• Q > K…………………….
• Q = K…………………….
DGo < 0, LnK > 0,
K>1
DGo > 0,
LnK < 0,
DGo = 0,
LnK = 0,
•
•
•
•
REACTION SHIFTS RIGHT
REACTION SHIFTS LEFT
REACTION AT EQUILIBRIUM
EQUILIBRIUM MIXTURE IS
MOSTLY PRODUCTS
K<1 • EQUILIBRIUM MIXTURE IS
MOSTLY REACTANTS
K=1 • EQUILIBRIUM MIXTURE HAS
COMPARABLE AMOUNTS OF
REACTANTS AND PRODUCTS
EXAMPLE:
At the start of a reaction there are the following species in a 3.50 L
reaction vessel at 430 oC:
0.0218 mole H2 , 0.0145 mole I2 , 0.0783 mole HI.
Kc = 54.3. Determine if the system is at equilibrium and if not,
determine the direction in which it will proceed.
H2(g) + I2(g) D 2 HI(g)
[H2] = 0.0218 mol / 3.50 L = 0.00623 M
[I2] = 0.0145 mol / 3.50 L = 0.00414 M
[HI] = 0.0783 mol / L = 0.0224 M
Q=
0.02242
= 19.5
0.00414 x 0.00623
Q < K therefore the reaction shifts
RIGHT
GIVEN THE REACTION; N2 (g) + O2 (g) D 2 NO(g)
Keq =
[NO]2
= 1 x 10-30 at 25oC
[N2] [O2]
IN ORDER FOR Keq TO BE THAT SMALL, THE NUMERATOR
MUST BE SMALL COMPARED TO THE DENOMINATOR.
THIS MEANS THAT [NO] << [N2] [O2].
THEREFORE, THIS Keq SHOWS VERY LITTLE PRODUCTION
OF PRODUCTS (THE NO).
IT IS NOT A FEASABLE REACTION.
WE CAN CALCULATE THE [NO] , GIVEN [N2] = 0.040 M AND
[O2] = 0.010 M
Keq = 1 x 10 –30 = [NO]2
(0.010)(0.040)
= 4 x 10-34 = [NO]2
2 x 10-12 = [NO]
Example:
FOR THE FOLLOWING REACTION, WHERE Keq = 0.64 AT
900K, AND TO START THE REACTION, CO2 AND H2 ARE
BOTH 0.10 M, WHAT ARE THE EQUILIBRIUM
CONCENTRATIONS OF ALL SPECIES?
CO2 (g) + H2 (g) D CO(g) + H2O(g)
CHOOSE A SPECIES TO CALL ‘X’ AND DETERMINE INITIAL
AND FINAL CONCENTRATIONS.
CO2
H2
CO
H2O
INITIAL
0.10
0.10
0
0
CHANGE
-X
-X
+X
+X
FINAL
0.10-X
0.10-X
X
X
continued…………..
0.64 =
x2
(0.64)1/2 =
(0.10 – x)2
x
0.10 – x
X = 0.044 M
[CO2] = 0.100 – 0.044 = 0.056 = [H2]
[CO] = 0.044 = [H2O]
USING THE SAME EQUATION AND Keq, BUT WITH CO2 =
0.100 M AND H2 = 0.200 M, CALCULATE THE EQUILIBRIUM
CONCENTRATIONS.
0.64 =
x2
(0.200 – x) (0.100 – x)
=
x2
(0.0200 – 0.200x – 0.100x + x2)
Using the quadratic equation, [ x ]= 0.0524 M = [CO] = [H2O]
0.10 – 0.0524 = 0.0476 = [CO2] , 0.200 – 0.0524 = 0.148 = [H2]
Example:
N2 O4 (g) D 2NO2 (g)
Kc = 0.36 at 100.oC
And starting concentration of N2 O4 = 0.100 M
What are the equilibrium concentrations of [NO2] and [N2O4] ?
N2 O4
NO2
INITIAL
0.100
0.000
CHANGE
-X
+2X
FINAL
0.100 – X
2X
Kc =
(2X)2
0.100 - X
=
4X2
0.100 – X
4X2 – 0.35 X + 0.036 = 0 , USING THE QUADRATIC EQUATION;
X = 0.060,
2X = 0.120 M = [NO2], 0.100 – 0.060 = 0.040 = [N2O4]
THE EFFECT OF CHANGES IN CONDITIONS ON
A SYSTEM AT EQUILIBRIUM
WHEN CONDITIONS ARE CHANGED ON A SYSTEM AT
EQUILIBRIUM, THE EQUILIBRIUM IS DISRUPTED AND
THE CONCENTRATIONS MAY CHANGE.
SOME THINGS THAT CAN DISRUPT EQUILIBRIUM:
1. ADDING OR REMOVING REACTANT OR PRODUCT.
2. CHANGING THE VOLUME OF THE SYSTEM
3. CHANGING THE TEMPERATURE
To determine the direction the equilibrium will shift, we apply
LeChatelier’s principle. We calculate Kc to determine the
concentrations, as before.
Example:
GIVEN THAT Kc = 0.016 at 520 oC AND
CONCENTRATION OF [HI] = 0.080 M AND
[ I2] = [H2] = 0.010 M At EQUILIBRIUM, WHAT
WOULD THE NEW CONCENTRATIONS BE
WHEN EQUILIBRIUM IS RESTORED, IF THE [HI]
IS TEMPRORARILY RAISED TO 0.096 M?
INITIAL CONCENTRATIONS ARE THOSE IMMEDIATELY
FOLLOWING THE DISRUPTION IN EQUILIBRIUM:
2HI D H2 + I2
[HI]
[I2]
INITIAL
0.096 M
0.010 M
CHANGE
-2X
FINAL
0.096 – 2X
+X
0.010 + X
[H2]
0.010 M
+X
0.010 + X
Kc = 0.016 = (0.010 + X)(0.010 + X) = (0.010 + X)2
(0.096 – 2X)2
(0.096 – 2X)2
(0.016)1/2 = 0.010 + X
0.096 – 2X
0.126 = 0.010 + X
ALGEBRA TIME 

0.096 – 2X
X = 0.0017 M
[H2] = 0.010 + 0.0017 = 0.0117 M
[I2] = 0.010 + 0.0017 = 0.0117 M
[HI] = 0.096 – 2(0.0017) = 0.0926 M
NOTE: THIS VALUE IS BETWEEN THE STARTING CONC.,
0.080 AND HIGH VALUE OF 0.096. THIS MAKES SENSE!
ALWAYS CHECK THAT ANSWERS MAKE SENSE WHEN YOU
COMPLETE A PROBLEM!
EQUILIBRIUM OF A SYSTEM CAN ONLY BE DISRUPTED BY
ADDING OR REMOVING SPECIES IF THAT SPECIES
APPEARS IN THE EQUILIBRIUM EXPRESSION.
REMEMBER, SOLIDS AND PURE LIQUIDS ARE NOT
INCLUDED IN EQUILIBRIUM EXPRESSIONS.
example
CaCO3 (s) D CaO(s) + CO2(g)
Kc = [CO2]
THEREFORE, ADDING AND REMOVING
CaCO3(s) or CaO(s) DOES NOT AFFECT EQUILIBRIUM.
VOLUME CHANGES AND THE
EFFECT ON EQUILIBRIUM
N2O4 (g) D 2NO2 (g)
NOTE THAT THERE ARE TWO MOLES OF GAS ON THE
RIGHT AND ONE MOLE OF GAS ON THE LEFT.
LETS DETERMINE WHAT WILL HAPPEN IF THE VOLUME
OF THE CONTAINER IS DECREASED, (PRESSURE
INCREASED), OR THE VOLUME OF THE CONTAINER IS
INCREASED, (PRESSURE DECREASED). WE WILL USE
LeCHATELIER’S PRINCIPLE TO DETERMINE THE SHIFT.
WE TAKE INTO CONSIDERATION THE NUMBER OF GAS
PARTICLES ON THE LEFT COMPARED WITH THE
NUMBER ON THE RIGHT.
N2O4 (g) D 2NO2 (g)
IF THE PRESSURE IS INCREASED, (VOLUME
DECREASED), THERE WILL BE AN INCREASE IN
GAS PARTICLES PER UNIT VOLUME.
THEREFORE THE REACTION WILL SHIFT IN
THE DIRECTION TO DECREASE GAS
PARTICLES. SINCE THE RIGHT HAS TWICE AS
MANY PARTICLES AS THE LEFT, THE
REACTION WILL SHIFT LEFT (TOWARD THE
LESSER MOLES OF GAS. THUS, NO2 WILL FORM
MORE N2O4.
THE OPPOSITE WILL OCCUR IF THE PRESSURE
IS DECREASED AND THE VOLUME INCREASED.
AFFECTS OF PRESSURE ON
EQUILIBRIUM
 = N2O4  = NO2
2NO2



 
 
 
 
 
N2O4
  
 

THESE CHANGES IN CONCENTRATION CAN BE
CALCULATED AS BEFORE.
Example
Predict the direction of the shift in the reaction:
a. When the volume is increased
b. When the pressure is increased
•C(s) + H2O(g) D CO(g) + H2(g)
•SO2(g) + ½ O2(g) D SO3(g)
1a. 2 MOLES OF GAS ON THE RIGHT AND ONE ON THE
LEFT, THEREFORE, SHIFTS RIGHT.
1b. AN INCREASE IN PRESSURE RESULTS IN A SHIFT TO
THE LEFT, TOWARD THE LESSER MOLES OF GAS.
2a. A DECREASE IN PRESSURE, SO A SHIFT TO THE LEFT
2b. AN INCREASE IN PRESSURE CAUSES A RIGHT SHIFT.
PRESSURE INCREASE = SHIFT TOWARD
LESSER MOLES OF GAS. (TO DECREASE
THE NUMBER OF PARTICLES)
PRESSURE DECREASE = SHIFT TOWARD
GREATER MOLES OF GAS. (INCREASES
THE NUMBER OF PARTICLES)
WE CAN INCREASED PRESSURE TO A SYSTEM
WE
CAN INCREASE
TO A SYSTEM
AT EQUILIBRIUM
BYPRESSURE
ADDING ANOTHER
GAS AT
EQUILIBRIUM
BY ADDING
ANOTHER GAS
WHILE AT CONSTANT
VOLUME.
WHILE AT CONSTANT VOLUME.
IF WE ADD A GAS, AT CONSTANT VOLUME,
THAT IS UNREACTIVE TO OUR SYSTEM AT
EQUILIBRIUM, THE PRESSURE WILL INCREASE,
BUT THERE WILL BE NO SHIFT IN
EQUILIBRIUM.
HOW WOULD YOU INCREASE THE YIELD OF
NO2 IN THE FOLLOWING REACTION:
NO(g) + ½ O2(g) D NO2(g)
a. INCREASE PRESSURE BY COMPRESSION?
b. INCREASE VOLUME?
c. ADD AN INERT GAS?
THE ANSWER IS ‘a’. AN INCREASE IN
PRESSURE WILL RESULT IN A SHIFT TOWARD
THE LESSER MOLES OF GAS.
THE EFFECTS OF
TEMPERATURE CHANGE ON A
SYSTEM AT EQUILIBRIUM.
ACCORDING TO LeCHATELIER’S PRINCIPLE, IF
TEMPERATURE IS INCREASED ON A SYSTEM
AT EQUILIBRIUM, THE REACTION WILL SHIFT
IN A DIRECTION TO COUNTERACT AND THUS
ABSORB THE HEAT.
IF THE REACTION IS ENDOTHERMIC, THAT IS,
ONE THAT ABSORBS HEAT, INCREASING
TEMPERATURE WILL CAUSE A SHIFT FARTHER
TO THE RIGHT.
IF THE REACTION IS EXOTHERMIC, THAT IS,
ONE THAT EXPELLS HEAT, AN INCREASE IN
TEMPERATURE WILL CAUSE THE REACTION TO
SHIFT IN THE DIRECTION THAT ABSORBS HEAT,
WHICH IS A SHIFT TO THE LEFT.
N2O4(g) D 2NO2(g)
DH = +58.2KJ
WHAT EFFECT WILL AN INCREASE IN
TEMPERATURE HAVE ON THIS SYSTEM?
THE REACTION WILL SHIFT RIGHT TO
ABSORB THE EXCESS HEAT.
N2O4(g) D
2NO2(g)
DH = -58.2KJ
HOW WOULD A TEMPERATURE INCREASE
AFFECT THIS SYSTEM AT EQUILIBRIUM?
SINCE THIS REACTION IS EXOTHERMIC, A
TEMPERATURE INCREASE WILL CAUSE A
SHIFT TO THE LEFT.
IF THE REACTION IS ENDOTHERMIC, Kc
BECOMES LARGER WITH A TEMPERATURE
INCREASE. (A SHIFT TO THE RIGHT CAUSES
AN INCREASE IN THE PRODUCTION OF
PRODUCTS.
IF THE REACTION IS EXOTHERMIC, THE
REACTION WILL SHIFT LEFT WITH A
TEMPERATURE INCREASE, THEREBY
DECREASING THE PRODUCTS AND SO Kc
ALSO DECREASES.
Example
I2(g) D 2I(g)
DH = + 151 KJ
If this system is at equilibrium at 1000 oC, what
directional shifts would occur when:
a. I atoms are added
b. The system is compressed
c. The temperature is increased
d. Which of these would affect Kc if any, and what
would be the affect?
a. If I atoms were added the reaction would shift left to
use up the excess I atoms.
b. If the system was compressed the reaction would
shift left toward the lesser moles of gas.
c. If the temperature was increased the reaction would
shift in the direction that absorbs heat, which in this
case is right because it is an endothermic reaction.
d. The temperature increase would cause an increase
in Kc because the rightward shift increases the
production of products.
The Relationship of Kc and Kp
Kc refers to solutions with concentration expressed
in Molarity.
Kp refers to gases with concentration expressed as
partial pressures. For example: Kp = (PNO2)2
PN2O4
Terms for pure liquids or solids do not appear in the
expressions for Kp or Kc.
Kp and Kc have different numeric values.
For example;
For the reaction:
N2O4(g) D 2NO2(g)
At 100 oC
Kc = 0.36
Kp = 11
At 150 oC
Kc = 3.2
Kp = 110
TO RELATE Kp TO Kc WE HAVE TO HAVE A
RELATIONSHIP FOR PARTIAL PRESSURE AND
MOLARITY.
MOLARITY = n / V
AND FROM THE IDEAL
GAS LAW, PA = n RT
V
THEREFORE, AT EQUILIBRIUM, PA =[A] x RT
(because [A ] = n/V)
Example
N2O4(g) D 2NO2(g)
Kp = (PNO2)2
PN2O4
SO, Kp = [NO2]2 x (RT)2
[N2O4] x (RT)
= [NO2]2 x RT
[N2O4]
And since, Kc = [NO2]2
[N2O4]
THEN, Kp = Kc x RT
FOLLOWING IS A GENERAL EQUATION
WHICH IS VALID FOR ALL SYSTEMS:
Kp = Kc x (RT) Dng
Where Dng = the change in moles of gas from
products to reactants.
(moles gaseous products – moles gaseous reactants).
Example
N2O4(g) D 2NO2(g
N2(g) + 3 H2(g) D 2 NH3(g)
Dng = 2 – 1 = 1
Dng = 2 – 4 = -2
At 300 oC Kc = 9.5 for the following reaction:
N2(g) + 3 H2(g) D 2 NH3(g)
Calculate Kp.
T = 573 K,
Dng = 2 – 4 = -2
Kp = Kc (RT)-2 , Kp =
9.5
= 4.3 x 10 –3
(0.0821 x 573)2
Example
Calculate Kp at 520 oC for the following:
Dng = 2 – 2 = 0
2HI(g) D H2(g) + I2(g)
T = 793 K
Kc = 0.016
Solution:
Kp = 0.016 (RT)0 = 0.016
THE RELATIONSHIP OF FREE
ENERGY (DG), AND K
THE STANDARD, GIBBS FREE ENERGY IS
REPRESENTED BY THE SYMBOL, DG0. (TO BE
COVERED IN DEPTH IN THE THERMODYNAMICS
CHAPTER).
THE EQUATIONS THAT RELATE FREE ENERGY
TO K ARE,
DGO = - RT(ln K) AND
DG = DGO + RT(ln K)
THE STANDARD FREE ENERGY, DGO CAN BE
CALCULATED IN MUCH THE SAME WAY AS DH,
USING THERMODYNAMIC TABLES.
DGo = S DGOproducts - S DGOreactants
VALUES FOR THE STANDARD FREE ENERGY,
DGO, ARE TAKEN TO BE AT 1 atm PRESSURE
AND R= 8.314 J/MOLE K, AND TEMP. IN
KELVIN.
A NEGATIVE VALUE FOR DGO INDICATES A
SPONTANEOUS REACTION.
USING THE EQUATION, DGO = - RT (lnK) , YOU
CAN SEE THAT IF DGO IS NEGATIVE, THEN lnK IS
POSITIVE AND K>1. THE REACTION PROCEEDS
IN THE FORWARD DIRECTION.
IF DGO IS POSITIVE, THEN lnK IS NEGATIVE AND
K<1 SO THE REVERSE REACTION PROCEEDS
SPONTANEOUSLY.
IF DGO = 0 , lnK = 0 AND K = 1. THE REACTION IS
AT EQUILIBRIUM.
THEREFORE A LARGE +K MEANS A FORWARD
REACTION THAT WILL GO TO COMPLETION.
Example
Calculate K at 25 oC given the following:
a. DGO = -40.0 KJ/mol, T = 298 K
b. DGO = + 40.0 KJ/mol, T = 298 K
a. -40.0 KJ/mol =( - 0.008314 KJ/mol K) (298K) lnk
16.1 = ln k,
e16.1 = k,
1.03 x 107 = k
Note the large + k, and the negative DGO
b. + 40.0 = - 0.008314 ( 298K) ln k
9.7 x 10 –8 = k
note the + DGO and the k<1
THE FOLLOWING ARE MORE EXAMPLES:
CALC. DGo IF K = 1.0 x 10 10 AT 100 oC
DGo = - 8.314 j/mol K ( 373K)ln(1.0 x 1010)
DGo = -7.1 x 104 j = - 7.1 x 101 kj
For CaCO3(s) D
CaO(s) + CO2(g)
DGo = 0 at 1110 K
CALCULATE Kp AND Kc.
0 = -RT(lnK),
0
0 = -8.314 (1110) lnK
= lnK = 0, \K= 1 = Pco2(g) = 1atm.
-8.314 (1110)
Continued…..
…..and since Kp = Kc (RT)Dng
Kp
= Kc
(0.0821 x 1110)1
= 1.1 x 10 -2 M = [CO2]
A 1L REACTION VESSEL IS CHARGED WITH
0.500 mole H2 AND 0.500 mole I2 AT 430 oC. CALC.
THE CONCENTRATIONS OF H2 , I2 , AND HI AT
EQUILIBRIUM. Kc = 54.3
H2 (g) + I2 (g) D
2HI(g)
Always make a table of before and after concentrations:
H2
I2
HI
Initial
0.500
0.500
0
Change
-X
-X
+2X
0.500-X
2X
Final
0.500-X
Kc = [HI]2
54.3 =
[H2] [I2]
\ (54.3)1/2 =
(2X)2
(0.500-X)(0.500-X)
\ 7.37 = 2X
2X
0.500 – X
0.500 – X
7.37 (0.500) – 7.37X = 2X
3.68 = 9.37X
.393 = X


\ H2 = 0.500 – 0.393 = 0.107M
I2 = 0.500 – 0.393 = 0.107M
HI = 2(0.393) = 0.786 M
A 9.60 L REACTION VESSEL AT 430 oC is charged
with 4.20 mole HI. CALCULATE ALL
EQUILIBRIUM CONCENTRATIONS. Kc = 54.3
(Same equation as the last example)
4.20 mole HI / 9.60 L = 0.438 M
H2
I2
HI
Initial
0
0
0.438
Change
+x
+x
-2x
Final
x
x
0.438 – 2x
54.3 = [HI]2
[H2] [I2]
ans. HI=0.345 M
I2=H2=0.0467 M
From an earlier example we calculated a value for Q for
the reaction of H2 + I2 D 2HI
The species were in a 3.50 L vessel at 430 oC and
Kc = 54.3. The concentrations were as follows:
H2= 0.00623 M, I2 = 0.00414 M, HI = 0.0224 M
We determined that at these conditions, the reaction
would proceed right because Q<K. Calculate the
equilibrium concentrations.
H2
I2
HI
Initial
0.00623
0.00414
0.0224
Change
-x
-x
+2x
Final
0.00623 -x
0.00414 -x 0.0224 + 2x
54.3 =
(0.0224 + 2x)
(0.00623 –x)(0.00414 – x)
a= 50.3, b= 0.653, c = 8.98 x 10-4
Using algebra and the quadratic equation the solution
is : either x = 0.0114 M (more than we started with)
or x = 0.00156 M
At 350 oC , Kc = 2.37 x 10 –3 for
N2(g) + 3H2(g) D 2 NH3(g)
The equilibrium concentrations were as follows:
[N2] = 0.683 M, [H2] = 8.80 M , [NH3] = 1.05 M
If the [NH3] is quickly decreased to 0.774 M by
removing some of it, a) Predict the direction of
shift. b) Prove this by calculating Qc and
comparing with Kc.
a) Shift right to produce more NH3
b) (0.774)2
(0.683)(8.80)3
=
1.29 x 10 –3 = Qc
Qc < Kc \ shifts right