Download Chapter 5 Chemical Equilibrium 1 State whether each of the

Chapter 5 Chemical Equilibrium
1
State whether each of the following processes is endothermic or exothermic:
(a) C8H18(l) + 12.5 O2(g)  8 CO2(g) + 9 H2O(g)
exothermic
(b) CO2(s)  CO2(g)
endothermic
(c) H2O(g)  H2O(l)
exothermic
+
(d) NaBr(s)  Na (g) + Br¯(g)
endothermic
(e) Hg(l) at 25°C  Hg(l) at 50°C
endothermic
2
For the reaction Fe(s) + 2 Ag+(aq)
Fe+2(aq) + 2 Ag(s) initially at equilibrium, will the
equilibrium shift left, right, or neither if we dilute the solution with water? Why?
For this reaction, Q 
[Fe2(aq) ]
[Ag  (aq) ]2
. Diluting the reaction will cause [Fe+2(aq)] to decrease,
but [Ag+(aq)]2 will decrease even faster, so Q will increase. The reaction will therefore
shift left so as to reduce Q according to le Chatelier’s principle.
3
To maximize the production of CO(g) using the reaction C(s) + CO2(g) = 2 CO(g), should the
reaction be run at high or low pressure? Explain.
In this reaction, Δn = +1. According to Le Chatelier’s principle, decreasing the pressure
will cause the equilibrium position to move to the right, i.e. the products. Thus, the
reaction should be run at lower pressure to maximize the production of CO(g).
4
Why does a small amount of a chlorofluorocarbon injected into the atmosphere cause the
destruction of a much larger amount of ozone?
The Cl atoms enter in a cyclic destruction process with ozone, i.e. the Cl atoms are
catalytic, and are not consumed by these reactions. Specifically,
5.
Cl + O3  ClO + O2
ClO + O  Cl + O2
State whether each of the following processes is endothermic or exothermic:
(a) C8H18(l) + 12.5 O2(g)  8 CO2(g) + 9 H2O(g) exothermic
(b) CO2(s)  CO2(g)
endothermic
(c) H2O(g)  H2O(l)
exothermic
+
(d) NaBr(s)  Na (g) + Br¯(g)
endothermic
(e) Hg(l) at 25°C  Hg(l) at 50°C
endothermic
6.
At 298 K, Kp = 0.0900 for the reaction H2O(g) + Cl2O(g)
2 HOCl(g).
(a) Calculate the equilibrium partial pressure of each gas if initially pH2O = pCl2O = 100 Torr
and pHOCl = 0.
H2O(g)
Cl2O(g)
HOCl(g)
100
100
0
-x
-x
+2x
100 - x
100 - x
2x
Initial, Torr
Change, Torr
Equilibrium, Torr
Thus, at equilibrium
p 2 HOCl
( 2 x) 2
 K  0.090 
p H 2O p Cl 2O
(100  x)(100  x)
This happens to be a perfect square, so we can take the square root of both sides:
2x
 0.09  0.3
100  x
2 x  0.3(100  x)
2.3 x  30
x  13.0
Thus, pH2O = 100 – x = 87.0 Torr
pCl2 = 100 – x = 87.0 Torr
pHOCl = 2x = 26.0 Torr
(2x) 2
2(13.0) 2
Check:

 0.090
(100  x)(100  x) (100  13.0) 2
(b) Calculate the value and units of Kc at 100oC.
Kc = Kp(RT)-Δn
For this reaction, Δn = 0, thus Kc = Kp = 0.090
And since Kp is dimensionless, so is Kc.
7. Liquid water is always in equilibrium with H+(aq) and OH¯(aq) ions according to the reaction
H2O(l)
H+(aq) + OH-(aq). Using the data in the table below,
Hof, kJ mol-1
So, J K-1 mol-1
H2O(l)
-286
70
H+(aq)
0
0
OH-(aq)
-230
-11
(a) Calculate H° and S° for the reaction.
H° = Hf°(H+(aq)) + Hf°(OH-(aq)) - Hf°(H2O(l))
= 0 + (-230) – (-286)
= +56 kJ mol-1
S° = S°(H+(aq)) + S°(OH-(aq)) - S°(H2O(l))
= 0 + (-11) – (70)
= -81 J K-1 mol-1
(b) Calculate G° at 25°C for the reaction.
G° = H° - TS°
= 56000 J mol-1 – (25+273)K(-81 J K-1 mol-1)
= +80138 J mol-1
(c) Calculate the value of Kc at 25°C.
Kc = exp (-G°/RT)
= exp(-80138 J mol-1/(8.314 J K-1 mol-1 x 298 K))
= exp(-32.3)
= 8.97 x 10-15
(d) Calculate the concentrations of H+(aq) and OH¯(aq) at equilibrium.
Kc = [H+(aq)][OH-(aq)] (since H2O(l) is a liquid, it does not appear in the expression)
thus, [H+(aq)] = [OH-(aq)] = Kc1/2 = (8.97 x 10-15)1/2 = 9.47 x 10-8 mol L-1
8.
The vapor pressure of water at 34.1°C is 40.1 Torr. The enthalpy of vaporization of water
is
40.7 kJ mol-1. Calculate the vapor pressure of water at 85.5°C.
ln( p 2 )  ln( p1 ) 
 ln(401
. )
 H vap  1 1 
  
R  T1 T2 
40700 J mol 1  1
1 


1
1 
8.314 J K mol  307.2 K 358.6 K 
 5.98
Thus, p2 = exp(5.98) = 394 Torr.
9. Name three factors that affect the rate of a chemical reaction.
Any three of rate of collisions (or concentrations of reagents or pressure), orientation of collisions,
energy of collisions (or temperature), presence of a catalyst, reaction pathway.
10. For the reaction H2(g) + ½ O2(g)
H2O(g), which way will the equilibrium shift if
pressure is decreased? Why?
The equilibrium will shift to the left to increase the pressure, since thee are 1.5 moles of gas on the
left and only 1 mole of gas on the right.
11. Graphite is the standard state of carbon. Why do diamonds exist at 25cC and 1 atm?
Diamond is thermodynamically unstable, but the conversion to graphite is so slow that diamonds
will exist for a very long time under these conditions.
12. (a) Calculate the standard free energy change (kJ mol-1) for the following reaction at
25°C:
CO(g) + 3 H2(g)
CO(g)
H2(g)
CH4(g)
H2O(g)
-1
-110.5
0
-74.8
-241.8
-1
197.6
130.6
186.2
188.7
ΔH°f, kJ mol
-1
CH4(g) + H2O(g)
S°, J K mol
ΔHo = ΔHof(CH4(g)) + ΔHof(H2O(g)) - ΔHof(CO(g)) - 3ΔHof(H2(g))
= -74.8 + (-241.8) – (-110.5) – 3(0)
= -206.1 kJ mol-1
ΔSo = So(CH4(g)) + So(H2O(g)) - So(CO(g)) – 3So(H2(g))
= 186.2 + 188.7 – 197.6 – 3(130.6)
-214.5 J K-1 mol-1
ΔGo = ΔHo - TΔSo
= -206100 J mol-1 – 298 K(-214.5 J K-1 mol-1)
= -142179 J mol-1
=-142.2 kJ mol-1
(b) Calculate the value of the equilibrium constant Kp at 1000°C.
   Go 

K p  exp
 RT 


 142179 J mol 1

 exp
1
1
 8.314 J K mol (1273) K 
 6.83  105
13.
(a) Consider the following half-reaction:
CrO4-2(aq) + 4 H2O(l) + 3 e¯  Cr(OH)3(s) + 5 OH¯ (aq)
E° = -0.13 V
Calculate the potential of the reaction under physiological conditions, i.e. at 37.0°C
and pH = 7.40 if [CrO4-2] = 1.00 x 10-6 M.
pH = 7.40, thus pOH = 14 – pH = 6.6.
Thus, [OH-(aq)] = 10-6.60 = 2.51 x 10-7 mol L-1
Thus,
Q
[ OH  ( aq ) ]5
[ CrO4 2 ( aq ) ]
( 2.51  107 )5

100
.  106
 100
.  1027
RT
ln( Q)
nF
8.314JK 1mol 1 ( 37  273) K
  013
. V
ln(100
.  1027 )
3( 96487 C mol 1 )
E  Eo 
  013
. V  0.00890(  62.2)
  0.42 V
14 recisely define each symbol in the term Hof

 means change
H means enthalpy
o
means standard state
f means formation
The term means the standard enthalpy of formation.
15.
Define “state function” and give three examples of one.
Is a function whose value does not depend on the path taken or on the rate.
e.g.  H, S and G.
16. Consider the reaction 2 C6H14(l) + 19 O2(g)
12 CO2(g) + 14 H2O(g).
(a) Is the reaction endothermic or exothermic?
Exothermic
(b) Is So for the reaction positive or negative?
Positive
(c) Is this a spontaneous reaction? How can you tell?
Spontaneous because the free energy will always be less than zero, since <0 and
S>0
17.
A reaction has a value of Go very close to zero. At equilibrium, will there be mostly
reactants, mostly products, or about the same quantities of each? What equation tells
you so?
There will be about the same quantities of reactants and products.
Keq = exp [-Go/RT]
18.
For the reaction SiCl4(g) + 2 H2(g)
equilibrium shift if we:
(a) Add some H2(g)?
Reaction shifts right to use up the H2(g)
Si(s) + 4 HCl(g), which direction will the
(b) Add some Si(s)?
19.
Reaction will not shift since Si is a solid and does not enter into the equilibrium
expression.
Some data for the vapor pressure of water are presented in the table below:
Temperature, oC
5.0
10.0
90.0
95.0
Vapor Pressure, mm Hg
6.54
9.21
525.8
633.9
(a) Calculate the enthalpy of vaporization, Hvap (kJ mol-1) at low temperature, i.e.
use the data at 5.0oC and 10.0oC.
ln(p 2 )  ln(p1 ) 
thus, H vap
H vap  1 1 
  
R  T1 T2 


 ln(p )  ln(p ) 
2
1

 R 
1
1





T
T
2
1




 ln(9.21)  ln(6.54) 
  44,800 J  44.8 kJ
 8.314 JK 1mol 1 
1
1


 283K  278 K 


(b) Calculate the enthalpy of vaporization, Hvap (kJ mol-1) at high temperature, i.e.
use the data at 90.0oC and 95.0oC.
H vap


 ln(p )  ln(p ) 
2
1

 R 
1 1





T2 T1




 ln(633.9)  ln(525.8) 
  41,500 J  41.5 kJ
 8.314 JK 1mol 1 
1
1



 368 K 363K 


(c) You should have found that Hvap is smaller at the higher temperature. Why is this
so?
This is because at higher temperature, the water molecules already have higher energy,
so less is required to vaporize them from the liquid into the gas phase.
20. For the reaction MgCO3(s)
MgO(s) + CO2(g), which way will the equilibrium shift if
pressure is increased? Why?
The equilibrium will shift left so as to reduce the pressure. Note that there is only one
gaseous species in the reaction.
21. Define the term “state function” and give three examples of one.
A state function is one whose value is the same no matter what the rate of reaction and no
matter what path is taken from reactants to products. Some examples are S, H, G, p, V, T.
22 Given the following data:
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
3 Fe2O3(s) + CO(g)  2 Fe3O4(s) + CO2(g)
Fe3O4(s) + CO(g)  3 FeO(s) + CO2(g)
Ho = -23 kJ
Ho = -39 kJ
Ho = +18 kJ
calculate Ho for the reaction FeO(s) + CO(g)  Fe(s) + CO2(g)
6 FeO(s) + 2 CO2(g)  2 Fe3O4(s) + 2 CO(g)
Ho = -2(+18 kJ)
2 Fe3O4(s) + CO2(g)  3 Fe2O3(s) + CO(g)
Ho = -(-39 kJ)
3 Fe2O3(s) + 9 CO(g)  6 Fe(s) + 9 CO2(g)
Ho = 3(-23 kJ)
_________________________________________________________
6 FeO(s) + 6 CO(g)  6 Fe(s) + 6 CO2(g)
Ho = -36 + 39 - 69 = -66 kJ/mol
Thus,
FeO(s) + CO(g)  Fe(s) + CO2(g)
Ho -66 kJ/mol / 6 = -11 kJ/mol