1. Chemical Equilibrium Chapter 14
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2. Contents and Concepts Describing Chemical Equilibrium
Chemical Equilibrium—A Dynamic Equilibrium
The Equilibrium Constant
Heterogeneous Equilibria; Solvents in Homogeneous Equilibria
Using the Equilibrium Constant
Qualitatively Interpreting the Equilibrium Constant
Predicting the Direction of Reaction
Calculating Equilibrium Concentrations
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3. Changing Reaction Conditions: Le Châtelier’s Principle
Removing Products or Adding Reactants
Changing the Pressure and Temperature
Effect of a Catalyst
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4. Chemical reactions often seem to stop before they are complete.
Actually, such reactions are reversible. That is, the original reactants form products, but then the products react with themselves to give back the original reactants.
When these two reactions—forward and reverse—occur at the same rate, a chemical equilibrium exists.
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5. CO(g) + 3H2(g)  CH4(g) + H2O(g)
The graph shows how the amounts of reactants and products change as the reaction approaches equilibrium.
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6. Equilibrium is a state in which there are no observable changes as time goes by.
Chemical equilibrium is achieved when:
the rates of the forward and reverse reactions are equal and
the concentrations of the reactants and products remain constant
Physical equilibrium
H2O (l)
H2O (g)
NO2
Chemical equilibrium
N2O4 (g)
2NO2 (g)

7. CO(g) + 3H2(g)  CH4(g) + H2O(g)
This graph shows how the rates of the forward reaction and the reverse reaction change as the reaction approaches equilibrium.
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8. N2O4 (g) 2NO2 (g) equilibrium equilibrium equilibrium Start with NO2
Start with NO2 & N2O4

9. constant

10. N2O4 (g) NO2 (g)
K =
[NO2]2
[N2O4]
= 4.63 x 10-3
aA + bB cC + dD
K =
[C]c[D]d
[A]a[B]b
Law of Mass Action

11. Write the equilibrium expressions, Kc, for each of the following reactions:
2H2S(g)  2H2 (g) + S2 (g)
2HCl (g) + 1/2O2 (g)  Cl2(g) + H2O(g)

12. Equilibrium Will K = [C]c[D]d [A]a[B]b aA + bB cC + dD K >> 1
Lie to the right
Favor products
K << 1
Lie to the left
Favor reactants

13. We can apply stoichiometry to compute the content of the reaction mixture at equilibrium.
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14. When heated PCl5, phosphorus pentachloride, forms PCl3 and Cl2 as follows:
PCl5(g)  PCl3(g) + Cl2(g)
When 1.00 mol PCl5 in a 1.00-L container is allowed to come to equilibrium at a given temperature, the mixture is found to contain mol PCl3. What is the molar composition of the mixture?
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15. Initially we had 1.00 mol PCl5 and no PCl3 or Cl2.
We will organize this problem by using the chemical reaction to set up a table of initial, change, and equilibrium amounts.
Initially we had 1.00 mol PCl5 and no PCl3 or Cl2.
The change in each is stoichiometric: If x moles of PCl5 react, then x moles of PCl3 and x moles of Cl2 are produced.
For reactants, this amount is subtracted from the original amount; for products, it is added to the original amount.
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16. We can now find the amounts of the other substances.
PCl5(g) 
PCl3(g) +
Cl2(g)
Initial
1.00 mol
Change –x +x
Equilibrium
1.00 – x x
We were told that the equilibrium amount of PCl3 is mol. That means x = mol.
We can now find the amounts of the other substances.
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17. (given with 3 significant figures)
Moles PCl5 = 1.00 – = 0.87 mol
(2 decimal places)
Moles PCl3 = mol
(given with 3 significant figures)
Moles Cl2 = mol
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18. The Equilibrium Constant, Kc
The equilibrium constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration term to a power equal to its coefficient in the balanced chemical equation.
The equilibrium constant, Kc, is the value obtained for the Kc expression when equilibrium concentrations are substituted.
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19. Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.
N2O4 (g) NO2 (g)
Kp =
NO2
P 2
N2O4 P
Kc =
[NO2]2
[N2O4]
In most cases
Kc  Kp
aA (g) + bB (g) cC (g) + dD (g)
Kp = Kc(RT)Dn
Dn = moles of gaseous products – moles of gaseous reactants
= (c + d) – (a + b)

20. Homogeneous Equilibrium
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)
[CH3COO-][H3O+]
[CH3COOH][H2O]
Kc = ′
[H2O] = constant
[CH3COO-][H3O+]
[CH3COOH] =
Kc [H2O] ′
Kc =
General practice not to include units for the equilibrium constant.

21. What is the Kc expression for this reaction?
Methanol (also called wood alcohol) is made commercially by hydrogenation of carbon monoxide at elevated temperature and pressure in the presence of a catalyst:
2H2(g) + CO(g)  CH3OH(g)
What is the Kc expression for this reaction?
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22. When we are given some information about equilibrium amounts, we are able to calculate the value of Kc.
We need to take care to remember that the Kc expression uses molar concentrations.
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23. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = M, [Cl2] = M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g) COCl2 (g)
[COCl2]
[CO][Cl2] =
0.14
0.012 x 0.054
Kc =
= 220
Kp = Kc(RT)Dn
Dn = 1 – 2 = -1
R =
T = = 347 K
Kp = 220 x ( x 347)-1 = 7.7

24. The equilibrium constant Kp for the reaction
is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = atm and PNO = atm? 2
2NO2 (g) NO (g) + O2 (g)
Kp = 2
PNO PO
PNO PO 2
= Kp
PNO PO 2
= 158 x (0.400)2/(0.270)2
= 347 atm

25. What is the value of Kc at this temperature?
Carbon dioxide decomposes at elevated temperatures to carbon monoxide and oxygen:
2CO2(g)  2CO(g) + O2(g)
At 3000 K, 2.00 mol CO2 is placed into a 1.00-L container and allowed to come to equilibrium. At equilibrium, 0.90 mol CO2 remains.
What is the value of Kc at this temperature?
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26. We can find the value of x. 2.00 – 2x = 0.90 1.10 = 2x x = 0.55 mol
2CO2(g) 
2CO(g) +
O2(g)
Initial
2.00 mol
Change
–2x
+2x +x
Equilibrium
2.00 – 2x 2x x
0.90 mol
1.10 mol
0.55 mol
We can find the value of x.
2.00 – 2x = 0.90
1.10 = 2x
x = 0.55 mol
2x = 2(0.55) = 1.10 mol
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27. 2CO2(g)  2CO(g) + O2(g)
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28. Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.
CaCO3 (s) CaO (s) + CO2 (g)
[CaO][CO2]
[CaCO3]
Kc = ′
[CaCO3] = constant
[CaO] = constant
[CaCO3]
[CaO]
Kc x ′
Kc = [CO2] =
Kp = PCO 2
The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

29. In a heterogeneous equilibrium, in the Kc expression, the concentrations of solids and pure liquids are constant (due to these substances’ constant density).
As a result, we incorporate those constants into the value of Kc, thereby making a new constant, Kc. In other words, equilibrium is not affected by solids and pure liquids as long as some of each is present.
More simply, we write the Kc expression by replacing the concentration of a solid or pure liquid with 1.
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30. H2O(g) + C(s)  CO(g) + H2(g)
Write the Kc expression for the following reaction:
H2O(g) + C(s)  CO(g) + H2(g)
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31. does not depend on the amount of CaCO3 or CaO
CaCO3 (s) CaO (s) + CO2 (g)
PCO 2
= Kp
PCO 2
does not depend on the amount of CaCO3 or CaO

32. Consider the following equilibrium at 295 K:
The partial pressure of each gas is atm. Calculate Kp and Kc for the reaction?
NH4HS (s) NH3 (g) + H2S (g)
Kp = P
NH3
H2S P
= x =
Kp = Kc(RT)Dn
Kc = Kp(RT)-Dn
Dn = 2 – 0 = 2
T = 295 K
Kc = x ( x 295)-2 = 1.20 x 10-4

33. ′ ′ ′ ′ ′′ Kc = [C][D] [A][B] Kc = [E][F] [C][D] A + B C + D Kc Kc ′′
C + D E + F
[E][F]
[A][B]
Kc =
A + B E + F Kc
Kc = Kc ′′ ′ x
If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

34. ′ N2O4 (g) 2NO2 (g) 2NO2 (g) N2O4 (g) = 4.63 x 10-3 K = [NO2]2 [N2O4]
= 216
When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

35. When the reaction is doubled: 2aA + 2bB  2cC + 2dD; K2
Given:
aA + bB  cC + dD; K1
When the reaction is doubled:
2aA + 2bB  2cC + 2dD; K2
The equilibrium constant expression, K2 , is the square of the equilibrium constant expression, K1:
K2 =
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36. aA(g) + bB(g)  cC(g) + dD(g)
For the reaction
aA(g) + bB(g)  cC(g) + dD(g)
The equilibrium constant expressions are
Kc = and Kp =
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37. From the ideal gas law, we know that
How are these related?
We know
From the ideal gas law, we know that
So,
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38. In general, the value of Kp is different from that of Kc.
When you express an equilibrium constant for a gaseous reaction in terms of partial pressures, you call it the equilibrium constant, Kp.
In general, the value of Kp is different from that of Kc.
Recall the ideal gas law and the relationship between pressure and molarity of a gas:
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39. CO(g) + 3H2(g) CH4(g) + H2O(g)
For catalytic methanation,
CO(g) + 3H2(g) CH4(g) + H2O(g)
the equilibrium expression in terms of partial pressures becomes
and
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40. Writing Equilibrium Constant Expressions
The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm.
The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.
The equilibrium constant is a dimensionless quantity.
In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.
If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

41. Chemical Kinetics and Chemical Equilibrium
ratef = kf [A][B]2
A + 2B AB2 kf kr
rater = kr [AB2]
Equilibrium
ratef = rater
kf [A][B]2 = kr [AB2] kf kr
[AB2]
[A][B]2 =
Kc =

42. The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. IF
Qc > Kc system proceeds from right to left to reach equilibrium
Qc = Kc the system is at equilibrium
Qc < Kc system proceeds from left to right to reach equilibrium

43. Calculating Equilibrium Concentrations
Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.
Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.
Having solved for x, calculate the equilibrium concentrations of all species.

44. ICE At 1280oC the equilibrium constant (Kc) for the reaction
Is 1.1 x If the initial concentrations are [Br2] = M and [Br] = M, calculate the concentrations of these species at equilibrium.
Br2 (g) Br (g)
Let x be the change in concentration of Br2
Br2 (g) Br (g)
Initial (M)
0.063
0.012
ICE
Change (M) -x
+2x
Equilibrium (M) x x
[Br]2
[Br2]
Kc =
Kc =
( x)2 x
= 1.1 x 10-3
Solve for x

45. Kc =
( x)2 x
= 1.1 x 10-3
4x x = – x
4x x = 0
-b ± b2 – 4ac  2a
x =
ax2 + bx + c =0
x =
x =
Br2 (g) Br (g)
Initial (M)
Change (M)
Equilibrium (M)
0.063
0.012 -x
+2x x x
At equilibrium, [Br] = x = M
or M
At equilibrium, [Br2] = – x = M

46. Nitrogen and oxygen form nitric oxide. N2(g) + O2(g)  2NO(g)
If an equilibrium mixture at 25°C contains M N2 and M O2, what is the concentration of NO in this mixture? The equilibrium constant at 25°C is 1.0 × 10−30.
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47. N2(g) + O2(g)  2NO(g)
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48. When the initial concentration and the value of Kc are known, we return to the stoichiometric chart of initial, change, and equilibrium (ICE) amounts or concentrations to find the equilibrium concentrations.
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49. Hydrogen iodide decomposes to hydrogen gas and iodine gas.
2HI(g)  H2(g) + I2(g)
At 800 K, the equilibrium constant, Kc, for this reaction is
If 0.50 mol HI is placed in a 5.0-L flask, what will be the composition of the equilibrium mixture in molarities?
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50. 2HI(g)  H2(g) + I2(g) Initial 0.10 M Change +x Equilibrium 0.10 – 2x
Change
–2x +x
Equilibrium
0.10 – 2x x
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51. The Kc expression is, Kc =
Substituting:
Because the right side of the equation is a perfect square, we can take the square root of both sides.
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52. Solving: 0.126(0.10 – 2x) = x 0.0126 – 0.252x = x 0.0126 = 1.252x
x = M
Substituting:
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53. When the Kc expression is not a perfect square, the equation must be rearranged to fit the quadratic format:
ax2 + bx + c = 0
The solution is
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54. Le Châtelier’s Principle
When a system in chemical equilibrium is disturbed by a change in
temperature,
pressure, or
concentration,
the system shifts in equilibrium composition in a way that tends to counteract this change of variable.
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55. Le Châtelier’s Principle
If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.
Changes in Concentration
N2 (g) + 3H2 (g) NH3 (g)
Add
NH3
Equilibrium shifts left to offset stress

56. Le Châtelier’s Principle
Changes in Concentration continued
Remove
Add
Add
Remove
aA + bB cC + dD
Change
Shifts the Equilibrium
Increase concentration of product(s)
left
Decrease concentration of product(s)
right
Increase concentration of reactant(s)
right
Decrease concentration of reactant(s)
left

57. The following reaction is at equilibrium: COCl2(g)  CO(g) + Cl2(g)
a. Predict the direction of reaction when chlorine gas is added to the reaction mixture.
b. Predict the direction of reaction when carbon monoxide gas is removed from the mixture.
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58. Note: forward = right = .
COCl2(g)  CO(g) + Cl2(g)
When we add Cl2, the reaction will shift in the reverse direction to use it.
Note: reverse = left = .
When we remove CO, the reaction will shift in the forward direction to produce it.
Note: forward = right = .
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59. Le Châtelier’s Principle
Changes in Volume and Pressure
A (g) + B (g) C (g)
Change
Shifts the Equilibrium
Increase pressure
Side with fewest moles of gas
Decrease pressure
Side with most moles of gas
Increase volume
Side with most moles of gas
Decrease volume
Side with fewest moles of gas

60. c. COCl2(g)  CO(g) + Cl2(g)
In which direction will each reaction shift when the volume of the reaction container is increased?
a. CO(g) + 2H2(g)  CH3OH(g)
b. 2SO2(g) + O2(g)  2SO3(g)
c. COCl2(g)  CO(g) + Cl2(g)
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61. This reaction shifts reverse = left =  2SO2(g) + O2(g)2SO3(g)
When the container volume is increased, the total pressure is decreased. Each system will shift to produce more gas by shifting toward the side with more moles of gas.
CO(g) + 2H2(g)  CH3OH(g)
This reaction shifts reverse = left = 
2SO2(g) + O2(g)2SO3(g)
COCl2(g)  CO(g) + Cl2(g)
This reaction shifts forward = right = 
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62. Le Châtelier’s Principle
Changes in Temperature
Change
Exothermic Rx
Endothermic Rx
Increase temperature
K decreases
K increases
Decrease temperature
K increases
K decreases
N2O4 (g) NO2 (g)
colder
hotter

63. Decreasing the temperature has the opposite effect.
For an endothermic reaction, increasing the temperature increases the value of Kc.
For an exothermic reaction, increasing the temperature decreases the value of Kc.
Decreasing the temperature has the opposite effect.
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64. In addition to the value of Kc, we can consider the direction in which the equilibrium will shift.
When heat is added (temperature increased), the reaction will shift to use heat.
When heat is removed (temperature decreased), the reaction will shift to produce heat.
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65. Heat + 2H2O(g)  2H2(g) + O2(g)
Given:
2H2O(g)  2H2(g) + O2(g); DH° = 484 kJ
Would you expect this reaction to be favorable at high or low temperatures?
We rewrite the reaction to include heat:
Heat + 2H2O(g)  2H2(g) + O2(g)
When heat is added,
the reaction shifts forward = right = .
The reaction is favorable at high temperatures.
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66. 8CO(g) + 17H2(g)  C8H18(g) + 8H2O(g)
The Fischer–Tropsch process for the synthesis of gasoline consists of passing a mixture of carbon monoxide and hydrogen over an iron–cobalt catalyst.
A typical reaction that occurs in the process is
8CO(g) + 17H2(g)  C8H18(g) + 8H2O(g)
Suppose the reaction is at equilibrium at 200°C, then is suddenly cooled to condense the octane, and then the remaining gases are reheated to 200°C. In which direction will the equilibrium shift?
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67. 8CO(g) + 17H2(g) C8H18(g) + 8H2O(g)
The Fischer–Tropsch process for the synthesis of gasoline consists of passing a mixture of carbon monoxide and hydrogen over an iron–cobalt catalyst.
A typical reaction that occurs in the process is
8CO(g) + 17H2(g) C8H18(g) + 8H2O(g)
Suppose the reaction is at equilibrium at 200°C, then is suddenly cooled to condense the octane, and then the remaining gases are reheated to 200°C. In which direction will the equilibrium shift?
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68. This is essentially removing octane, a product
This is essentially removing octane, a product. This change causes the reaction to produce octane by shifting forward = right = .
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69. Le Châtelier’s Principle
Adding a Catalyst
does not change K
does not shift the position of an equilibrium system
system will reach equilibrium sooner
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift equilibrium.

70. Life at High Altitudes and Hemoglobin Production
Chemistry In Action
Life at High Altitudes and Hemoglobin Production
Hb (aq) + O2 (aq) HbO2 (aq)
Kc =
[HbO2]
[Hb][O2]

71. Chemistry In Action: The Haber Process
N2 (g) + 3H2 (g) NH3 (g) DH0 = kJ/mol

72. Le Châtelier’s Principle - Summary
Change
Shift Equilibrium
Change Equilibrium
Constant
Concentration
yes no
Pressure
yes* no
Volume
yes* no
Temperature
yes
yes
Catalyst no no
*Dependent on relative moles of gaseous reactants and products

73. N2O4 decomposes to NO2. The equilibrium reaction in the gas phase is
N2O4(g) 2NO2(g)
At 100°C, Kc = 0.36.
If a 1.00-L flask initially contains M N2O4, what will be the equilibrium concentration of NO2?
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74. Again, we begin with the table: N2O4(g)  2NO2(g)
Initial
0.100 M
Change –x
+2x
Equilibrium
0.100 – x 2x
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75. The Kc expression is, Kc =
Substitute:
Rearrange:
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76. Substitute to find the equilibrium concentration of NO2:
Solve:
We eliminate the negative value because it is impossible to have a negative concentration.
Substitute to find the equilibrium concentration of NO2:
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77. H2(g) + F2(g)  2HF(g); Kc = 1.15 × 102
Given:
H2(g) + F2(g)  2HF(g); Kc = 1.15 × 102
3.000 mol of each species is put in a L vessel. What is the equilibrium concentration of each species?
First calculate the initial concentrations:
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78. H2(g) + F2(g)  2HF(g) Initial 2.000 M Change +2x Equilibrium
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80. Now compute the equilibrium concentrations:
Double-check by substituting these equilibrium concentrations into the Kc expression and solving. The answer should be the value of Kc.
This is within round-off error.
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81. The value of Kc at 227°C is 0.0952 for the following reaction:
CH3OH(g)  CO(g) + 2H2(g)
What is Kp at this temperature?
Kp = (RT)Dn
T = = 500. K
R = L  atm/(mol  K)
Dn = 2
Kp = 1.60 × 102
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