Download Ch. 16 Study Guide

Chemistry 201: General Chemistry II - Lecture
Dr. Namphol Sinkaset
Chapter 16 Study Guide
Concepts
1. Reversible reactions can proceed in both the forward and reverse directions.
2. equilibrium: situation that exists when a reaction appears to have stopped and the
concentrations of the reactants and products no longer change. At equilibrium, the
forward and backward reaction rates are equal.
3. Equilibrium does not mean that all concentrations are equal!!
4. equilibrium constant, K: the ratio at equilibrium of the concentrations of products
raised to their stoichiometric coefficients divided by the concentrations of reactants
raised to their stoichiometric coefficients.
5. The law of mass action relates a balanced equation to its equilibrium constant
expression.
6. The numerical value of K gives an indication of the relative amounts of products and
reactants that exist at equilibrium.
7. Another way of interpreting the value of K is that it tells you how far the equilibrium
goes to formation of products.
8. When an equilibrium reaction is reversed, its equilibrium constant is the inverse of the
original (1/K).
9. When an equilibrium equation is multiplied by a factor, the equilibrium constant is
raised to that power.
10. When chemical equilibrium equations are added, the overall equilibrium constant is
the product of the individual equilibrium constants.
11. Since partial pressures are proportional to concentration, equilibrium constants can be
expressed in terms of partial pressures.
12. Conversion between Kc and Kp relies on determining the change in moles of gas in a
reaction.
13. Pure solids and pure liquids are not included in the equilibrium constant expression.
14. Equilibrium constants are unitless even though molarity concentrations or partial pressures are used to calculate them.
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15. No matter how a reaction is set up, the value of the equilibrium will be the same if the
temperature is kept constant.
16. The reaction quotient, Qc , is expressed as concentration of products over concentration of reactants, with each substance raised to its stoichiometric power. There is also
a Qp , of course. Q is for a specific set of reaction conditions that may or may not be
at equilibrium.
17. If Q > K (concentration or pressure), then the reaction is not at equilibrium. Furthermore, the reaction will shift to the left.
18. If Q < K (concentration or pressure), then the reaction is not at equilibrium; it will
shift to the right.
19. If Q = K (concentration or pressure), then the reaction is at equilibrium.
20. A concentration table (or pressure table for a Kp problem) is a good way to organize
information given in an equilibrium problem.
21. Key points to keep in mind when performing equilibrium calculations: (1)
can only use equilibrium concentrations in the Kc /Kp expression, (2) initial concentrations should be in molarity if using Kc , (3) changes in concentrations always occur in
the same ratio as the coefficients in the balanced equation, and (4) all reactants should
change in one direction while all products change in the opposite direction.
22. When K is really small or really big, simplifications can be made in the equilibrium
expression. Unknown x values can be discarded in any addition/subtraction operation
but not from any multiplication/division operation. When checking whether or not
the assumption is valid, the percentage difference should be 5% or less.
23. Successive approximations can be used if the 5% rule is not met or if the problem
involves evaluating a higher order equation.
24. Le Châtelier’s Principle is used to make qualitative predictions about changes in
chemical equilibria. It states: “When a chemical system at equilibrium is disturbed,
the system shifts in the direction that minimizes the disturbance.”
25. There are multiple ways to disturb an equilibrium: (1) add/remove a reactant or
product; (2) change the volume/pressure in reactions that involve a gas; (3) change
the temperature.
26. An equilibrium will shift in a direction that will partially consume a reactant or product
that is added or partially replace a reactant or product that has been removed.
27. Reducing the volume of a gaseous reaction mixture shifts the equilibrium in whichever
direction will, if possible, decrease the number of moles of gas.
28. Temperature affects an equilibrium based on whether the reaction is exo- or endothermic.
29. Catalysts do not affect the position of the equilibrium.
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Equations
products] (Concentration equilibrium constant expression)
1. Kc = [[reactants
]
P
2. Kp = P products (Pressure equilibrium constant expression)
reactants
3. For the expressions above, make sure you use the coefficient in the balanced
equation as the power for each individual compound!! Examples are shown
below.
4. Kc =
5. Kp =
[C]c [D]d
[A]a [B]b
g h
PG
PH
ePf
PE
F
(Kc for the generic reaction aA + bB *
) cC + dD)
(Kp for the generic gas-phase reaction eE + fF *
) gG + hH)
6. Kp = Kc (RT )∆ng (Conversion between Kp and Kc )
products] (Reaction quotient, again use coefficients as powers)
7. Q = [[reactants
]
8. x =
√
−b± b2 −4ac
2a
(Quadratic equation for exact equilibrium calculations)
Representative Problems
R26. Use the following equilibria to calculate Kc for the reaction 2CH3 OH(g) +
H2(g) ↔ C2 H6(g) + 2H2 O(g) .
CH4(g)
2CH4(g)
+ H2 O(g)
*
) C2 H6(g) + H2(g)
*
) CH3 OH(g) + H2(g)
Kc = 9.5 × 10−13
Kc = 2.8 × 10−21
Somehow, we need to have these two reactions add up to the overall reaction given to
us. The only way we can do this is by reversing the second reaction and multiplying by 2.
When we do that, we need to take the inverse of Kc and then square it. After manipulating
the equations, we multiply the individual Kc ’s to get the overall Kc .
2CH4(g)
2CH3 OH(g) + 2H2(g)
2CH3 OH(g) + H2(g)
*
) C2 H6(g) + H2(g)
*
) 2CH4(g) + 2H2 O(g)
*
) C2 H6(g) + 2H2 O(g)
Kc = 9.5 × 10−13
Kc = 1.3 × 1041
Kc = 1.2 × 1029
R46. At 460 ◦ C, the reaction, SO2(g) + NO2(g) ↔ NO(g) + SO3(g) , has Kc = 85.0
A reaction flask at 460 ◦ C contains these gases at the following concentrations:
[SO2 ] = 0.00250 M, [NO2 ] = 0.00350 M, [NO] = 0.0250 M, [SO3 ] = 0.0400 M.
(a) Is the reaction at equilibrium? (b) If not, which way will the reaction have
to proceed to arrive at equilibrium?
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This is a problem where we need to calculate the reaction quotient, Q, for the reaction
and compare its value to Kc .
[NO][SO3 ]
[SO2 ][NO2 ]
(0.0250 M)(0.0400 M)
=
(0.00250 M)(0.00350)
= 114
Q =
We see that Q is larger than Kc . Therefore, the reaction will need to shift to the left to
go towards equilibrium.
R60. At 25 ◦ C, Kc = 0.145 for the following reaction in the solvent CCl4 : 2BrCl
↔ Br2 + Cl2 . If the initial concentration of each substance in a solution is 0.0400
M, what will their equilibrium concentrations be?
Well, it’s an equilibrium problem, so first we write the Kc expression.
Kc =
[Br2 ][Cl2 ]
[BrCl]2
Since we’re given that every species has an initial concentration of 0.0400 M, we need to
calculate Q to find out which way the equilibrium is going.
[Br2 ][Cl2 ]
[BrCl]2
(0.0400 M)(0.0400 M)
=
(0.0400 M)2
= 1
Q =
We see that Q is larger than Kc , so the equilibrium will shift to the left. When we set up
our chart, we know that we will need to have the product concentrations decrease and the
reactant concentration increase. Additionally, we need to take the reaction stoichiometry
into account.
*
Reaction:
2BrCl
Br2
+
Cl2
)
Initial
0.0400
0.0400
0.0400
Change
+2x
−x
−x
Equil. 0.0400 + 2x
0.0400 − x
0.0400 − x
We place the equilibrium row into the Kc expression and solve.
Kc =
[Br2 ][Cl2 ]
[BrCl]2
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0.145 =
0.3808 =
0.01523 + 0.7616 x =
1.7616 x =
x =
(0.0400 − x)(0.0400 − x)
(0.0400 + 2x)2
(0.0400 − x)
(0.0400 + 2x)
0.0400 − x
0.02477
0.01406
Now we just make the necessary addition/subtraction from each compound in the reaction. [BrCl] = 0.0681 M, [Br2 ] = 0.0259 M, [Cl2 ] = 0.0259 M.
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