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CHAPTER 13
CHEMICAL EQUILIBRIUM
Questions
- Because of the 2 : 1 mole ratio between NH 3 and N 2 in the balanced equation, NH 3 will disappear at a rate that is twice as fast as the rate that N 2 appears. Because of the 3 : 1 mole ratio between H 2 and N 2 in the balanced equation, H 2 will appear at a rate that is three times the rate that N 2 appears. At equilibrium, however, all rates of appearance and disappearance will be equal to each other. This always occurs when a reaction reaches equilibrium.
- No, equilibrium is a dynamic process. Both reactions:
H 2 O + CO → H 2 + CO 2 and H 2 + CO 2 → H 2 O + CO
are occurring at equal rates. Thus 14 C atoms will be distributed between CO and CO 2.
- No, it doesn't matter from which direction the equilibrium position is reached. Both experiments will give the same equilibrium position because both experiments started with stoichiometric amounts of reactants or products.
- A K value much greater than one (K >> 1) indicates there are relatively large concentrations of product gases/solutes as compared with the concentrations of reactant gases/solutes at equilibrium. A reaction with a very large K value is a good source of products.
- A K value much less than one (K >> 1) indicates that there are relatively large concentrations of reactant gases/solutes as compared with the concentrations of product gases/solutes at equilibrium. A reaction with a very small K value is a very poor source of products.
15. H 2 O(g) + CO(g) ⇌ H 2 (g) + CO 2 (g) K =
[H O][CO]
[H ][CO ]
2
K is a unitless number because there is an equal number of moles of product gases as moles of reactant gases in the balanced equation. Therefore, we can use units of molecules per liter instead of moles per liter to determine K.
We need to start somewhere, so let’s assume 3 molecules of CO react. If 3 molecules of CO react, then 3 molecules of H 2 O must react, and 3 molecules each of H 2 and CO 2 are formed. We would have 6! 3 = 3 molecules of CO, 8! 3 = 5 molecules of H 2 O, 0 + 3 = 3 molecules of H 2 , and 0 + 3 = 3 molecules of CO 2 present. This will be an equilibrium mixture if K = 2.0:
K =
L
3 moleculesCO L
5 moleculesHO
L
3 moleculesCO L
3 moleculesH
2
2 2
Because this mixture does not give a value of K = 2.0, this is not an equilibrium mixture. Let’s try 4 molecules of CO reacting to reach equilibrium. Molecules CO remaining = 6! 4 = 2 molecules of CO Molecules H 2 O remaining = 8! 4 = 4 molecules of H 2 O Molecules H 2 present = 0 + 4 = 4 molecules of H 2 Molecules CO 2 present = 0 + 4 = 4 molecules of CO 2
K = 2. 0
L
2 moleculesCO L
4 moleculesHO
L
4 moleculesCO L
4 moleculesH
2
2 2
Because K = 2.0 for this reaction mixture, we are at equilibrium.
- When equilibrium is reached, there is no net change in the amount of reactants and products present because the rates of the forward and reverse reactions are equal to each other. The first diagram has 4 A 2 B molecules, 2 A 2 molecules, and 1 B 2 molecule present. The second diagram has 2 A 2 B molecules, 4 A 2 molecules, and 2 B 2 molecules. Therefore, the first diagram cannot represent equilibrium because there was a net change in reactants and products. Is the second diagram the equilibrium mixture? That depends on whether there is a net change between reactants and products when going from the second diagram to the third diagram. The third diagram contains the same numbers and types of molecules as the second diagram, so the second diagram is the first illustration that represents equilibrium.
The reaction container initially contained only A 2 B. From the first diagram, 2 A 2 molecules and 1 B 2 molecule are present (along with 4 A 2 B molecules). From the balanced reaction, these 2 A 2 molecules and 1 B 2 molecule were formed when 2 A 2 B molecules decomposed. Therefore, the initial number of A 2 B molecules present equals 4 + 2 = 6 molecules A 2 B.
- K and Kp are equilibrium constants, as determined by the law of mass action. For K, concentration units of mol/L are used, and for Kp, partial pressures in units of atm are used (generally). Q is called the reaction quotient. Q has the exact same form as K or Kp , but instead of equilibrium concentrations, initial concentrations are used to calculate the Q value. The use of Q is when it is compared with the K value. When Q = K (or when Q (^) p = Kp ), the reaction is at equilibrium. When Q ≠ K, the reaction is not at equilibrium, and one can deduce the net change that must occur for the system to get to equilibrium.
- H 2 (g) + I 2 (g) → 2 HI(g) K = [H ][I ]
[HI]
2 2
2
H 2 (g) + I 2 (s) → 2 HI(g) K = [H ]
[HI]
2
2 (Solids are not included in K expressions.)
23. K = 1.3 × 10 −^2 = 3
2 2
2 3 [N ][H ]
[ NH]
for N 2 (g) + 3 H 2 (g) ⇋ 2 NH 3 (g).
When a reaction is reversed, then Knew = 1/Koriginal. When a reaction is multiplied through by a value of n, then Knew = (Koriginal )n^.
a. 1/2 N 2 (g) + 3/2 H 2 (g) ⇌ NH 3 (g) KN= (^3) / 2 2
1 / 2 2
2 3 [N] [H ]
[ NH]
= K1/2^ = (1.3 × 10 −^2 )1/2^ = 0.
b. 2 NH 3 (g) ⇌ N 2 (g) + 3 H 2 (g) KO = (^22) 3
3 2 2
- 3 10
K
[NH]
[N ][H ]
= = × − = 77
c. NH 3 (g) ⇌ 1/2 N 2 (g) + 3/2 H 2 (g) KNNN = [NH]
[N] [H ]
3
3 / 2 2
1 / 2 2
1 / 2
2
1 / 2
- 3 10
K
×
d. 2 N 2 (g) + 6 H 2 (g) ⇌ 4 NH 3 (g) K = (^6) 2
2 2
4 3 [N ][H]
[NH ]
= (K)^2 = (1.3 × 10 −^2 )^2 = 1.7 × 10 −^4
- H 2 (g) + Br 2 (g) ⇌ 2 HBr(g) Kp = (P )(P )
P
H 2 Br 2
2 HBr (^) = 3.5 × 10^4
a. HBr ⇌ 1/2 H 2 + 1/2 Br 2
1 / 2 4
1 / 2
HBr p
1 / 2 Br
1 / 2 ' H p (^3). 5 10
K
P
(P ) (P )
K 22 ⎟⎟
⎟ ×
= 5.3 × 10−^3
b. 2 HBr ⇌ H 2 + Br (^2 ) p
2 HBr
'' H Br p (^3). 5 10
K
P
(P )(P )
K 2 2
×
= = = = 2.9 × 10−^5
c. 1/2 H 2 + 1/2 Br 2 ⇌ HBr (K ) 190 (P ) (P )
P
K (^1) / 2 p^1 /^2 Br
1 / 2 H
''' HBr p 2 2
- 2 NO(g) + 2 H 2 (g) ⇌ N 2 (g) + 2 H 2 O(g) K = (^2) 2
2
2 2 2 [NO][H ]
[N][HO]
K = 3252
2 32
( 8. 1 10 ) ( 4. 1 10 )
− −
− −
× ×
× ×
= 4.0 × 10 6
26. K =
[N ][O ]
[ NO]^42
2 2
2
M M
= × − M = 6.9 × 10−^4
27. [NO] =
3. 0 L
5 × 10 −^3 mol = 1.5 × 10 −^3 M ; [Cl 2 ] =
0 L
4 mol = 0.80 M
[NOCl] =
0 L
0 mol = 0.33 M ; K = (^2)
32 2
2
2
( 0. 33 )
[NOCl]
[ NO][Cl ] × −
= = 1.7 × 10 −^5
28. [N 2 O] =
2. 00 L
00 × 10 −^2 mol ; [N 2 ] =
00 L
80 × 10 −^4 mol ; [O 2 ] =
00 L
50 × 10 −^5 mol
K =
[N][O]
[NO]
4 2 5
2 2
4 2 5
2 2
2
2 2
2 2 − −
−
− −
−
× ×
×
⎛ ×
⎛ ×
⎛ ×
= 4.08 × 10 8 L/mol
If the given concentrations represent equilibrium concentrations, then they should give a value of K = 4.08 × 10 8.
4 2
2
× − = 4.08 × 10^
8
Because the given concentrations when plugged into the equilibrium constant expression give a value equal to K (4.08 × 10 8 ), this set of concentrations is a system at equilibrium.
- Kp = (^2) NO
O
2 NO
2
2 P
P × P
52 5
( 0. 55 )
( 6. 5 × 10 −^ ) ( 4. 5 × 10 − )
= 6.3 × 10 −^13
- Kp = (^3) N H
2 NH 2 2
3 P P
P
×
2 2
( 0. 85 )( 3. 1 10 )
−
−
×
×
= 3.8 × 10 4 atm−^2
3
2
( 0. 525 )( 0. 00761 )
= 1.21 × 10 3
When the given partial pressures in atmospheres are plugged into the Kp expression, the value does not equal the K (^) p value of 3.8 × 10 4. Therefore, one can conclude that the given set of partial pressures does not represent a system at equilibrium.
- Kp = K(RT)∆n^ , where ∆n = sum of gaseous product coefficients! sum of gaseous reactant coefficients. For this reaction, ∆n = 3! 1 = 2.
K =
( (^0). 15 )
[CHOH]
[CO ][H ]^2
3
2
2 = = 1.9 mol^2 /L^2
Kp = K(RT)^2 = 1.9(0.08206 L atm/KCmol × 600. K) 2 = 4.6 × 10^3
Use the reaction quotient Q to determine which way the reaction shifts to reach equilibrium. For the reaction quotient, initial concentrations given in a problem are used to calculate the value for Q. If Q < K, then the reaction shifts right to reach equilibrium. If Q > K, then the reaction shifts left to reach equilibrium. If Q = K, then the reaction does not shift in either direction because the reaction is at equilibrium.
a. Q = = =
1. 0 L
10 mol
0 L
10 mol
1. 0 L
- 0 mol
[HO][ClO]
[HOCl]
2
2 0 2 0
2 (^0) 1.0 × 10^2
Q > K, so the reaction shifts left to produce more reactants at equilibrium.
b. Q = =
2. 0 L
080 mol
0 L
98 mol
2. 0 L
- 084 mol
2
0.090 = K; at equilibrium
c. Q = =
3. 0 L
0010 mol
0 L
56 mol
3. 0 L
- 25 mol
2
110 > K
Reaction shifts to the left to reach equilibrium.
- As in Exercise 39, determine Q for each reaction, compare this value to Kp (= 0.0900), and then determine which direction the reaction shifts to reach equilibrium. Note that for this reaction, K = Kp because ∆n = 0.
a. = = =
× ( 1. 00 atm)( 1. 00 atm)
( 1. 00 atm) P P
P 2
HO ClO
2 HOCl 2 2
Q 1.
Q > Kp , so the reaction shifts left to reach equilibrium.
b. Q = ( 200 .torr)( 49. 8 torr)
( 21. 0 torr)^2 = 4.43 × 10−^2 < Kp
The reaction shifts right to reach equilibrium. Note : Because Q and Kp are unitless, we can use any pressure units when determining Q.
c. Q = ( 296 torr)( 15. 0 torr)
( 20. 0 torr)^2 = 0.0901 ≈ Kp ; at equilibrium
- CaCO 3 (s) ⇌ CaO(s) + CO 2 (g) Kp = PCO 2 = 1.
a. Q = PCO 2 ;we only need the partial pressure of CO 2 to determine Q because solids do not appear in equilibrium expressions (or Q expressions). At this temperature, all CO 2 will be in the gas phase. Q = 2.55, so Q > K (^) p ; the reaction will shift to the left to reach equilibrium; the mass of CaO will decrease.
b. Q = 1.04 = K (^) p , so the reaction is at equilibrium; mass of CaO will not change.
c. Q = 1.04 = K (^) p , so the reaction is at equilibrium; mass of CaO will not change.
d. Q = 0.211 < K (^) p ; the reaction will shift to the right to reach equilibrium; mass of CaO will increase.
- CH 3 CO 2 H + C 2 H 5 OH ⇌ CH 3 CO 2 C 2 H 5 + H 2 O K = [CHCOH][CHOH]
[CHCOCH ][HO]
3 2 2 5
a. Q = ( 0. 010 )( 0. 010 )
= 220 > K; reaction will shift left to reach equilibrium, so the concentration of water will decrease.
b. Q = ( 0. 0020 )( 0. 10 )
= 2.2 = K; reaction is at equilibrium, so the concentration of water will remain the same.
c. Q = ( 0. 044 )( 6. 0 )
= 0.40 < K; because Q < K, the concentration of water will in- crease because the reaction shifts right to reach equi- librium.
d. Q = ( 0. 88 )( 10. 0 )
= 2.2 = K; at equilibrium, so the water concentration is un- changed.
e. K = 2.2 = ( 0. 10 )( 5. 0 )
( 2. 0 )[H 2 O]
, [H 2 O] = 0.55 M
f. Water is a product of the reaction, but it is not the solvent. Thus the concentration of water must be included in the equilibrium expression because it is a solute in the reaction. When water is the solvent, then it is not included in the equilibrium expression.
43. K = 2 2
22 3 2 2
2
2 2 ( 0. 11 )
( 1. 9 10 )[O ]
[HO]
[H ][O ] − × −
× = , [O 2 ] = 0.080 M
44. KP =
P 0. 0159
P P
P
2 NO
2
Br
2 NO
2 NOBr × 2 ×
= , P NO = 0.0583 atm
2 SO 3 (g) ⇌ 2 SO 2 (g) + O 2 (g) K = 2
3
2
2 2 [SO ]
[SO ][O]
Initial 12.0 mol/3.0 L 0 0 Let x mol/L of SO 3 react to reach equilibrium. Change − x → + x + x / Equil. 4.0 − x x x /
From the problem, we are told that the equilibrium SO 2 concentration is 3.0 mol/3.0 L = 1.0 M ([SO 2 ]e = 1.0 M ). From the ICE table setup, [SO 2 ]e = x, so x = 1.0. Solving for the other equilibrium concentrations: [SO 3 ]e = 4.0 − x = 4.0 − 1.0 = 3.0 M ; [O 2 ] = x /2 = 1.0/2 = 0.50 M.
K = 2
3
2
2 2 [SO]
[SO ][O ]
2
( 3. 0 )
Alternate method: Fractions in the change column can be avoided (if you want) be defining x differently. If we were to let 2 x mol/L of SO 3 react to reach equilibrium, then the ICE table setup is:
2 SO 3 (g) ⇌ 2 SO 2 (g) + O 2 (g) K = 2
3
2
2 2 [SO ]
[SO ][O]
Initial 4.0 M 0 0 Let 2 x mol/L of SO 3 react to reach equilibrium. Change − 2 x → +2 x + x Equil. 4.0 − 2 x 2 x x
Solving: 2 x = [SO 2 ]e = 1.0 M , x = 0.50 M ; [SO 3 ]e = 4.0 − 2(0.50) = 3.0 M ; [O 2 ]e = x = 0.50 M
These are exactly the same equilibrium concentrations as solved for previously, thus K will be the same (as it must be). The moral of the story is to define x in a manner that is most comfortable for you. Your final answer is independent of how you define x initially.
- When solving equilibrium problems, a common method to summarize all the information in the problem is to set up a table. We commonly call this table the ICE table because it summarizes i nitial concentrations, c hanges that must occur to reach equilibrium, and e quilibrium concentrations (the sum of the initial and change columns). For the change column, we will generally use the variable x , which will be defined as the amount of reactant (or product) that must react to reach equilibrium. In this problem, the reaction must shift right since there are no products present initially. The general ICE table for this problem is:
2 NO 2 (g) ⇌ 2 NO(g) + O 2 (g) K = 2
2
2
2
[NO ]
[NO][O ]
Initial 8.0 mol/1.0 L 0 0 Let x mol/L of NO 2 react to reach equilibrium Change − x → + x + x / Equil. 8.0 − x x x /
Note that we must use the coefficients in the balanced equation to determine the amount of products produced when x mol/L of NO 2 reacts to reach equilibrium. In the problem, we are told that [NO]e = 2.0 M. From the set up, [NO]e = x = 2.0 M. Solving for the other concen- trations: [NO]e = 8.0 − x = 8.0 − 2.0 = 6.0 M ; [O 2 ]e = x /2 = 2.0/2 = 1.0 M. Calculating K:
K = 2
2 2 2
2
2
( 6. 0 )
[NO ]
[NO][O ]
M
M M
= = 0.11 mol/L
Alternate method: Fractions in the change column can be avoided (if you want) by defining x differently. If we were to let 2 x mol/L of NO 2 react to reach equilibrium, then the ICE table set-up is:
2 NO 2 (g) ⇌ 2 NO(g) + O 2 (g) K = 2
2
2
2
[NO ]
[NO][O ]
8.0 M 0 0
Let 2 x mol/L of NO 2 react to reach equilibrium Change − 2 x → +2 x + x Equil. 8.0 − 2 x 2 x x
Solving: 2 x = [NO]e = 2.0 M , x = 1.0 M ; [NO 2 ]e = 8.0 − 2(1.0) = 6.0 M ; [O 2 ]e = x = 1.0 M
These are exactly the same equilibrium concentrations as solved for previously; thus K will be the same (as it must be). The moral of the story is to define x in a manner that is most comfortable for you. Your final answer is independent of how you define x initially.
49. 3 H 2 (g) + N 2 (g) ⇌ 2 NH 3 (g)
Initial [H 2 ] 0 [N 2 ] 0 0 x mol/L of N 2 reacts to reach equilibrium
Change! 3 x! x → +2 x
Equil [H 2 ] 0! 3 x [N 2 ] 0! x 2 x
From the problem:
[NH 3 ]e = 4.0 M = 2 x , x = 2.0 M ; [H 2 ]e = 5.0 M = [H 2 ] 0! 3 x ; [N 2 ]e = 8.0 M = [N 2 ] 0! x
5.0 M = [H 2 ] 0! 3(2.0 M ), [H 2 ] 0 = 11.0 M ; 8.0 M = [N 2 ] 0! 2.0 M , [N 2 ] 0 = 10.0 M
50. N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g); with only reactants present initially, the net change that must
occur to reach equilibrium is a conversion of reactants into products. At constant volume and temperature, n ∝ P. Thus, if x atm of N 2 reacts to reach equilibrium, then 3 x atm of H 2 must also react to form 2 x atm of NH 3 (from the balanced equation). Let’s summarize the problem in a table that lists what is present initially, what change in terms of x that occurs to reach equilibrium, and what is present at equilibrium (initial + change). This table is typically called an ICE table for i nitial, c hange, and e quilibrium.
x
x −
, 10.0! (10.0) x = 1.00 + 2 x , (12.0) x = 9.0, x = 0.75 M
[H 2 ] = [I 2 ] = 1.00! 0.75 = 0.25 M ; [HI] = 1.00 + 2(0.75) = 2.50 M
- Because only reactants are present initially, the reaction must proceed to the right to reach equilibrium. Summarizing the problem in a table:
N 2 (g) + O 2 (g) ⇌ 2 NO(g) Kp = 0.
Initial 0.80 atm 0.20 atm 0 x atm of N 2 reacts to reach equilibrium Change − x − x → +2 x Equil. 0.80 − x 0.20 − x 2 x
Kp = 0.050 = N 2 O 2
2 NO P P
P
×
( 2 )^2
x x
x − −
, 0.050[0.16 − (1.00) x + x^2 ] = 4 x^2
4 x^2 = 8.0 × 10−^3 − (0.050) x + (0.050) x^2 , (3.95) x^2 + (0.050) x − 8.0 × 10−^3 = 0 Solving using the quadratic formula (see Appendix 1 of the text):
x = 2 a
− b ±(b^2 − 4 ac)^1 /^2
2 ( 3. 95 )
− 0. 050 ±[( 0. 050 )]^2 − 4 ( 3. 95 )(− 8. 0 × 10 −^3 )]^1 /^2
x = 3.9 × 10−^2 atm or x = −5.2 × 10−^2 atm; only x = 3.9 × 10−^2 atm makes sense ( x cannot be negative), so the equilibrium NO concentration is:
PNO = 2 x = 2(3.9 × 10−^2 atm) = 7.8 × 10−^2 atm
- H 2 O(g) + Cl 2 O(g) ⇌ 2 HOCl(g) K = 0.090 = [HO][Cl O]
[HOCl] 2 2
2
a. The initial concentrations of H 2 O and Cl 2 O are:
- 02 g
1 mol
0 L
0 gH 2 O × = 5.5 × 10 −^2 mol/L;
90 g
1 mol
0 L
0 gCl 2 O × = 2.3 × 10 −^2 mol/L
H 2 O(g) + Cl 2 O(g) ⇌ 2 HOCl(g)
Initial 5.5 × 10 −^2 M 2.3 × 10 −^2 M 0 x mol/L of H 2 O reacts to reach equilibrium
Change! x! x → +2 x
Equil. 5.5 × 10 −^2! x 2.3 × 10 −^2! x 2 x
K = 0.090 =
2 2
2
x x
x × − − × − −
1.14 × 10 −^4! (7.02 × 10 −^3 ) x + (0.090) x^2 = 4 x^2
(3.91) x^2 + (7.02 × 10 −^3 ) x! 1.14 × 10 −^4 = 0 (We carried extra significant figures.)
Solving using the quadratic formula:
− 7. 02 × 10 −^3 ± ( 4. 93 × 10 −^5 + 1. 78 × 10 −^3 )^1 /^2
= 4.6 × 10 −^3 or !6.4 × 10 −^3
A negative answer makes no physical sense; we can't have less than nothing. Thus x = 4.6 × 10 −^3 M.
[HOCl] = 2 x = 9.2 × 10 −^3 M ; [Cl 2 O] = 2.3 × 10 −^2! x = 0.023! 0.0046 = 1.8 × 10 −^2 M
[H 2 O] = 5.5 × 10 −^2! x = 0.055! 0.0046 = 5.0 × 10 −^2 M
b. H 2 O(g) + Cl 2 O(g) ⇌ 2 HOCl(g)
Initial 0 0 1.0 mol/2.0 L = 0.50 M 2 x mol/L of HOCl reacts to reach equilibrium
Change + x + x ←! 2 x
Equil. x x 0.50! 2 x
K = 0.090 = 2
2
2 2
[HO][ClO]
[HOCl] x
− x
The expression is a perfect square, so we can take the square root of each side:
x
- 50 − 2 x , ( 0.30) x = 0.50! 2 x , (2.30) x = 0.
x = 0.217 (We carried extra significant figures.)
x = [H 2 O] = [Cl 2 O] = 0.217 = 0.22 M ; [HOCl] = 0.50! 2 x = 0.50! 0.434 = 0.07 M
- 2 SO 2 (g) + O 2 (g) ⇋ 2 SO 3 (g) Kp = 0.
Initial 0.50 atm 0.50 atm 0 2 x atm of SO 2 reacts to reach equilibrium Change − 2 x − x → +2 x Equil. 0.50 − 2 x 0.50 − x 2 x
Kp = 0.25 = (^22)
3 O
2 SO
2 SO P P
P
×
2
2
x x
x − −
This will give a cubic equation. Graphing calculators can be used to solve this expression. If you don’t have a graphing calculator, an alternative method for solving a cubic equation is to use the method of successive approximations (see Appendix 1 of the text). The first step is to guess a value for x. Because the value of K is small (K < 1), not much of the forward
b. The reaction must shift to reactants (shifts left) to reach equilibrium.
N 2 O 4 (g) ⇌ 2 NO 2 (g)
Initial 0 9.0 atm
Change + x ←! 2 x
Equil. x 9.0! 2 x
Kp = x
( 9. 0 − 2 x )^2 = 0.25, 4 x^2! (36.25) x + 81 = 0 (carrying extra sig. figs.)
Solving: x = 2 ( 4 )
− ( − 36. 25 )±[(− 36. 25 )^2 − 4 ( 4 )( 81 )]^1 /^2
, x = 4.0 atm
The other value, 5.1, is impossible. PN (^2) O 4 = x = 4.0 atm; PNO 2 = 9.0! 2 x = 1.0 atm
c. No, we get the same equilibrium position starting with either pure N 2 O 4 or pure NO 2 in stoichiometric amounts.
- a. The reaction must proceed to products to reach equilibrium because only reactants are present initially. Summarizing the problem in a table:
2 NOCl(g) ⇌ 2 NO(g) + Cl 2 (g) K = 1.6 × 10 −^5
Initial
0 L
0 mol = 1.0 M 0 0
2 x^ mol/L of NOCl reacts to reach equilibrium
Change! 2 x → +2 x + x
Equil. 1.0! 2 x 2 x x
K = 1.6 × 10 −^5 = 2
2 2
2
2
( 1. 0 2 )
[NOCl]
[NO][Cl ] x
x x −
If we assume that 1.0! 2 x ≈ 1.0 (from the small size of K, we know that the product concentrations will be small), then:
1.6 × 10 −^5 = 2
3
- 0
4 x , x = 1.6 × 10 −^2 ; now we must check the assumption.
1.0! 2 x = 1.0! 2(0.016) = 0.97 = 1.0 (to proper significant figures)
Our error is about 3%; that is, 2 x is 3.2% of 1.0 M. Generally, if the error we introduce by making simplifying assumptions is less than 5%, we go no further; the assumption is said to be valid. We call this the 5% rule. Solving for the equilibrium concentrations:
[NO] = 2 x = 0.032 M ; [Cl 2 ] = x = 0.016 M ; [NOCl] = 1.0! 2 x = 0.97 M ≈ 1.0 M
Note : If we were to solve this cubic equation exactly (a longer process), we get x = 0.016. This is the exact same answer we determined by making a simplifying assumption. We saved time and energy. Whenever K is a very small value ( K << 1), always make the
assumption that x is small. If the assumption introduces an error of less than 5%, then the answer you calculated making the assumption will be considered the correct answer.
b. 2 NOCl(g) ⇌ 2 NO(g) + Cl 2 (g)
Initial 1.0 M 1.0 M 0 2 x mol/L of NOCl reacts to reach equilibrium Change! 2 x → +2 x + x Equil. 1.0! 2 x 1.0 + 2 x x
1.6 × 10 −^5 = 2
2 2
2
( 1. 0 )
( 1. 0 2 )( ) x x
x x
(assuming 2 x << 1.0)
x = 1.6 × 10 −^5 ; assumptions are great (2 x is 3.2 × 10−^3 % of 1.0).
[Cl 2 ] = 1.6 × 10 −^5 M and [NOCl] = [NO] = 1.0 M
c. 2 NOCl(g) ⇌ 2 NO(g) + Cl 2 (g)
Initial 2.0 M 0 1.0 M 2 x mol/L of NOCl reacts to reach equilibrium Change! 2 x → +2 x + x Equil. 2.0! 2 x 2 x 1.0 + x
1.6 × 10 −^5 =
( 2 ) ( 1. 0 )^2
2
(^2) x
x
x x
(assuming x << 1.0)
Solving: x = 4.0 × 10 −^3 ; assumptions good ( x is 0.4% of 1.0 and 2 x is 0.4% of 2.0).
[Cl 2 ] = 1.0 + x = 1.0 M ; [NO] = 2(4.0 × 10 −^3 ) = 8.0 × 10 −^3 M ; [NOCl] = 2.0 M
- N 2 O 4 (g) ⇌ 2 NO 2 (g) K = [NO]
[NO]
2 4
2 (^2) = 4.0 × 10 − 7
Initial 1.0 mol/10.0 L 0 x mol/L of N 2 O 4 reacts to reach equilibrium Change! x → +2 x Equil. 0.10! x 2 x
K =
[NO]
[NO]
2 4
2 (^2) = x
x
- 10 −
( 2 )^2
= 4.0 × 10 −^7 ; because K has a small value, assume that x is small compared to 0.10, so that 0.10! x ≈ 0.10. Solving:
4.0 × 10 −^7 ≈
4 x^2 , 4 x^2 = 4.0 × 10 −^8 , x = 1.0 × 10 −^4 M
NH 4 OCONH 2 (s) ⇌ 2 NH 3 (g) + CO 2 (g) Kp = 2.9 × 10 −^3
Initial 0 0 Some NH 4 OCONH 2 decomposes to produce 2 x atm of NH 3 and x atm of CO 2. Change → +2 x + x Equil. 2 x x
Kp = 2.9 × 10 −^3 = PNH^2 3 × PCO 2 = (2 x )^2 ( x ) = 4 x^3
x =
3 1 /^3
4
⎛ × −
= 9.0 × 10 −^2 atm; PNH 3 = 2 x = 0.18 atm; PCO 2 = x = 9.0 × 10 −^2 atm
P (^) total = PNH 3 + PCO 2 = 0.18 atm + 0.090 atm = 0.27 atm
- NH 4 Cl(s) ⇌ NH 3 (g) + HCl(g) Kp = PNH 3 × P (^) HCl
For this system to reach equilibrium, some of the NH 4 Cl(s) decomposes to form equal moles of NH 3 (g) and HCl(g) at equilibrium. Because mol HCl produced = mol NH 3 produced, the partial pressures of each gas must be equal to each other.
At equilibrium: Ptotal = PNH 3 + P (^) HCl and PNH 3 = P (^) HCl
P (^) total = 4.4 atm = 2 P (^) NH 3 ,2.2 atm = PNH 3 = PHCl ; Kp = (2.2)(2.2) = 4.
Le Chatelier's Principle
- a. No effect; adding more of a pure solid or pure liquid has no effect on the equilibrium position.
b. Shifts left; HF(g) will be removed by reaction with the glass. As HF(g) is removed, the reaction will shift left to produce more HF(g).
c. Shifts right; as H 2 O(g) is removed, the reaction will shift right to produce more H 2 O(g).
- When the volume of a reaction container is increased, the reaction itself will want to increase its own volume by shifting to the side of the reaction that contains the most molecules of gas. When the molecules of gas are equal on both sides of the reaction, then the reaction will remain at equilibrium no matter what happens to the volume of the container.
a. Reaction shifts left (to reactants) because the reactants contain 4 molecules of gas compared with 2 molecules of gas on the product side.
b. Reaction shifts right (to products) because there are more product molecules of gas (2) than reactant molecules (1).
c. No change because there are equal reactant and product molecules of gas.
d. Reaction shifts right.
e. Reaction shifts right to produce more CO 2 (g). One can ignore the solids and only concentrate on the gases because gases occupy a relatively huge volume compared with solids. We make the same assumption when liquids are present (only worry about the gas molecules).
- a. Right b. Right c. No effect; He(g) is neither a reactant nor a product.
d. Left; because the reaction is exothermic, heat is a product:
CO(g) + H 2 O(g) → H 2 (g) + CO 2 (g) + heat
Increasing T will add heat. The equilibrium shifts to the left to use up the added heat.
e. No effect; because the moles of gaseous reactants equals the moles of gaseous products (2 mol versus 2 mol), a change in volume will have no effect on the equilibrium.
- a. The moles of SO 3 will increase because the reaction will shift left to use up the added O 2 (g).
b. Increase; because there are fewer reactant gas molecules than product gas molecules, the reaction shifts left with a decrease in volume.
c. No effect; the partial pressures of sulfur trioxide, sulfur dioxide, and oxygen are unchanged, so the reaction is still at equilibrium.
d. Increase; heat + 2 SO 3 ⇌ 2 SO 2 + O 2 ; decreasing T will remove heat, shifting this endothermic reaction to the left.
e. Decrease
- a. Left b. Right c. Left
d. No effect (reactant and product concentrations are unchanged)
e. No effect; because there are equal numbers of product and reactant gas molecules, a change in volume has no effect on this equilibrium position.
f. Right; a decrease in temperature will shift the equilibrium to the right because heat is a product in this reaction (as is true in all exothermic reactions).
- a. Shift to left
b. Shift to right; because the reaction is endothermic (heat is a reactant), an increase in temperature will shift the equilibrium to the right.
c. No effect d. Shift to right
