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Chapter 14ISM - 화학 교제 14판 답지입니다.
화학및실험
Kyung Hee University
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CHAPTER 14
CHEMICAL EQUILIBRIUM
Problem Categories Biological: 14. Conceptual: 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14. Descriptive: 14, 14, 14. Environmental: 14, 14. Industrial: 14, 14, 14, 14.
Difficulty Level Easy: 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14. Medium: 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14. Difficult: 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14.
14 c
[B]
[A]
K =
(1) With Kc = 10, products are favored at equilibrium. Because the coefficients for both A and B are one, we expect the concentration of B to be 10 times that of A at equilibrium. Choice (a) is the best choice with 10 B molecules and 1 A molecule.
(2) With Kc = 0, reactants are favored at equilibrium. Because the coefficients for both A and B are one, we expect the concentration of A to be 10 times that of B at equilibrium. Choice (d) is the best choice with 10 A molecules and 1 B molecule.
You can calculate Kc in each case without knowing the volume of the container because the mole ratio
between A and B is the same. Volume will cancel from the Kc expression. Only moles of each component are needed to calculate Kc.
14 Note that we are comparing similar reactions at equilibrium – two reactants producing one product, all with coefficients of one in the balanced equation.
(a) The reaction, A + C U AC has the largest equilibrium constant. Of the three diagrams, there is the
most product present at equilibrium.
(b) The reaction, A + D U AD has the smallest equilibrium constant. Of the three diagrams, there is the
least amount of product present at equilibrium.
14 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.
34
11
4 10−
== = ×
×
'2 1033
K
K
14 The problem states that the system is at equilibrium, so we simply substitute the equilibrium concentrations
into the equilibrium constant expression to calculate Kc.
Step 1: Calculate the concentrations of the components in units of mol/L. The molarities can be calculated by simply dividing the number of moles by the volume of the flask.
2
2 mol [H ] 0. 12 L
==M
5 6 2
1 10 mol [S ] 1 10 12 L
− ==× ×− M
2
8 mol [H S] 0. 12 L
==M
Step 2: Once the molarities are known, Kc can be found by substituting the molarities into the equilibrium constant expression. 22 2 226 22
[H S] (0)
[H ] [S ] (0) (1 10 )−
== =
×
7 Kc 1× 10
If you forget to convert moles to moles/liter, will you get a different answer? Under what circumstances will the two answers be the same?
14 Using Equation (14) of the text: KP = Kc(0 T)Δn
where, Δn = 2 − 3 = − 1
and T = (1273 + 273) K = 1546 K
KP = (2 × 1022 )(0 × 1546)− 1 = 1 × 1020
14 Strategy: The relationship between Kc and KP is given by Equation (14) of the text. What is the change in the number of moles of gases from reactant to product? Recall that
Δn = moles of gaseous products − moles of gaseous reactants
What unit of temperature should we use?
Solution: The relationship between Kc and KP is given by Equation (14) of the text.
KP = Kc(0 T) Δn
Rearrange the equation relating KP and Kc, solving for Kc.
c (0 )Δ
= P n
K
K
T
Because T = 623 K and Δn = 3 − 2 = 1, we have:
1 10 5 (0 ) (0)(623 K)
− Δ
×
c== =P n 3 107
K
T
K × −
14 We substitute the given pressures into the reaction quotient expression.
32 5
PCl Cl PCl
(0)(0)
0.
(0)
P== =
PP
Q
P
The calculated value of QP is less than KP for this system. The system will change in a way to increase QP
until it is equal to KP. To achieve this, the pressures of PCl 3 and Cl 2 must increase, and the pressure of PCl 5 must decrease.
Could you actually determine the final pressure of each gas?
14 Strategy: Because they are constant quantities, the concentrations of solids and liquids do not appear in the equilibrium constant expressions for heterogeneous systems. The total pressure at equilibrium that is given is due to both NH 3 and CO 2. Note that for every 1 atm of CO 2 produced, 2 atm of NH 3 will be produced due to the stoichiometry of the balanced equation. Using this ratio, we can calculate the partial pressures of NH 3 and CO 2 at equilibrium.
Solution: The equilibrium constant expression for the reaction is
2 3
2 KPPP= NH CO
The total pressure in the flask (0 atm) is a sum of the partial pressures of NH 3 and CO 2.
PP PTN=+=HC 32 O0 atm
Let the partial pressure of CO 2 = x. From the stoichiometry of the balanced equation, you should find that
PPNH 32 =2 Therefore, the partial pressure of NH 3 = 2 x. Substituting into the equation for total pressure gives: PTN=+=+=PP xxxHC 32 O 23
3 x = 0 atm
CO 2
xP==0 atm
NH 3
Px== 20 .242atm
Substitute the equilibrium pressures into the equilibrium constant expression to solve for KP.
2 3
22 ==NH CO (0) (0) = PP 7× 10 − 3 KP
14 Of the original 1 moles of Br 2 , 1% has dissociated. The amount of Br 2 dissociated in molar concentration is:
2 1 mol [Br ] 0 0. 0 L
=× =M
Setting up a table:
Br 2 (g) U 2Br(g)
Initial (M): 1 mol 1. 0 L
= M 0
Change (M): −0 +2(0) Equilibrium (M): 1 0.
22
2
[Br] (0) [Br ] 1.
== =×− 4
Kc 6 10
14 If the CO pressure at equilibrium is 0 atm, the balanced equation requires the chlorine pressure to have the same value. The initial pressure of phosgene gas can be found from the ideal gas equation.
(3 10 2 mol)(0 L atm/mol K)(800 K) 1 atm (1. 5 0 L )
×⋅− ⋅
== =
nRT P V
Since there is a 1:1 mole ratio between phosgene and CO, the partial pressure of CO formed (0 atm) equals the partial pressure of phosgene reacted. The phosgene pressure at equilibrium is:
CO(g) + Cl 2 (g) U COCl 2 (g)
Initial (atm): 0 0 1. Change (atm): +0 +0 −0. Equilibrium (atm): 0 0 0.
The value of KP is then found by substitution.
2
2
COCl 2 CO Cl
0.
(0)
===3.
P
P PP
K
14 Let x be the initial pressure of NOBr. Using the balanced equation, we can write expressions for the partial pressures at equilibrium.
PNOBr = (1 − 0)x = 0
PNO = 0
PBr 2 =0
The sum of these is the total pressure.
0 + 0 + 0 = 1 = 0 atm
x = 0 atm
The equilibrium pressures are then
PNOBr = 0(0) = 0 atm
PNO = 0(0) = 0 atm
PBr 2 ==0(0) 0 atm
We find KP by substitution.
2
22 NO Br 3 22 NOBr
() (0) (0)
9 10
( ) (0)
== =×−
P
PP
K
P
The relationship between KP and Kc is given by
KP = Kc(RT)Δn
What happens in the special case when the two component reactions are the same? Can you generalize this relationship to adding more than two reactions? What happens if one takes the difference between two reactions?
14 K = KK'"
K = (6 × 10 − 2 )(6 × 10 − 5 )
K = 4 × 10
− 6
14 Given:
2
2 '1CO 4 CO
P==×1 10
P
K
P
2
2
'' COCl 3 CO Cl
P ==6 10×−
P
K
PP
For the overall reaction:
2
22
2 COCl '"2 14 32 2 CO Cl
===PP( ) (1 10 )(6 10 )××− =4 109
P
KK
PP
KP ×
14 To obtain 2SO 2 as a reactant in the final equation, we must reverse the first equation and multiply by two.
For the equilibrium, 2SO 2 (g) U 2S(s) + 2O 2 (g)
2 2 ''' 106 c '5 2 c
11
5 10
4 10
==⎛⎞⎜⎟⎛⎞⎜⎟=×−
⎜⎟⎜⎟×
⎝⎠⎝⎠
K
K
Now we can add the above equation to the second equation to obtain the final equation. Since we add the two equations, the equilibrium constant is the product of the equilibrium constants for the two reactions.
2SO 2 (g) U 2S(s) + 2O 2 (g) Kc'''=×5 10− 106
2S(s) + 3O 2 (g) U 2SO 3 (g) Kc''=×9 10 128
2SO 2 (g) + O 2 (g) U 2SO 3 (g) Kc=×=KKcc''' '' 5× 1023
14 (a) Assuming the self-ionization of water occurs by a single elementary step mechanism, the equilibrium constant is just the ratio of the forward and reverse rate constants. 5 f 11 r
2 10
1 10
− − −
×
== = =×
×
kk 1 1016 kk
K
(b) The product can be written as:
[H + ][OH − ] = K[H 2 O]
What is [H 2 O]? It is the concentration of pure water. One liter of water has a mass of 1000 g
(density = 1 g/mL). The number of moles of H 2 O is:
1mol 1000 g 55 mol 18 g
×=
The concentration of water is 55 mol/1 L or 55 M. The product is:
[H+][OH−] = (1 × 10 − 16 )(55) = 1 × 10 − 14
We assume the concentration of hydrogen ion and hydroxide ion are equal.
[H + ] = [OH − ] = (1 × 10 − 14 ) 1/ = 1 × 10 − 7 M
14 At equilibrium, the value of Kc is equal to the ratio of the forward rate constant to the rate constant for the reverse reaction.
c ff 2 r
12.
5 10−
== =
×
kk K k
kf = (12)(5 × 10 − 2 ) = 0.
The forward reaction is third order, so the units of kf must be:
rate = kf[A] 2 [B]
f 33 2 rate /s 1/ s (concentration)
===⋅
M
kM M
kf = 0 /M 2 ⋅s
14 Given:
3
2 2
2 SO 4 2 SO O
P==5 10×
P
K
PP
Initially, the total pressure is (0 + 0) atm or 1 atm. As the reaction progresses from left to right toward equilibrium there will be a decrease in the number of moles of molecules present. (Note that 2 moles of SO 2 react with 1 mole of O 2 to produce 2 moles of SO 3 , or, at constant pressure, three atmospheres of reactants forms two atmospheres of products.) Since pressure is directly proportional to the number of molecules present, at equilibrium the total pressure will be less than 1 atm.
14 Strategy: We are given the initial concentrations of the gases, so we can calculate the reaction quotient (Qc). How does a comparison of Qc with Kc enable us to determine if the system is at equilibrium or, if not, in which direction the net reaction will proceed to reach equilibrium?
Solution: Recall that for a system to be at equilibrium, Qc = Kc. Substitute the given concentrations into
the equation for the reaction quotient to calculate Qc. 2 2 30 c 33 20 20
[NH ] [0]
0.
[N ] [H ] [0][0]
Q == =
Comparing Qc to Kc, we find that Qc < Kc (0 < 1). The ratio of initial concentrations of products to reactants is too small. To reach equilibrium, reactants must be converted to products. The system proceeds from left to right (consuming reactants, forming products) to reach equilibrium.
Therefore, [NH 3 ] will increase and [N 2 ] and [H 2 ] will decrease at equilibrium.
The equilibrium constant relationship is given by:
2 c 22
[HBr] [H ][Br ]
K =
Substitution of the equilibrium concentration expressions gives
2 6 c 2
(0 2 )
2 10
−
==×
x K x
Taking the square root of both sides we obtain:
3
0 2
1 10
−
=×
x x
x = 1 × 10 − 4
The equilibrium concentrations are:
[H 2 ] = [Br 2 ] = 1 × 10
− 4 M
[HBr] = 0 − 2(1 × 10 − 4 ) = 0 M
If the depletion in the concentration of HBr at equilibrium were defined as x, rather than 2x, what would be
the appropriate expressions for the equilibrium concentrations of H 2 and Br 2? Should the final answers be different in this case?
14 Strategy: We are given the initial amount of I 2 (in moles) in a vessel of known volume (in liters), so we can calculate its molar concentration. Because initially no I atoms are present, the system could not be at equilibrium. Therefore, some I 2 will dissociate to form I atoms until equilibrium is established.
Solution: We follow the procedure outlined in Section 14 of the text to calculate the equilibrium concentrations.
Step 1: The initial concentration of I 2 is 0 mol/2 L = 0 M. The stoichiometry of the problem
shows 1 mole of I 2 dissociating to 2 moles of I atoms. Let x be the amount (in mol/L) of I 2 dissociated. It follows that the equilibrium concentration of I atoms must be 2x. We summarize the changes in concentrations as follows:
I 2 (g) U 2I(g)
Initial (M): 0 0. Change (M): −x + 2 x Equilibrium (M): (0 − x) 2 x
Step 2: Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. 22 5 c 2
[I] (2 )
3 10
[I ] (0 )
== =×−
−
x K x
4 x 2 + (3 × 10 − 5 )x − (7 × 10 − 7 ) = 0
The above equation is a quadratic equation of the form ax 2 + bx + c = 0. The solution for a quadratic equation is b b4ac 2 2a
−± −
x =
Here, we have a = 4, b = 3 × 10 − 5 , and c = −7 × 10 − 7. Substituting into the above equation,
(3) (3) 4(4)(7) 5527 2(4)
−× ± × − −×−− −
x=
( 3 10 ) 53 (3 10 )
8
−× ± ×−−
x =
x = 4 × 10 − 4 M or x = −4 × 10 − 4 M
The second solution is physically impossible because you cannot have a negative concentration. The first solution is the correct answer.
Step 3: Having solved for x, calculate the equilibrium concentrations of all species.
[I] = 2 x = (2)(4 × 10 − 4 M) = 8 × 10 − 4 M
[I 2 ] = (0 − x) = [0 − (4 × 10 − 4 )] M = 0 M
Tip: We could have simplified this problem by assuming that x was small compared to 0. We could then assume that 0 − x ≈ 0. By making this assumption, we could have avoided solving a quadratic equation.
14 Since equilibrium pressures are desired, we calculate KP.
KP = Kc(0 T)Δn = (4 × 10 − 3 )(0 × 800) 1 = 0.
COCl 2 (g) U CO(g) + Cl 2 (g)
Initial (atm): 0 0 0. Change (atm): −x +x +x
Equilibrium (atm): (0 − x) x x
2 0. (0 )
=
−
x x
x 2 + 0 − 0 = 0
x = 0 atm
At equilibrium:
PCOCl 2 =− =(0 0)atm 0 atm
PCO = 0 atm
PCl 2 =0 atm
Substitution gives the equation:
2
2 2 CO CO
()
1.
(4 )
== =
P −
P x K Px
This can be rearranged to the quadratic:
x 2 + 1 − 6 = 0
The solutions are x = 1 and x = −3; only the positive result has physical significance (why?). The equilibrium pressures are
PCO = x = 1 atm
PCO 2 =(4−=1) 2 atm
14 The initial concentrations are [H 2 ] = 0 mol/5 L = 0 M and [CO 2 ] = 0 mol/5 L = 0 M.
H 2 (g) + CO 2 (g) U H 2 O(g) + CO(g)
Initial (M): 0 0 0 0. Change (M): −x −x +x +x
Equilibrium (M): 0 − x 0 − x x x
2 2 c 2 22
[H O][CO]
4.
[H ][CO ] (0 )
===
−
x K x
Taking the square root of both sides, we obtain:
2.
0.
=
−
x x
x = 0 M
The equilibrium concentrations are:
[H 2 ] = [CO 2 ] = (0 − 0) M = 0 M
[H 2 O] = [CO] = 0 M
14 (a) Addition of more Cl 2 (g) (a reactant) would shift the position of equilibrium to the right.
(b) Removal of SO 2 Cl 2 (g) (a product) would shift the position of equilibrium to the right.
(c) Removal of SO 2 (g) (a reactant) would shift the position of equilibrium to the left.
14 (a) Removal of CO 2 (g) from the system would shift the position of equilibrium to the right.
(b) Addition of more solid Na 2 CO 3 would have no effect. [Na 2 CO 3 ] does not appear in the equilibrium constant expression.
(c) Removal of some of the solid NaHCO 3 would have no effect. Same reason as (b).
14 (a) This reaction is endothermic. (Why?) According to Section 14, an increase in temperature favors an endothermic reaction, so the equilibrium constant should become larger.
(b) This reaction is exothermic. Such reactions are favored by decreases in temperature. The magnitude of Kc should decrease.
(c) In this system heat is neither absorbed nor released. A change in temperature should have no effect on the magnitude of the equilibrium constant.
14 Strategy: A change in pressure can affect only the volume of a gas, but not that of a solid or liquid because solids and liquids are much less compressible. The stress applied is an increase in pressure. According to Le Châtelier's principle, the system will adjust to partially offset this stress. In other words, the system will adjust to decrease the pressure. This can be achieved by shifting to the side of the equation that has fewer moles of gas. Recall that pressure is directly proportional to moles of gas: PV = nRT so P ∝ n.
Solution: (a) Changes in pressure ordinarily do not affect the concentrations of reacting species in condensed phases because liquids and solids are virtually incompressible. Pressure change should have no effect on this system.
(b) Same situation as (a).
(c) Only the product is in the gas phase. An increase in pressure should favor the reaction that decreases the total number of moles of gas. The equilibrium should shift to the left, that is, the amount of B should decrease and that of A should increase.
(d) In this equation there are equal moles of gaseous reactants and products. A shift in either direction will have no effect on the total number of moles of gas present. There will be no change when the pressure is increased.
(e) A shift in the direction of the reverse reaction (left) will have the result of decreasing the total number of moles of gas present.
14 (a) A pressure increase will favor the reaction (forward or reverse?) that decreases the total number of moles
of gas. The equilibrium should shift to the right, i., more I 2 will be produced at the expense of I.
(b) If the concentration of I 2 is suddenly altered, the system is no longer at equilibrium. Evaluating the
magnitude of the reaction quotient Qc allows us to predict the direction of the resulting equilibrium shift. The reaction quotient for this system is:
20 c 2 0
[I ]
[I]
Q =
Increasing the concentration of I 2 will increase Qc. The equilibrium will be reestablished in such a way
that Qc is again equal to the equilibrium constant. More I will form. The system shifts to the left to establish equilibrium.
(c) The forward reaction is exothermic. A decrease in temperature at constant volume will shift the system to the right to reestablish equilibrium.
14 Strategy: (a) What does the sign of ΔH° indicate about the heat change (endothermic or exothermic) for the forward reaction? (b) The stress is the addition of Cl 2 gas. How will the system adjust to partially offset the stress? (c) The stress is the removal of PCl 3 gas. How will the system adjust to partially offset the stress? (d) The stress is an increase in pressure. The system will adjust to decrease the pressure. Remember, pressure is directly proportional to moles of gas. (e) What is the function of a catalyst? How does it affect a reacting system not at equilibrium? at equilibrium?
(b) Assuming that the amount of added solid CaO is not so large that the volume of the system is altered significantly, there should be no change at all. If a huge amount of CaO were added, this would have the effect of reducing the volume of the container. What would happen then?
(c) Assuming that the amount of CaCO 3 removed doesn't alter the container volume significantly, there
should be no change. Removing a huge amount of CaCO 3 will have the effect of increasing the container volume. The result in that case will be the same as in part (a).
(d) The pressure of CO 2 will be greater and will exceed the value of KP. Some CO 2 will combine with
CaO to form more CaCO 3. (Shift left)
(e) Carbon dioxide combines with aqueous NaOH according to the equation
CO 2 (g) + NaOH(aq) → NaHCO 3 (aq)
This will have the effect of reducing the CO 2 pressure and causing more CaCO 3 to break down to CO 2 and CaO. (Shift right)
(f) Carbon dioxide does not react with hydrochloric acid, but CaCO 3 does.
CaCO 3 (s) + 2HCl(aq) → CaCl 2 (aq) + CO 2 (g) + H 2 O(l)
The CO 2 produced by the action of the acid will combine with CaO as discussed in (d) above. (Shift left)
(g) This is a decomposition reaction. Decomposition reactions are endothermic. Increasing the temperature will favor this reaction and produce more CO 2 and CaO. (Shift right)
14 (i) The temperature of the system is not given. (ii) It is not stated whether the equilibrium constant is KP or Kc (would they be different for this reaction?). (iii) A balanced equation is not given. (iv) The phases of the reactants and products are not given.
14 (a) Since the total pressure is 1 atm, the sum of the partial pressures of NO and Cl 2 is
1 atm − partial pressure of NOCl = 1 atm − 0 atm = 0 atm
The stoichiometry of the reaction requires that the partial pressure of NO be twice that of Cl 2. Hence,
the partial pressure of NO is 0 atm and the partial pressure of Cl 2 is 0 atm.
(b) The equilibrium constant KP is found by substituting the partial pressures calculated in part (a) into the equilibrium constant expression.
2
22 NO Cl 22 NOCl
(0) (0)
(0)
== =0.
PP
P
KP
14 (a) 22
22 NO NO 11 NO
2 10
(3)(0)
== =×−
P
PP
K
PP
PNO = 1 × 10 − 6 atm
(b)
2 4 10 31 NO (0)(0)
×=− P
PNO = 2 × 10 − 16 atm
(c) Since KP increases with temperature, it is endothermic.
(d) Lightening. The electrical energy promotes the endothermic reaction.
14 The equilibrium expression for this system is given by:
KPPP= CO 22 H O
(a) In a closed vessel the decomposition will stop when the product of the partial pressures of CO 2 and H 2 O equals KP. Adding more sodium bicarbonate will have no effect.
(b) In an open vessel, CO 2 (g) and H 2 O(g) will escape from the vessel, and the partial pressures of CO 2 and H 2 O will never become large enough for their product to equal KP. Therefore, equilibrium will never be established. Adding more sodium bicarbonate will result in the production of more CO 2 and H 2 O.
14 The relevant relationships are:
2 c
[B]
[A]
K = and
2 B A
P
P
KP=
KP = Kc(0 T) Δn = Kc(0 T) Δn = + 1
We set up a table for the calculated values of Kc and KP.
T (°C) Kc KP
200
(0) 2
56.
(0)
= 56(0 × 473) = 2 × 10
3
300
(0) 2
3.
(0)
= 3(0 × 573) = 1 × 102
400
(0) 2
2.
(0)
= 2(0 × 673) = 116
Since Kc (and KP) decrease with temperature, the reaction is exothermic.
14 (a) The equation that relates KP and Kc is:
KP = Kc(0 T)Δn
For this reaction, Δn = 3 − 2 = 1
21042 (0 ) (0 298)
× −
== =
×
44 c 810
KP
T
K × −
(b) Because of a very large activation energy, the reaction of hydrogen with oxygen is infinitely slow without a catalyst or an initiator. The action of a single spark on a mixture of these gases results in the explosive formation of water.
From the balanced equation the percent decomposed is
0 mol 100% 0 mol
×=48%
(c) If the temperature does not change, KP has the same value. The total pressure will still be 0 atm at equilibrium. In other words the amounts of ammonia and hydrogen sulfide will be twice as great, and the amount of solid ammonium hydrogen sulfide will be:
[0 − 2(0)]mol = 0 mol NH 4 HS
14 Total number of moles of gas is:
0 + 0 + 0 = 1 mol of gas
You can calculate the partial pressure of each gaseous component from the mole fraction and the total pressure.
NO NO T
0.
0 atm 0 atm 1.
PP==× =Χ
OO 22 T
0.
0 atm 0 atm 1.
PP==× =Χ
NO 22 NO T
0.
0 atm 0 atm 1.
PP==Χ ×=
Calculate KP by substituting the partial pressures into the equilibrium constant expression.
2
2
22 NO 22 NO O
(0)
(0) (0)
== =1 105
P
PP
KP ×
14 Since the reactant is a solid, we can write:
KPPP=()NH 322 CO
The total pressure is the sum of the ammonia and carbon dioxide pressures.
PPPtotal=+NH 32 CO
From the stoichiometry,
PPNH 32 = 2 CO
Therefore: PPPPtotal=+== 2 CO 22 CO 3 CO 2 0 atm
PCO 2 =0 atm
PNH 3 =0 atm
Substituting into the equilibrium expression:
KP = (0) 2 (0) = 4 × 10 − 3
14 Set up a table that contains the initial concentrations, the change in concentrations, and the equilibrium concentration. Assume that the vessel has a volume of 1 L.
H 2 + Cl 2 U 2HCl
Initial (M): 0 0 3. Change (M): +x +x − 2 x
Equilibrium (M): (0 + x) x (3 − 2 x)
Substitute the equilibrium concentrations into the equilibrium constant expression, then solve for x. Since Δn = 0, Kc = KP. 22 c 22
[HCl] (3 2 ) 193 [H ][Cl ] (0 )
−
== =
+
x K xx
Solving the quadratic equation,
x = 0.
Having solved for x, calculate the equilibrium concentrations of all species.
[H 2 ] = 0 M [Cl 2 ] = 0 M [HCl] = 3 M
Since we assumed that the vessel had a volume of 1 L, the above molarities also correspond to the number of moles of each component.
From the mole fraction of each component and the total pressure, we can calculate the partial pressure of each component.
Total number of moles = 0 + 0 + 3 = 4 mol
PH 2 =×=0 atm
PCl 2 =×=0 atm
PHCl=×=1 atm
14 Set up a table that contains the initial concentrations, the change in concentrations, and the equilibrium
concentrations. The initial concentration of I 2 (g) is 0 mol/0 L = 0 M. The amount of I 2 that dissociates is (0)(0 M) = 0 M. We carry extra significant figures throughout this calculation to minimize rounding errors.
I 2 U 2I
Initial (M): 0 0 Change (M): −0 +(2)(0) Equilibrium (M): 0 0.
Substitute the equilibrium concentrations into the equilibrium constant expression to solve for Kc.
22 4 2
[I] (0)
2 10
[I ] 0.
== =×=− 4
c 2 10
K × −
KP = Kc(0 T)Δn
KP = (2 × 10 − 4 )(0 × 860) 1
= 0.
Chapter 14ISM - 화학 교제 14판 답지입니다.
Course: 화학및실험
University: Kyung Hee University
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