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Resolution for Practice Problems - General Chemistry | CHEM 142, Assignments of Chemistry

Jackson State University (JSU)Chemistry
Prof. John D. Watts

Material Type: Assignment; Professor: Watts; Class: GENERAL CHEMISTRY; Subject: Chemistry; University: Jackson State University; Term: Spring 2009;

Typology: Assignments

Pre 2010

Uploaded on 08/08/2009

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Solutions to Chapter 14 Practice Problems (Watts). Spring 2009.
Note: these are my solutions, warts and all. I have worked them out and typed them from
scratch, which takes me a lot of time and effort. They have not been copied from the
textbook publisher’s materials. I don’t guarantee correctness. If an instructor or tutor
wishes to make use of these solutions for teaching, that is OK, but please provide proper
acknowledgement of my work.
Problem 4
From the information given, at equilibrium
0.10
]][[
]][[
BA
DC
. Therefore, [A][B] =
0.100[C][D]. The answer is (c).
Problem 10
CO (g) + H2O (g) CO2 (g) + H2 (g)
OHCO
HCO
p
PP
PP
K
2
22
N2 (g) + 3H2 (g) 2NH3 (g)
3
22
2
3
)(
)(
HN
NH
p
PP
P
K
NH4HS (s) NH3 (g) + H2S (g)
]][[
23
SHNHK
p
Problem 13
4HCl (g) + O2 (g) 2H2O (g) + 2Cl2 (g) ΔHHo = -114.4 kJ
Changes to increase the amount of Cl2 (g) at equilibrium:
Add HCl or O2. [These will cause the forward reaction to occur].
Decrease the volume of the container. [This favors the side with the smaller number of
moles of gas molecules, i.e. forward reaction will occur].
Decrease the temperature [The (forward) reaction is exothermic, so heat is like a
product].
If H2O is removed, more Cl2 will be produced.
Problem 14
Yes for (a) and (c). No for (b).
In (a) and (c) there is a smaller number of gas molecules on the product (right) side,
hence increasing the pressure from 1 to 10 atm will cause more product to form. In (b)
there are equal numbers of gas molecules on both sides, hence there will be no effect.
pf3
pf4
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Solutions to Chapter 14 Practice Problems (Watts). Spring 2009. Note: these are my solutions, warts and all. I have worked them out and typed them from scratch, which takes me a lot of time and effort. They have not been copied from the textbook publisher’s materials. I don’t guarantee correctness. If an instructor or tutor wishes to make use of these solutions for teaching, that is OK, but please provide proper acknowledgement of my work. Problem 4 From the information given, at equilibrium 10.^0 [ ][ ] [ ][ ]  A B C D

. Therefore, [A][B] = 0.100[C][D]. The answer is (c). Problem 10 CO (g) + H 2 O (g)  CO 2 (g) + H 2 (g) CO HO CO H p P P

P P

K

2

^22

N 2 (g) + 3H 2 (g)  2NH 3 (g) (^3) 2 2 2 3 ( )

N H NH p P P

P

K 

NH 4 HS (s)  NH 3 (g) + H 2 S (g) K^ p [^ NH 3 ][ H 2 S ] Problem 13 4HCl (g) + O 2 (g)  2H 2 O (g) + 2Cl 2 (g) ΔHHo^ = -114.4 kJ Changes to increase the amount of Cl2 (g) at equilibrium: Add HCl or O 2. [These will cause the forward reaction to occur]. Decrease the volume of the container. [This favors the side with the smaller number of moles of gas molecules, i.e. forward reaction will occur]. Decrease the temperature [The (forward) reaction is exothermic, so heat is like a product]. If H 2 O is removed, more Cl 2 will be produced. Problem 14 Yes for (a) and (c). No for (b). In (a) and (c) there is a smaller number of gas molecules on the product (right) side, hence increasing the pressure from 1 to 10 atm will cause more product to form. In (b) there are equal numbers of gas molecules on both sides, hence there will be no effect.

Problem 16 (a) 2NO (g)  N 2 (g) + O 2 (g) ΔHHo^ = -180.5 kJ (b) 2SO 3 (g)  2SO 2 (g) + O 2 (g) ΔHHo^ = 197.8 kJ Decrease in dissociation with increase in temperature: (a). It is an exothermic reaction, so heat is like a product. Raising the temperature is like adding a product, hence reverse reaction occurs. The opposite is true for (b), since (b) is endothermic (i.e.heat is like a reactant). Problem 20 The key formula is ngas K (^) p K c RT  ( ) (Eqn. 14.3 on page 583). (a) 1.^210 (^0.^0821668 )^21.^9 (or^22 to^2 s.f.)  ^3  ^1  K p (b)  1. 32  10 ^2 ( 0. 0821  1000 )^1  1. 61  10 ^4 Kp (c) Since ΔHngas = 0, Kp = Kc = 2. Problems 25, 26, 37, 42 --- later. Problem 44 N 2 O 4 (g) 2NO 2 (g) ΔHH = 57.2 kJ We are told that 12.5% of the N 2 O 4 molecules are dissociated at equilibrium. An example of what this means: initially suppose [N 2 O 4 ] = 1 M (and [NO 2 ] = 0). Then at equilibrium [N 2 O 4 ] = 0.875 M and [NO 2 ] = 0.250 M. (a) If the volume of the reaction vessel is doubled, there will be a shift to the side with more gas molecules, i.e. the forward reaction will occur (L → R shift), more N 2 O 4 molecules will dissociate and more NO 2 molecules will form. (b) If the temperature is raised, the endothermic reaction (i.e. the forward reaction) will occur. [Heat is like a reactant in an endothermic reaction]. (c) There will be no effect if a catalyst is added. Catalysts only affect the rate at which equilibrium is reached, not the position of the equilibrium. (d) There will be no effect if neon gas is added to the mixture in a constant-volume flask: the concentrations and partial pressures of the reactants and products do not change.

E (4.20 x 10-3) – 2x x x Initial number of moles of ICl: nICl = 0.682 g / 162.35 g mol-1 = 4.20 x 10-3^ mol. We are told that 0.0383 g of I 2 is present at equilibrium. From this we can find the number of moles: nI2 = 0.0383 g / 253.80 g mol-1^ = 1.51 x 10-4^ mol. This is x, the number of moles of I 2 at equilibrium. Equilibrium number of moles of ICl: nICl = (4.20 x 10-3) – 2(1.51 x 10-4) = 3.898 x 10- mol. Now we can calculate Kc: [Note that we don’t actually need to obtain equilibrium concentrations for this example since we have the same number of moles on both sides, so we don’t need to use the volume of the container]. Kc = 3 3 2 4 2 2 (^2 21). 50 10 ( 3. 898 10 ) ( 1. 51 10 ) [ICl ] [I ][Cl ]         Problem 60 Set up ICE table. I am doing this based on numbers of moles [As in #57, you don’t need to use the volume of the container to solve the problem]. Butane  Isobutane I 0.0860 mol 0 C -x x E 0.0860 – x x 0.0860 mol is the initial number of moles of butane: 5.00 g / 58.14 g mol-1^ = 0.0860 mol. From Kc, we can solve for x: Kc = [Butane] ( 0. 0860 ) [Isobutane ] x x   This leads to: 0.^0764 mol

  1. 94

1 K

0. 0860 K

c c (^) ^    x  The problem asks for the mass of isobutane at equilibrium: 0.0764 mol x 58.14 g mol-1^ = 4.44 g. Problem 66. Later.

Problem 70 3 2 3 2 ( )

H H S p P

P

K  0. 265 atm

( ) 0. 200 atm 3 3 3 2 3 2 2     p H S p H S H K P K P P The total gas pressure is the sum of the partial pressures of H 2 S and H 2 , i.e. 0.465 atm.