key.bansal.chemicalEquilibrium.pdf

7/28/2019 key.bansal.chemicalEquilibrium.pdf

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ANSWER KEY

EXERCISEI

Q.l (a) 25, shifts left, (b) 0.22, shiftsright,(c) oo, shifts left, (d) 1, shifts right, (e) 0, shift right, (f) 4, shifts left

Q.2 (a) K = [Ag+] [CI

-] is less than 1. AgCl is insoluble thus the concentration ofions are much less than 1M

(b) K= 1 /[Pb2+

] [Cl~]2

is greater than one because PbCl2 is insoluble and formation of the solid will

reduce the concentration ofions to a low levelQ.4 K about 10 Q.6p (a) incomplete (b) almost complete Q.7 c Q.8 ~ 9x 10"

32mol/L

Q. 9 The reaction is not an equilibrium because Qc > Kc. The reactionwill proceedfromright to left to reach

equilibrium

Q.ll 5.9 x 10~3M Q.12 [NO] = 0.056 M, [N2] = [02] = 1.37 M

Q.13 [PC13] = [C y = 0.071 M,[PC15] = 0.089

Q. 14 PCIF = PP2 = 0.389 atm, PCIF3 = 1 -08 atm

Q.15 Kp =0.4, a - 0 . 1 Q.16 50%

Q.17 (a) 6.667 x 10"3 mol L_1; (b) n (N204) = 0.374 mol; n (N02) = 0.052 mol ;

(c) 10.49 atm (d) 6.44%

Q.18 0.97 atm Q.19 Kp = 1.3 x 10"3

atm"2

Q.20 Kp= 2.5 atm, P = 15 atm Q.21 53.33%

Q.22 K = 4 Q.23 31/27 Q.24 22.4 mg

Q.25 PH20 =5 x 10-I5atm Q.26 0.821 atm

Q. 27 add N2, add H2, increase the pressure, heat the reaction

Q.28 (a) shiftright, shiftleft, (b) shift right, no effect, (c) shift left, shiftleft, (d) shift left, shiftright

Q.29 (a) K = [CH30H]/[H2]2[C0],

(b) 1. [H2] increase, [CO] decrease, [CH3OH] increase; 2. [H2] increase, [CO] decrease, [CH3OH]decrease ; 3. [H2] increase, [CO] increase, [CH3OH] increase ; 4. [H2] increase, [CO] increase,

[CHgOH] increase; 5. [H2] increase, [CO] increase, [CHgOH] decrease; 6. no change

Q.30 (a) K = [C0][H2]/[H20];(b) in each of the following cases the mass of carbon will change, butits concentration (activity) will notchange. 1. [H20] no change, [CO] no change, [H2] no change; 2. [H20] decrease, [CO] decrease,

[H2] decrease; 3. [H20] increase, [CO] increase, [H2] decrease; 4. [H20] increase, [CO] increase,

[H2] increase; 5. [H20] decrease, [CO] increase, [H2] increase

Q.31 bQ.32 Add NaCl or some other salt that produces Cl~ in the solution. Cool the solution.

Q.33 a

kf [C]

Q.34 kf[A][B] = kr[ C] ; -^ = j ^ = k c Q.36 216

Q.38 (i) 2; (ii) 1.2 mol/L; (iii) 0.1moles/hr

Q.39 krincrease more than k^ this means that Ea (reverse) is greater than Ea (forward). The reaction is

exothermic when Ea (reverse) > Ea (forward).

Q.43 (a)-9.574 J/mol,(b)A=1010

,(c) 9.96 xlO9, (d) 9.98 x 10

9

Q.44 16.06 kJ Q.45 -810 J/mol;-5872 J/mol and 41.3 kJ/mol

Q.46 1.3 x 108

Q.47 0.058

Q.48 29.0 Q.49 Kp = 0.0313 atm, Kc = 1.28 x 10"3

[C02]3

(Pco2)3

1 _ J _ Q 5 0 ( a ) K c = I c o f ' Kp = ' ^ K C

= [ 0 j ' K P = (P0, )3 ' = KP =

Kc = [Ba2+

] [S042

-]


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Q.51 Kc= 1.51k Kp = 49.6 Q.52 l J x l O ^ M

Q.53 [CO] = [H2] = 0.18 M ; [H20] = 1.02 M

Q.54 (a) Kc= 0.573 and Kp= 23.5; (b) to the right, [PC15] =0.365 M; [PC13] = 0.285 M, ; [Cl2] = 0.735 M

Q.56 -1.005 kJ/mol Q.57 AG = 0 ; K = 1

Q.58 AH= 9.07 kJ/mol; AS=-8.92 J/moHK"1

1.

5.

9.

12.

15.

18.

21.

25.

29.

33.

37.

C

exothermic

Kolog

-2

Ki

AH

2.303R

temperature

backward

decreases

T

T

F

F

T

PROFICIENCY TEST

12. high 3.

104.

VK

6. KP = KC(RT) 7. backward 8. AG0 = - RT INK

T 2 - V

T2T! J10. zero 11. mol

2L

-2

13. Guldberg and Waage 14. decreases

16. high 17. KP = KC(RT)AN

19. same amount of 20. largervalue of

22. T 23. F 24. F

26. F 27. T 28. T

30. T 31. T 32. T

34. F 35. F 36. F

38. F 39.

EXERCISEII

T 40. F

Q.l

Q.2

Q.5

Q.6

Q.7

Q.8

Q.9

Q.ll

Q.14

Q.16

Q.18

Q.20

Q.22

Q.23

Q.25

Q.27

Q.29

Q.31

Q.32

Q.34

Kp(atm)= 1,13

2.4 mole Q.3 Kp = 0.01 atm Q.4 0.379 atm

PCS2 = 1.284 atm, PS2 = 0.1365 atm

300L

(i) xco = 0.765, xC02 = 0.235; p(C02) = 0.938 atm (ii) PTotal = 0.68 atm

(K = 3), nS02= 0.92, nS03= 0.48, nNO=l .28, % 0 2 = 0.22

PNO= 0.64 atm, P ^ =0.095 atmNO22.7 g/ lit Q.12 K = 1.337, K = 0.0263

Q.10

Q.13

Q.15

a = 0.415 and 0.2

Kp= 1.862 x 1012 atm -1/2

V = 144 mL

= 0.3 molQ. 17 Kc=54, nm=0.9 mol, nI2= 0.05 mol, nH2:

Q.19 nCQ2 = 0.938, nH2= 1.938, n c o = 0.062, nH2Qg= 4.062

Q.21 (a) 400mm2,900mm2 (b) 4: 9, (c) 72.15 mm Hg

pNa = 0.843 M Pa; pNa2= 0.170 M Pa; kp =0.239

9.34 g

48 atm

6.71 x io~4

a = 0.5

Kc= 1/12, [R] = 4 (initial), = 1.5 (final) Q.24 dissociation = 48.5%, 80.05%

ArH = 75.5 kJ mol"1 Q.26 B -> NH4N02; Total pressure = 84.34 atm

K = 2.58 Q.28 K=707.2, backward reaction is favoured

Kc = 480 Q.30 1.32 xlO"3

KA= 779.4, KB = 6.074 x 105;KC= 1.283 x 10~

3

To be proved Q.33 CuS04.5H20 = 9.2 x 10-4mol, CuS04 = 8 x 10~

5 moles

kc =0.111;kc =0.14 Q.35 (a) 1.05 atm, (b) 3.43 atm"1 Q.36 314.1 atm


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Q.37 pC02 =202.65 kPa; pH20 =3.16kPa;pc o = T).124kPa

Q.38 103.47 kJ/mol

Q.39 (a)Kp=7.563 x 10"2, v = 35.62, p(H20)=p(CO)=0.05atm,p(H2)=0.1167atm,p(C02)=0.2833atm

(b) Kj=9, K2=119

Q.40 K = 1.17 x 10-3

EXERCISE III

Q.l B Q.2 A Q.3 A Q.4 D

Q.5 B Q.6 B Q.7 D Q.8 A

Q.9 C,D,E Q.10 D Q.ll C Q.12 C,D

Q.13 A Q.14 AB,C,D Q.15 A Q.16 AQ.17 A Q.18 D Q. 19 (i) B ,(ii) C,(iii) B,(iv) A

Q-20 (i) A, (ii) B, (iii) A

EXERCISE IV

Q.l XN2= 0.79, X02= 0.21 Q.2 D

Q.3 p = 0.454 g L -1 Q.4 Fraction decomposed = 0.4

Q.5 4.54gdm-3

Q.6 (i)kc= 8.1 x 10"5mol

2L

2;kp = 4.19 x 1CT

2atm

2(ii)Noeffect;

Q.7 15991 J mol"1 , 12304 JmoH ; B > C > A

Q.8 (i) 5.705 x 103

kJmol-1

(ii) Since initial Gibbsfreeenergy change ofthe reaction is positive, so the reverse reaction will takeplace

Q.9 B

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key.bansal.chemicalEquilibrium.pdf