Chapter- 07
CHEMICAL EQUILIBRIUM
(ANSWER KEY & SOLUTION)
Class - XIII
CHEMISTRY
DPP - 01
ANSWER KEY
1.
(a) 25, shifts left, (b) 0.22, shifts right, (d) 1, shifts right, (e) 0, shift right
(b)
2NH3

3 atm
Qp = 2/9 < Kp
N2
+
2 atm
3H2
1 atm
 shifts in forward direction
2.
4.
K about 10
6 × 10–22
3.
5.
250
(i) 2; (ii) 1.2 mol/L
6.
8.
2 × 109
0.9
7.
9.
PCIF = PF2 = 0.1 atm, PClF3 = 05 atm
KC = 100
10.
12.
14.
Kp = 0.2463
11.
(a) incomplete (b) almost complete
–32
64 × 10
13.
2.4 mole
The reaction is not an equilibrium because Qc > Kc. The reaction will proceed from right to left
to reach equilibrium
Forward reaction.
16.
5.9 × 10–3 M
[NO] = 2M, [N2] = [O2] = 0.5M
18.
[PCl3] = [Cl2] = 0.027 M, [PCl5] = 0.073
[A] = 0.34 M, [B] = 1.16 M, [C] = 1.16 M 20.
KP = 0.4, a ~ 0.1
50%
22.
Kp = 0.01 atm
–3
–1
(a) 6.667 × 10 mol L ; (b) n (N2O4)=0.374 mol; n (NO2)=0.052 mol ; (c) 10.49 atm (d)
6.44 %
15.
17.
19.
21.
23.
24.
26.
28.
29.
31.
33.
0.97 atm
Kp= 2.5 atm, P = 15 atm
0.379 atm
PCS2 = 1.8 atm, PS2 = 0.2 atm
0.821 atm
P = 5 × 10–15 atm
35.
36
Less than 50 %
5 (a) decrease
(b) increase
(f)no change
(g) increase
(i) When decreasing temperature
(a) Forward
(b) Forward
(ii) Increasing the pressure
(a) Forward
(b) No change
37.
25.
27.
2.7 g / lit
53.33%
30.
32
34.
22.4 mg
4
5 × 10–3 atm.
(c) decrease
(h) increase
(d) increase (e) increase
(i) no change (j) no change
(c) Backward
(d) Forward
(c) Backward
(d) Forward
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48
48
38.
add N2, add H2, increase the pressure , heat the reaction
39.
(a) shift right (b) shift right (c) shift left (d) shift left
40.
(a) K = [CH3OH]/[H2]2[CO] ,
(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,[CH3OH] increase ; 5. [H2] increase, [CO] increase, [CH3OH] decrease ; 6. no change
41.
(a) K = [CO][H2]/[H2O] ;
(b) in each of the following cases the mass of carbon will change, but its concentration
(activity) will not change. 1. [H2O] no change, [CO] no change, [H2] no change ; 2. [H2O]
decrease, [CO] decrease, [H2] decrease ; 3. [H2O] increase, [CO] increase, [H2] decrease; 4.
[H2O] increase, [CO] increase, [H2] increase ; 5. [H2O] decrease , [CO] increase , [H2] increase
42.
2
43.
Add NaCl or some other salt that produces Cl– in the solution. Cool the solution.
44.
kr increase more than kf, this means that Ea (reverse) is greater than Ea (forward). Hence
exothermic reaction.
45.
Exothermic
46.
(a) 4 × 10–3 (b) (i) decrease (ii) no change (iii) increase (iv) increase (v) no change (vi)
increase
47.
100
48.
(a) 1.05 atm, (b) 3.43 atm–1
49.
27
50.
15
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49
49
SOLUTIONS
1.
11
1
=
= 25
2
(0.2)
0.04
QC > KC , backward shift
2  (1)2
2
(b) QP =
=
2
(3)
9
QP < KP , forward shift
(a) QC =
(c) QP =
11
=
1
QP < KP , forward shift
(d) Since initially only reactants are present, so reaction will shift to forward
2.
If K is large. Then tendency of reaction to convert in product will be more
3.
KC =
Kf
Kb
4.
KC =
Kf
Kb
5.
(i) Amount if  formed = 2 × amount of A reacted

n=2
 2B
(ii) A 
K=
6.
[B]2
(0.6)2
=
= 1.2
0.3
[A]
 2NO(g)
N2(g) + O2(g) 
KC = 0.5×10–9
 N2(g) + O2(g)
2NO 
KC =
1
= 2×109
9
0.5 10
 ClF(g) + F2(g)
ClF3(g) 
7.
Pi
Peq
0.15
0.15 – x
0.2 =
0
x
0
x
x2
0.15  x
x = 0.01
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 PCl3 (g) + Cl2 (g)
PCl5 (g) 
8.
0.5×10–3
Ceq
KC =
1.5 102  3 102
= 0.9
0.5 103
 2HI(g)
H2(g) + I2(g) 
9.
Ci
Ceq
4.8×10–3
0.4×10–3
0
0
–3
0.4×10
KC =
10.
1.5×10–2 3×10–2
4×10–3
(4 103 )2
= 100
0.4 103  0.4 103
KP = PH O = 0.2463
2
KP = KC(RT)2
KC =
11.
0.2463
= 0.01
0.0821 300
If KC is very large, extent of reaction is almost 100%.
 2O3 (g) KC = 8×10–55
3O2 (g) 
12.
8×10–3
Ceq
x
8×10–55 =
x2
(8 103 )3
x2 = 8×10–55 × (8)3 × 10–9
x2 = (8)4 × 10–64
x = 64 × 10–32
 2C
A + B 
13.
ni
neq
2
2–x
4=
3
3–x
0
2x
4x 2
(2  x)(3  x)
x = 1.2
14.
 CS2(g) + 4H2(g)
CH4(g) + 2H2S(g) 
3
4
3
3
0.3  (0.3)
0.3  (0.4)2
4
QC =
=
81
81 104
= ×10–2
2
16  10
16
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15.
 CO(g) + 3H2(g)
H2O(g) + CH4(g) 
0.03 M
0.05 M
0.15 M
0.2 M
0.15  (0.2)
= 0.8
0.03  0.05
3
QC =
QC < KC , forward shift
16.
 2NH3(g)
3H2(g) 
N2(g) +
0.036 M
0.15 M
x
2
0.29 =
x
0.036  (0.15)3
x = 5.9×10–3
 2NO(g)
N2(g) + O2(g) 
17.
Ci
Ceq
1.5
1.5–x
KC =
1.5
1.5–x
0
2x
(2x)2
= 16
(1.5  x)(1.5  x)
2x
=4
(1.5  x)
x=1
 PCl3(g) + Cl2(g)
PCl5 (g) 
18.
Ci
Ceq
0.1
0.1–x
0
x
0
x
x2
= 10–2
0.1  x
x = 0.027
19.
KC =
2 2
=4
1
 B(g) + C(g)
A(g) 
Ceq
0.5–x
1+x
1+x
(1  x)
=4
0.5  x
2
x = 0.16
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20.
KP =
4 2
.P
1  2
 
 
2
4  14
4 2
× 10
2 × 1.5 =
2
1


1
1 4
4  0.01
=
×10 = 0.4
1  0.01
21.
 2NO2(g)
N2O4 (g) 
2
1
4 2
=
×
2
1 
3
2
2
1

=
1  2
3
42 = 1
2 =
1
2
 2NO(g) + Br2(g)
2NOBr(g) 
a
0
0
22.
ni
neq
2a
3
a
3
a
6
Peq
0.16
0.8
0.4
(0.8)2  0.4
KP =
= 0.1
(0.16)2
24.
KP =
(0.1)2
4
×4=
2
1  (0.1)
99
1
25 × P = 4
99
1  1 25
96
P=
= 0.97 atm
99
25.
 PCl3(g) + Cl2(g)
PCl5 (g) 
2
×P
1  2
2
0.178 =
×1
1  2
KP =
 = 0.8
208.5
= 115.83
1.8
1115.83
d=
= 2.7 g/L
0.0821 523
Mmix =
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26.
92
1 
1
=
3
4 1
69 =
KP =
4
9 ×5 = 9 ×5
8
1  19
9
20
= 2.5
8
230
92
=
1 
3
=
 = 0.2
1 4
20
= 25
.P
8
1  125
P = 15 atm
27.
46
1 
46
1+=
30
30 =
 = 0.53
28.
KP = KC
 H2S(g)
H2(g) + S(s) 
ni
neq
0.2
0.2–x
1
1–x
6.8×10–2 =
0
x
x
0.2  x
x = 0.0127
 CS2(g)
S2(g) + C(s) 
29.
Pi
Peq
2
2–x
0
x
x
=9
2x
11x = 18
x=
18
11
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30.
 CaO(s) + CO2(g)
CaCO3(s) 
KP + PCO = 4×10–2
2
n CO2 =
4 102  0.521
0.08211000
= 4×10–4
nCaO = n CO = 4×10–4
2
WCaO = 4×10–4× 56 g
= 0.4×56 g
= 22.4 mg
 CaO(s) + CO2(g)
CaCO3(s) 
31.
neq
0.2×0.75
KP = PCO =
2
0.2  0.75  0.08211000
15
KP = 0.821
32.
 NiCO(g)
Ni(s) + 4CO(g) 
x
x
x
= 0.125
x4
1
= 0.125
x3
x=2
PT = 4 atm
33.
KP =
1
(PH2O )6
(PH2O )6 =
PH2O =
34.
= 6.4×1085
1
1
= 10–85 =
×10–84
6.4
64
1
×10–14 = 5×10–15
2
KP = (PH O )2 = 2.25×10–4
2
PH2O
= 15×10–3 = 1.5×10–2
R.H. =
1.5 102
×760×100
22.8
= 50 %
If RH < 50% reaction will shift forward
36.
(a) Forward shift
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No of moles of H2O will decrease
(b) For ward shift  moles of H2O will decrease
(c) Forward shift  moles of HCl will increase
(d) Backward shift  moles of Cl2 will increase
(e) Backward shift  partical pressure of each goes will increase
(f) No change
(g) H > 0, T, KC
(h) T, forward shift [HCl]
(i) At constant volume – no change
37.
On increasing temperature reaction will shift towards endothermic direction and on increasing
pressure reaction will more towards lesser number of gaseous moles.
38.
Decreasing volume of container
Increasing temperature
Addition of N2
Addition of H2
40.
(b)
(i)
 CH3OH (g)
2H2(g) + CO(g) 

conc.

(ii) conc. 
(iii) conc.
(iv) conc.
(v) conc. 





44.
Keq at 1400 K =










0.29
1.1106
2.9
=
× 105
1.1
= 2.6×105
1.3
1.4 105
1.3
=
× 105
1.4
Keq at 1500 K =
T, Keq
45.
= 0.92×105
e×0
Keq at 2000 K
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(iii) = –[2(i) + (ii) ]
K=
=
1
(4.4)  5.311010
2
1010
(4.4)2  5.31
Keq at 1000 K
K = 2.24 × 22
T, K
47.
e×0
 A(g) B(s) + C(g)
X(s) 
P1 + P2
P1
 D(g) + A(s) + E(s)
Y(s) 
P2
P2 + P1
P1 (P1 + P2) = 500
P2 (P1 + P2) = 2000
P1 + P2 = 50
P1 = 2×50 = 100 atm
49.
 NH3(g) + H2S (g)
NH4HS(s) 
1–x
1
  N2(g) + 3 H2(g)
NH3(g) 
2
2
2
2
= 0.25
3x
2
 x = 0.5
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