Ncert Solutions Chemistry Class 11th
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7.36. Which of the following are Lewis Acids?
H2O,BF3, H+ and NH4+
7.36. Which of the following are Lewis Acids?
H2O,BF3, H+ and NH4+
7.35. What is meant by conjugate acid-base pair? Find the conjugate acid/base for the following species: HNO2, CN–, HClO4F–, OH–, CO32-, and S2-.
7.35. What is meant by conjugate acid-base pair? Find the conjugate acid/base for the following species: HNO2, CN–, HClO4F–, OH–, CO32-, and S2-.
An acid-base pair which differs by a proton only is known as conjugate acid-base pair. For example, HCl, and Cl represents conjugate acid base pair.
The conjugate acid/base for the species HNO2, CN–, HClO4F–, OH–, CO32-, and S2- are NO2–, HCN, ClO4–, HF, H2O, HCO3– and HS–.respectively.
Upvote 
7.34. The reaction, CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g) is at equilibrium at 1300 K in a 1L flask. It also contains 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.
7.34. The reaction, CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g) is at equilibrium at 1300 K in a 1L flask. It also contains 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.
Kc = [CH4] [H2O] / [CO] [H2]3
3.90 = [CH4]×0.02 / (0.30) (0.10)3
[CH4] = 3.90×0.30× (0.10)3 / 0.02
=5.85×102 M
Thus, the concentration of methane in the mixture is 5.85×102 M.
Upvote 7.33. The value of Kc for the reaction 3O2(g) ⇌2O3(g) is 2.0 x 10-50 at 25°C. If equilibrium concentration of O2 in air at 25°C is 1.6 x 10-2, what is the concentration of O3?
7.33. The value of Kc for the reaction 3O2(g) ⇌2O3(g) is 2.0 x 10-50 at 25°C. If equilibrium concentration of O2 in air at 25°C is 1.6 x 10-2, what is the concentration of O3?
The equilibrium constant expression is
Kc = [O3]2 / [O2]3
2.0×1050 = [O3]2 / (1.6×102)3
[O3]2=2.0×1050× (1.6×102)3
=8.192×1056
Hence, the equilibrium concentration of ozone is
[O3]= 2.86×1028M
Upvote 7.32. Predict which of the following will have appreciable concentration of reactants and products:
a) Cl2 (g) ⇌2Cl (g) Kc = 5 ×10–39
b) Cl2 (g) + 2NO (g) ⇌2NOCl (g) Kc = 3.7 × 108
c) Cl2 (g) + 2NO2 (g) ⇌2NO2Cl (g) Kc = 1.8
7.32. Predict which of the following will have appreciable concentration of reactants and products:
a) Cl2 (g) ⇌2Cl (g) Kc = 5 ×10–39
b) Cl2 (g) + 2NO (g) ⇌2NOCl (g) Kc = 3.7 × 108
c) Cl2 (g) + 2NO2 (g) ⇌2NO2Cl (g) Kc = 1.8
Following conclusions can be drawn from the values of Kc .
(a) Since the value of Kc is very small, this means that the molar concentration of the products is very small as compared to that of the reactants.
(b) Since the value of Kc is quite large, this means that the molar concentra
Upvote 7.31. Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2 In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction.
CO (g) + H2O(g) ⇌ CO2(g) + H2(g)
If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam so that PCO = PH2O = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp = 10.1 at 400°C.
7.31. Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2 In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction.
CO (g) + H2O(g) ⇌ CO2(g) + H2(g)
If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam so that PCO = PH2O = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp = 10.1 at 400°C.
Let the partial pressures of CO and H2 be p.
CO (g) + H2?O(g) ? CO2?(g) + H2?(g)
Initial conc. 4 bar 4 bar &nb
Upvote 7.30. At 473 K, the equilibrium constant Kc for the decomposition of phosphorus pentachloride (PCl5) is 8.3 x 10-3. If decomposition proceeds as:
PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) ΔrHϴ = 124.0 kJ mol–1
(a) Write an expression for Kc for the reaction
(b) What is the value of Kc for the reverse reaction at the same temperature.
(c) What would be the effect on Kc if
(i) More of PCl5is added (ii) pressure is increased (iii) the temperature is increased.
7.30. At 473 K, the equilibrium constant Kc for the decomposition of phosphorus pentachloride (PCl5) is 8.3 x 10-3. If decomposition proceeds as:
PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) ΔrHϴ = 124.0 kJ mol–1
(a) Write an expression for Kc for the reaction
(b) What is the value of Kc for the reverse reaction at the same temperature.
(c) What would be the effect on Kc if
(i) More of PCl5is added (ii) pressure is increased (iii) the temperature is increased.
(a) The equilibrium constant expression is
Kc = [PCl3] [Cl2] / [PCl5]
(b) The equilibrium constant expression for the reverse reaction is
Kc (reverse) = 1/ (8.3×103) =120.48
(c) (i) By adding more of PCl5, value of Kc will remain constant because there is no change in temperature.
(ii) When pres
Upvote 7.29. Describe the effect of:
(i) Addition of H2 (ii) Addition of CH3OH
(iii) Removal of CO (iv) Removal of CH3OH
on the equilibrium of the reaction:
2H2(g) + CO (g) ⇌ CH3OH (g)
7.29. Describe the effect of:
(i) Addition of H2 (ii) Addition of CH3OH
(iii) Removal of CO (iv) Removal of CH3OH
on the equilibrium of the reaction:
2H2(g) + CO (g) ⇌ CH3OH (g)
(i) Equilibrium will be shifted in the forward direction.
(ii) Equilibrium will be shifted in the backward direction.
(iii) Equilibrium will be shifted in the backward direction.
(iv) Equilibrium will be shifted in the forward direction.
Upvote 7.28 Dihydrogen gas is obtained from natural gas by partial oxidation with steam as
per following endothermic reaction:
CH4 (g) + H2O (g) ⇌ CO (g) + 3H2 (g)
(a) Write as expression for Kp for the above reaction.
(b) How will the values of Kp and composition of equilibrium mixture be affected by
(i) Increasing the pressure
(ii) Increasing the temperature
(iii) using a catalyst?
7.28 Dihydrogen gas is obtained from natural gas by partial oxidation with steam as
per following endothermic reaction:
CH4 (g) + H2O (g) ⇌ CO (g) + 3H2 (g)
(a) Write as expression for Kp for the above reaction.
(b) How will the values of Kp and composition of equilibrium mixture be affected by
(i) Increasing the pressure
(ii) Increasing the temperature
(iii) using a catalyst?
(a) The expression for Kp for the reaction is:
Kp = (pCO) x (pH2) / (pCH4) x (pH2O)
(b) (i) By increasing the pressure, the number of moles per unit volume will increase. In order to decrease the same, the equilibrium gets shifted to the left or in the backward direction. As a result, more of
Upvote 7.27. The equilibrium constant for the following reaction is 1.6 ×105 at 1024K
H2(g) + Br2(g) ⇌ 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.
7.27. The equilibrium constant for the following reaction is 1.6 ×105 at 1024K
H2(g) + Br2(g) ⇌ 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.
For the equilibrium reaction
2HBr (g) ? H2(g) + Br2(g) , the equilibrium constant is K = 1/1.6×105.
Initial pressure &
Upvote