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Class 11th

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6 months ago

Class 11th Equilibrium

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-.       
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Vishal Baghel

Contributor-Level 10

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.

New answer posted

6 months ago

Class 11th Equilibrium

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.         
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Vishal Baghel

Contributor-Level 10

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.

New answer posted

6 months ago

Class 11th Equilibrium

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?
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Vishal Baghel

Contributor-Level 10

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

New answer posted

6 months ago

Class 11th Equilibrium

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
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Vishal Baghel

Contributor-Level 10

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 concentration of the products is very large as compared to that of the reactants.

(c) Since the value of Kc is 1.8, this means that both the products and reactants have appreciable concentration.

New answer posted

6 months ago

Class 11th Equilibrium

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) + H2​O(g) ⇌ CO2​(g) + H2​(g)
If a reaction ves

...more
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Vishal Baghel

Contributor-Level 10

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             0          0

At equilibrium            4-p          4-p                p         

...more

New answer posted

6 months ago

Class 11th Equilibrium

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.  
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Vishal Baghel

Contributor-Level 10

(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 pressure is increased, the value of the equilibrium constant remains unaffected.

(iii) By increasing the temperature, the forward reaction will be favoured since it is endothermic in nature. Therefore, the value of equilibrium constant will increase.

New answer posted

6 months ago

Class 11th Equilibrium

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)          
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Vishal Baghel

Contributor-Level 10

(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.

New answer posted

6 months ago

Class 11th Equilibrium

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?          
0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

(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 reactants will be formed and the value of Kp will decrease.

(ii) If the temperature is increased, according to Le Chatelier's principle, the forward reaction will be favoured as it is endothermic. Therefore, the equilibrium gets shifted to the right and the value of Kp will increase.

(iii) The addition of catalyst will not change t

...more

New answer posted

6 months ago

Class 11th Equilibrium

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.    
0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

For the equilibrium reaction

                            2HBr (g) ? H2(g) + Br2(g) , the equilibrium constant is K = 1/1.6*105.

Initial pressure     10               0           0

At equilibrium      10-p          p/2        p/2


The equilibrium constant exp

...more

New answer posted

6 months ago

Class 11th Equilibrium

7.26. Which of the following reactions will get affected by increasing the pressure? Also mention whether the change will cause the reaction to go into forward or backward direction.

(i) COCl2 (g) ​⇌ CO (g) + Cl2 (g)

(ii) CH4 (g) + 2S2 (g) ​⇌ CS2 (g) + 2H2S (g)

(iii) CO2 (g) + C (s) ​⇌ 2CO (g)

(iv) 2H2 (g) + CO (g) ​⇌ CH3OH (g)

(v) CaCO3 (s) ​⇌CaO (s) + CO2 (g)

(vi) 4NH3 (g) + 5O2 (g) ​⇌ 4NO (g) + 6H2O (g)   
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Vishal Baghel

Contributor-Level 10

Only those reactions will be affected by increasing the pressure in which the number of moles of the gaseous reactants and products are different (i.e. when np ≠ nr) (gaseous). In general,

  • The reaction will go to the left if np> nr.
  • The reaction will go to the right if nr> np.
    Keeping this in mind,

(i) Increase in pressure will favour backward reaction because np  (2) > nr (1)

(ii) Increase in pressure will not affect equilibrium because np = nr = 3.

(iii) Increase in pressure will favour backward reaction because np  (2) > nr  (1)

(iv) Increase in pressure will favour forward reaction because np&nb

...more

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