Ncert Solutions Chemistry Class 11th
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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)
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)
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 min
Upvote 
7.25. Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
(a) PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
(b) CaO (s) + CO2 (g) ⇌ CaCO3 (s)
(c) 3Fe (s) + 4H2O (g) ⇌ Fe3O4 (s) + 4H2 (g)
7.25. Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
(a) PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
(b) CaO (s) + CO2 (g) ⇌ CaCO3 (s)
(c) 3Fe (s) + 4H2O (g) ⇌ Fe3O4 (s) + 4H2 (g)
(i) Pressure will increase in the forward reaction and number of moles of products will increase.
(ii) Pressure will increase in backward reaction and number of moles of products will decrease.
(iii) The change in pressure will have no effect on the equilibrium constant and there will be
Upvote 7.24. Calculate (a) ∆Gϴ and (b) the equilibrium constant for the formation of NO2 from NO and O2 at 298 K
NO (g) + ½ O2 (g)⇌ NO2 (g)
where
ΔfGϴ (NO2) = 52.0 kJ/mol
ΔfGϴ (NO) = 87.0 kJ/mol
ΔfGϴ (O2) = 0 kJ/mol
7.24. Calculate (a) ∆Gϴ and (b) the equilibrium constant for the formation of NO2 from NO and O2 at 298 K
NO (g) + ½ O2 (g)⇌ NO2 (g)
where
ΔfGϴ (NO2) = 52.0 kJ/mol
ΔfGϴ (NO) = 87.0 kJ/mol
ΔfGϴ (O2) = 0 kJ/mol
(a) ΔG? = Δf?G?(NO2?) − [Δf?G?(NO) + ½ ?Δf?G?(O2?)]
= 52.0 − 87.0 −1/2 × 0
= −35 kJ/mol
(b) logK = − ΔG? / 2.30
Upvote 7.23. At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% CO by mass.
C (s) + CO2 (g) ⇌ 2CO (g)
Calculate Kc for this reaction at the above temperature.
7.23. At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% CO by mass.
C (s) + CO2 (g) ⇌ 2CO (g)
Calculate Kc for this reaction at the above temperature.
Let the mixture has 100g as total mass.
So, The masses of CO and CO2? are 90.55g and 100 − 90.55 = 9.45 g respectively.
Therefore, te number of moles of CO
&n
Upvote 7.22. Bromine monochloride (BrCl) decomposes into bromine and chlorine and reaches the equilibrium:
2BrCl (g) ⇌ Br2 (g) + Cl2 (g)
The value of Kc is 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 x10-3mol L-1, what is its molar concentration in the mixture at equilibrium?
7.22. Bromine monochloride (BrCl) decomposes into bromine and chlorine and reaches the equilibrium:
2BrCl (g) ⇌ Br2 (g) + Cl2 (g)
The value of Kc is 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 x10-3mol L-1, what is its molar concentration in the mixture at equilibrium?
Let x moles of BrCl decompose in order to attain the equilibrium. The initial molar concentration and the molar concentration at equilibrium point of different species may be represented as follows:
2BrCl2?(g) ?→?Ar2?(g)+Cl2?(g)
3.3×10−3m 0&nbs
Upvote 7.21 Equilibrium constant, Kc for the reaction
N2 (g) + 3H2 (g) ⇌ 2NH3 (g) at 500 Kc is 0.061
At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L–1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1 NH3. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?
7.21 Equilibrium constant, Kc for the reaction
N2 (g) + 3H2 (g) ⇌ 2NH3 (g) at 500 Kc is 0.061
At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L–1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1 NH3. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?
According to available data:
N2 = [3.0], H2 = [2.0], NH3 = [0.50]
Qc = [NH3]2 / [N2] [ H2]3
= [0.50]2 / [3.0] [ 2.0]3
= 0.25/24
= 0.0104.
Since the value of Qc is less than that of Kc (0.061), the reaction is not in a state of equilibrium. It will proceed in the forward direction till Qc becomes the
Upvote 7.20. One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO2.
FeO (s) + CO (g) Fe (s) + CO2 (g); Kp = 0.265 atm at 1050K
What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO= 1.4 atm and pCO2= 0.80 atm?
7.20. One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO2.
FeO (s) + CO (g) Fe (s) + CO2 (g); Kp = 0.265 atm at 1050K
What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO= 1.4 atm and pCO2= 0.80 atm?
The initial partial pressures of CO and CO2 are 1.40 atm and 0.8. atm respectively.
The expression for the reaction quotient is:
Qp=PCO2 / PCO=0.80 / 1.4
Upvote 7.19. A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was reached, the concentration of PCl5 was found to be 0.5 x 10-1 mol L-1. If Kc is 8.3 x 10-3 what are the concentrations of PCl3 and Cl2 at equilibrium?
7.19. A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was reached, the concentration of PCl5 was found to be 0.5 x 10-1 mol L-1. If Kc is 8.3 x 10-3 what are the concentrations of PCl3 and Cl2 at equilibrium?
Let the initial molar concentration of PCl5 per litre = x mol
Molar concentration of PCl5 at equilibrium = 0.05 mol
Therefore, moles of PCl5 decomposed = (x – 0.05) mol
Moles of PCl3 formed = (x – 0.05) mol
Moles of Cl2 formed = (x – 0.05) mol
The molar conc./litre of reactants a
Upvote 7.18. Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
CH3COOH (l) + C2H5OH (l) ⇌CH3COOC2H5 (l) + H2O (l)
(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
7.18. Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
CH3COOH (l) + C2H5OH (l) ⇌CH3COOC2H5 (l) + H2O (l)
(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
(i) The concentration ratio (Concentration quotient), Qc for the reaction is:
Qc = [CH3COOC2H5] [ H2O] / [CH3COOH] [C2H5OH]
(ii)
| CH3COOH | C2H5OH | CH3COOC2H5 | H2O |
Initial molar concentration | 1.0 mol | 0.18 mol | 0 | 0 |
Molar concentration at equilibrium | (1 – 0.171) = 0.829 mol | (0.18 – 0.171) = 0.009 mol | 0.171 mol | 0.171 mol |
Applying
Kc = [CH3COOC2H5] [H2O] / [CH3COOH] [C2H5OH]
= (0.171 mol) x (0.171 mol) / (0.829 mol) (0.009 mol)
= 3.92
(iii)
| CH3COOH | C2H5OH | CH3COOC2H5 | H2O |
Initial molar concentration | 1.0 mol | 0.5 mol | 0.214 mol | 0.214 mol |
Molar concentration at equilibrium | (1 – 0.214) = 0.786 mol | (0.5 – 0.214) = 0.286 mol | 0.214 mol | 0.214 mol
|
Qc = [CH3COOC2H5] [H2O] / [CH3COOH] [
Upvote 7.17. Kp =0.04 atm at 898 K for the equilibrium shown below. What is the equilibrium concentration ok C2H6 when it is placed in a flask at 4 atm pressure,and allowed to come to equilibrium.
C2H6(g) ⇌ C2H4(g) + H2(g).
7.17. Kp =0.04 atm at 898 K for the equilibrium shown below. What is the equilibrium concentration ok C2H6 when it is placed in a flask at 4 atm pressure,and allowed to come to equilibrium.
C2H6(g) ⇌ C2H4(g) + H2(g).
The equilibrium reaction is
C2?H6?(g) ? C2?H4?(g)+H2?(g).
Initial | 4 | 0 | 0 |
Change | −x | x | x |
Equilibrium | 4−x | x | x |
The expression for the equilibrium constant is Kp?= (?PC2?H4?) (PH2) / PC2?H6?.
Substituting the values in the above equation, we get
Upvote