Equilibrium
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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
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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 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
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