Equilibrium
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7.10. At 450 K, Kp = 2.0 × 1010/bar for the given reaction at equilibrium.
2SO2(g) + O2(g) ?? 2SO3 (g)
What is Kc at this temperature?
7.10. At 450 K, Kp = 2.0 × 1010/bar for the given reaction at equilibrium.
2SO2(g) + O2(g) ?? 2SO3 (g)
What is Kc at this temperature?
Kp = Kc (RT)? ng
=> Kc = Kp (RT)-? ng
Putting the values of Kp= 2.0x1010 bar-1, R= 0.083 L bar K-1 mol-1, T = 450 K, and? ng = 2-3= -1
=> Kc = (2.0 x 1010 bar-1) x [ (0.083 L bar K-1 mol-1) x (450 K)]- (-1)
= 7.47 x 1011 mol-1 L
= 7.47 x 1011 M-1
Upvote 
7.9. Nitric oxide reacts with Br2and gives nitrosyl bromide as per reaction given below:
2NO (g) + Br2 (g) ? 2NOBr (g)
When 0.087 mole of NO and 0.0437 mole of Br2 are mixed in a closed container at constant temperature, 0.0518 mole of NOBr is obtained at equilibrium. Determine the compositions of the equilibrium mixture.
7.9. Nitric oxide reacts with Br2and gives nitrosyl bromide as per reaction given below:
2NO (g) + Br2 (g) ? 2NOBr (g)
When 0.087 mole of NO and 0.0437 mole of Br2 are mixed in a closed container at constant temperature, 0.0518 mole of NOBr is obtained at equilibrium. Determine the compositions of the equilibrium mixture.
According to the equation, 2 moles of NO (g) react with 1 mole of Br2 (g) to form 2 moles of NOBr (g). The composition of the equilibrium mixture can be calculated as follows:
No. of moles of NOBr (g) formed at equilibrium = 0.0518 mol
No. of moles of NO (g) taking part in reac
Upvote 7.8. Reaction between nitrogen and oxygen takes place as follows:
2N2?(g) + O2?(g) ? 2N2?O(g)
If a mixture of 0.482 mol of N2 and 0.933 mol of O2 is placed in a reaction vessel of volume 10 L and allowed to form N2O at a temperature for which Kc = 2.0 x 10-37, determine the composition of the equilibrium mixture.
7.8. Reaction between nitrogen and oxygen takes place as follows:
2N2?(g) + O2?(g) ? 2N2?O(g)
If a mixture of 0.482 mol of N2 and 0.933 mol of O2 is placed in a reaction vessel of volume 10 L and allowed to form N2O at a temperature for which Kc = 2.0 x 10-37, determine the composition of the equilibrium mixture.
Let x moles of N2 (g) take part in the reaction. According to the equation, x/2 moles of O2 (g) will react to form x moles of N2O (g). The molar concentration per litre of different species before the reaction and at the equilibrium point is:
| [N2? ] | [O2] | [N2? O] |
Initial concentration | 0.482? / 10 | 0.933/ 10 | 0 |
Conc. at equilibrium | (0.482 – x)? / 10 | (0.933 – x/2) / 10 | x/ 10 |
The value of equilibrium constant (2.0 x 10-37
Upvote 7.7 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
7.7 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
This is because molar concentration of a pure solid or liquid is independent of the amount present.
Since density as well as molar mass of pure liquid or solid is fixed; their molar concentrations are constant.
The concentration of solid or liquid is = = =
At constant temperature, the de
Upvote 7.6. For the following equilibrium, K =6.3 x 1014 at 1000 K.
NO (g) + O3 —–> NO2 (g) + O2(g).
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for the reverse reaction?
7.6. For the following equilibrium, K =6.3 x 1014 at 1000 K.
NO (g) + O3 —–> NO2 (g) + O2(g).
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for the reverse reaction?
For reverse reaction, KC (reverse) = 1/ KC = 1 / 6.3 x 1014 = 1.59 x 10-15
Upvote 7.5. Find out the value of Kc for each of the following equilibria from the value of Kp:
(i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g); Kp= 1.8 × 10–2 at 500 K
(ii) CaCO3 (s) ⇌ CaO (s) + CO2(g); Kp= 167 at 1073 K
7.5. Find out the value of Kc for each of the following equilibria from the value of Kp:
(i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g); Kp= 1.8 × 10–2 at 500 K
(ii) CaCO3 (s) ⇌ CaO (s) + CO2(g); Kp= 167 at 1073 K
Using the equation Kp = Kc (RT)ng
(i) From the given equation, Δng = 3 – 2 = 1,
R = 0.0821 L atm K-1 mol-1
T = 500 K, Kp= 1.8 × 10–2
Thus, Kc = Kp / (RT)ng= (1.8 x 10-2) / (0.0821 L atm K-1 mol-1 x 500 K)
4.4 x 10-4 mol L-1
(ii) From the given equation, Δng =1,
R = 0.0821 L atm K-1 mol-1
T =
Upvote 7.4. Write the expression for the equilibrium constant, Kc for each of the following reactions
(i) 2NOCl (g) ?? 2NO (g) + Cl2 (g)
(ii) 2Cu(NO3)2 (s) ?? 2CuO (s) + 4NO2 (g) + O2 (g)
(iii) CH3COOC2H5 (aq) + H2O (l) ?? CH3COOH (aq) + C2H5OH (aq)
(iv) Fe3+ (aq) + 3OH– (aq) ?? Fe(OH)3(s)
(v) I2 (s) + 5F2?2IF5
7.4. Write the expression for the equilibrium constant, Kc for each of the following reactions
(i) 2NOCl (g) ?? 2NO (g) + Cl2 (g)
(ii) 2Cu(NO3)2 (s) ?? 2CuO (s) + 4NO2 (g) + O2 (g)
(iii) CH3COOC2H5 (aq) + H2O (l) ?? CH3COOH (aq) + C2H5OH (aq)
(iv) Fe3+ (aq) + 3OH– (aq) ?? Fe(OH)3(s)
(v) I2 (s) + 5F2?2IF5
(i) The expression for the equilibrium constant is Kc= [NO]2 [Cl2] / [NOCl]2.
(ii) The expression for the equilibrium constant is Kc= [NO2]4 [O2] / [ (2Cu (NO3)2]2 = [NO2]4 [O2].
(iii) The expression for the equilibrium constant is Kc= [CH3COOH] [C2H5OH] / [CH3COOC2H5].
(iv)&nb
Upvote 7.3. At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms
I2 (g) ? 2I (g)
Calculate Kp for the equilibrium.
7.3. At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms
I2 (g) ? 2I (g)
Calculate Kp for the equilibrium.
According to the given condition,
Total pressure of equilibrium mixture = 105 Pa
Partial pressure of iodine atoms (I), PI = 40 % of 105 Pa
= 0.4 x 105 Pa
Partial pressure of iodine molecules (I2), PI2 = 60 % of 105 Pa
= 0.6 x 105 Pa
Kp for the equilibrium = (PI)2/ PI2 = (0.4 x 105 Pa)2 / (0.6 x 105 Pa)
=
Upvote 7.1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(a) What is the initial effect of the change on the vapour pressure?
(b) How do the rates of evaporation and condensation change initially?
(c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
7.1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(a) What is the initial effect of the change on the vapour pressure?
(b) How do the rates of evaporation and condensation change initially?
(c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
(a) On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours is now distributed over a larger space.
(b) On increasing the volume of the container, the rate of evaporation will increase initially because now more space is available. S
Upvote