Ncert Solutions Chemistry Class 11th
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7.16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?
2ICl (g) ⇌ I2 (g) + Cl2 (g); Kc = 0.14
7.16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?
2ICl (g) ⇌ I2 (g) + Cl2 (g); Kc = 0.14
Suppose at equilibrium, the molar concentration of both I2 and Cl2 is x mol L-1.
Kc = [I2] [ Cl2] / [ICl]2= x2 / (0.78 – 2x)2
=>x/ (0.78 – 2x) = (0.14)1/2 = 0.374
=> x= 0.167
[ICl] = (0.78 – 2 x 0.167) = (0.78 – 0.334) = 0.446 M
[I2] = 0.167 M,
[Cl2] = 0.167 M
Upvote 
7.15. At 700 K, equilibrium constant for the reaction:
H2?(g) + I2 (g) ?2HI (g)
is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2?(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?
7.15. At 700 K, equilibrium constant for the reaction:
H2?(g) + I2 (g) ?2HI (g)
is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2?(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?
H2 (g) + I2 (g)? 2HI (g); K=64, T=700K
2HI? H2 + I2 K=1/64700K
a 0 0
a (1−α) aα/2 aα/2
0.5 x x
x2 / (0.5)
Upvote 7.14. One mole of H2O and one mole of CO are taken in a 10 litre vessel and heated to 725 K, at equilibrium point 40 percent of water (by mass) reacts with carbon monoxide according to equation.
H2O (g) + CO (g) ? H2 (g) + CO2 (g)
Calculate the equilibrium constant for the reaction.
7.14. One mole of H2O and one mole of CO are taken in a 10 litre vessel and heated to 725 K, at equilibrium point 40 percent of water (by mass) reacts with carbon monoxide according to equation.
H2O (g) + CO (g) ? H2 (g) + CO2 (g)
Calculate the equilibrium constant for the reaction.
Number of moles of water originally present = 1 mol
Percentage of water reacted= 40%
Number of moles of water reacted= 1 x 40/100 = 0.4 mol
Number of moles of water left= (1 – 0.4) = 0.6 mole
According to the equation, 0.4 mole of water will react with 0.4 mole of carbon monoxide to form 0.4 mole of hy
Upvote 7.13. The equilibrium constant expression for a gas reaction is,
Write the balanced chemical equation corresponding to this expression.
7.13. The equilibrium constant expression for a gas reaction is,
Write the balanced chemical equation corresponding to this expression.
Balanced chemical equation for the reaction is
4 NO (g) + 6 H2O (g)? 4 NH3 (g) + 5 O2 (g)
Upvote 7.12. A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant Kc for the reaction
N2?(g) + 3H2? (g) ? 2NH3? (g) is 1.7×102
Is this reaction at equilibrium? If not, what is the direction of net reaction?
7.12. A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant Kc for the reaction
N2?(g) + 3H2? (g) ? 2NH3? (g) is 1.7×102
Is this reaction at equilibrium? If not, what is the direction of net reaction?
According to the given equation, concentration quotient,
Qc = [NH3]2 / [N2] [ H2]3
= (8.13/20 mol L-1)2 / (1.57 / 20 mol L-1) x (1.92 / 20 mol L-1)3
= 2.38 x 103
The equilibrium constant (Kc) for the reaction = 1.7 x 10-2
As Qc ≠ Kc, the reaction is not in a state of equilibrium.
Upvote 7.11. A sample of HI (g) is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of HI (g) is 0.04 atm. What is Kp for the given equilibrium?
2HI (g) ? H2 (g) + I2 (g)
7.11. A sample of HI (g) is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of HI (g) is 0.04 atm. What is Kp for the given equilibrium?
2HI (g) ? H2 (g) + I2 (g)
pHI = 0.04 atm, pH2 = 0.08 atm, pI2 = 0.08 atm
Kp= (pH2 x pI2) / (pHI)2 = (0.08 x 0.08) / (0.04 x 0.04)
= 4.0
Upvote 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