Chemistry Chapter 7 Chemical Equilibrium Class 11 Notes

1st Year Chemistry Class 11 Notes for kpk Chapter no 7 Chemical Equilibrium for kpk, long question and short question 2021 notes.

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Chemical Equilibrium Class 11 Notes 2021 for kpk

Q.2 i) The Change in temperature changes the equilibrium position of the reaction

N2(g) + O2(g) ⇌ 2NO(g)
but a change in pressure does not, why?

Answer:

hange in temperature changes

As the reaction between N2(g) and O2(g) to form NO(g) is an endothermic reaction, so the change in temperature will shift the equilibrium position of the reaction. An increase in temperature will push it in forwarding direction while a decrease in temperature will push the equilibrium in a backward position. On the other hand, pressure has no effect on this reaction, because this reaction happens without a change in a number of moles. So there is no volume change during the reaction and pressure will have no effect on the equilibrium position of the reaction.


Q.2 ii) Give the concentration units for the following reversible reactions.

a) PCl5(g)    ⇌ PCI3(g) + Cl2(g)
b) N2(g)+3H2(g) ⇌ 2NH3(g)
c) H2(g) + CO2(g) ⇌ CO(g) + H2O(g)
d) 4NH3(g) + 5O2(g)  ⇌ 4NO(g) + 6H2O(g)

Answer:
a) PCl5(g)    ⇌ PCI3(g) + Cl2(g)
As we know that,

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From the given equation we can write,

Give the concentration units for the following reversible reactions.
Give the concentration units for the following reversible reactions.
chemistry class `` notes

Read more: Chemistry Class 11 Notes 2021 States of Matter-III Solids Chapter 6


Q.2 iii) Why the value of Kc falls with the rise in temperature for the synthesis of SO3?

 2SO2(g) + O2(g) ⇌ 2SO3(g) ΔH = -94.58 kJ mole-1

Answer:
As we know that, Kc expression for this reaction is,
2SO2(g) + O2(g) ⇌ 2SO3(g) ΔH = -94.58 kJ mole-1

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Since the reaction is exothermic in forwarding direction, its reverse is an endothermic reaction. So, a rise in temperature will shift the equilibrium in the backward direction. This will decrease the concentration of SO3 in the mixture and as a result, the value of Kc will fall.


Q.2 iv) There is a dynamic not static equilibrium present between liquid and vapour at constant temperature. Explain.

Answer:
At constant temperature, there is a dynamic equilibrium present between liquid and vapours. The reason is that the evaporation of liquid into vapours and the condensation of vapours to form liquid, both processes keep on taking place at exactly the same rate, i.e. the rate of evaporation becomes equal to rate of condensation. As the changes are taking place but the overall effect of changes is zero, it is said to be in dynamic equilibrium.

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Q.2 v) The change in concentration of reactants does not change the value of equilibrium constant permanently. Elaborate.

Answer:
When the concentration of reactants is changed at equilibrium position, the system is disturbed.  The reaction moves in forward or backward direction according to Le-Chatelier’s principle to restore the equilibrium. As a result the concentrations of various components are changed so that the ratio of products concentration to reactants concentration (i.e. equilibrium constant Kc) remains constant.

Q.2 vi) Discuss the equilibrium of a sparingly soluble salt.

Answer:
Equilibrium of Sparingly Soluble Salt:

When a sparingly soluble salt is shaken with water, it dissolves until a saturated solution is formed. The solution contains the dissolved ions of salt and undissolved salt. Equilibrium is established between ions of the salt and undissolved salt.
Example
For example, AgCl forms Ag+ and Cl- ions when dissolved in water to form a saturated solution. The equilibrium established between dissolved Ag+(aq), Cl(aq) and undissolved AgCl is represented as:

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Q.2 vii) Common ion effect operates best in the purification of certain substances. Explain.

Answer:
Common ion effect is the suppression of solubility of one electrolyte by the addition of another strong electrolyte having a common ion. A substance (salt) can be purified by suppressing its solubility with common ion. When solubility is suppressed, the ions combine and form the pure substance in solid form which is free of any impurities.
For example, NaCl is purified by passing HCl gas through a saturated solution of NaCl.

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Cl is a common ion due to which equilibrium shifts to the backward direction whereby NaCl precipitates.

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Q.2 viii)The ionization of calcium oxalate is suppressed by adding CaCl2 to it, why?

Answer:
Calcium oxalate is a weak electrolyte which ionizes up to little extent. When CaCl2 (a strong electrolyte) is added to the aqueous solution of calcium oxalate, the ionization of calcium oxalate is suppressed due to the presence of common Ca+2 ion.

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Q.2 ix)The solubility of a sparingly soluble substance is calculated from the solubility product data. Explain.

Answer:
Solubility of a sparingly soluble salt can be calculated by using solubility product data. As we know that solubility product (Ksp) is the product of molar concentrations of ions in the saturated solution. Calculation of the concentration of ions using the formula of solubility product gives the solubility of the salt.
Example:
Solubility product of PbSO4 is 1.6 x 10-8.

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Long Questions Chemistry Class 11 Notes for kpk 2021

Q.3) Write down Kc for the following reversible reactions.

i) PCl5(g) ⇌  PCl3(g) + Cl2(g)
ii) 2HI(g)   ⇌ H2(g) + l2(g)
iii) H2(g) + CO2(g)  ⇌ CO(g) + H2O(g)
iv) 3Fe(g) + 4H2O(g)  ⇌  Fe3O4(g) + 4H2(g)
v) 2N2O5 (g) ⇌ 4NO2(g) + O2(g)

Answer:
i) PCl5(g) ⇌  PCl3(g) + Cl2(g)

The Kc expression for this reaction is written is,


Q.4) Predict the effect of increasing pressure on the following gaseous equilibrium.


i) N2(g) + O2(g) ⇌ 2NO(g)
ii) PCl5(g) ⇌  PCl3 + Cl2
iii) 2NO2(g) ⇌ N2O4(g)
iv) 2SO2(g) + O2(g) ⇌ 2SO3(g)

Answer:
i) N2(g) + O2(g) ⇌ 2NO(g)

As the number of moles of products are equal to those of reactants, it means the reaction occurs without any change in number of moles and hence volume also does not change. Increasing pressure will have no effect on the equilibrium state of this gaseous reaction.
ii) PCl5(g) ⇌  PCl3 + Cl2
As the number of moles of products are greater than those of reactants, it means the reaction occurs with an increase in volume. So if pressure is increased, the volume will decrease and the equilibrium will be shifted in the backward direction i.e. in the direction of decreased volume according to Le-Chatelier’s principle.
iii) 2NO2(g) ⇌ N2O4(g)
As the number of moles of product are smaller than those of reactants, it means the reaction occurs with a decrease in number of moles. So by increasing pressure, the volume will decrease and the equilibrium will be shifted in forward direction i.e. in the direction of decreased volume.
iv) 2SO2(g) + O2(g) ⇌ 2SO3(g)
As the number of moles of product are smaller than those of reactants, it means the reaction occurs with a decrease in number of moles. So by increasing pressure, the volume will decrease and the equilibrium will be shifted in forward direction i.e. in the direction of decreased volume.


Q.7) The solubility of CaF2 in water at 25°C is found to be 2.05 x 10-4M. What is the Ksp at this temperature?

Answer:
Solubility of CaF2 in water = 2.05 x 10-4 M
Solubility value suggests that 2.05 x 10-4 moles of CaF2 will completely dissolve in one dm3 solution to form ions.

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So the concentrations of ions in saturated solution will be,
[Ca+2] = 2.05 x 10-4 M    and,
[F] = 2(2.05 x 10-4) = 4.1 x 10-4 M
Using the formula for solubility product (Ksp),
Ksp = [Ca+2][F]2
Ksp = (2.05 x 10-4)( 4.1 x 10-4)2 = 3.446 x 10-11


Q.8) The solubility product of AgCl is 1.8 x 10-10. Calculate the molar solubility of the salt.

Answer:
Ksp = 1.8 x 10-10
Solubility of AgCl = [AgCl] =?

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Ksp = [Ag+][Cl] = 1.8 x 10-10
From the above chemical equation,
[Ag+] = [Cl]
Let [Ag+] = x mole m-3, then [Cl] = x mole dm-3
So,
Ksp = (x)(x) = 1.8 x 10-10
x2 = 1.8 x 10-10

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According to chemical equation, AgCl and Ag+ are equimolar. So, 1.34 x 10-5 moles of AgCl will completely dissolve in one dm3 solution to produce 1.34 x 10-5 Ag+ ions. Hence the solubility of AgCl is 1.34 x 10-5 M.
Solubility of AgCl = [AgCl] = 1.34 x 10-5 M


Q.9) A mixture of 0.5 moles of H2 and I2 each was placed in a one litre flask at 400°C. Calculate the concentration of H2, l2 and HI at equilibrium. The Kc for the reaction is 54.3.

Answer:
Initial conc. of H2 = 0.5 moles dm-3
Initial conc. of I2 = 0.5 moles dm-3
Kc = 54.3

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As x represents the concentration of H2 and I2 each, and their concentration cannot exceed the value of initial concentration. Hence the correct value is (x=0.393).
So,
[H2]eq = 0.5 – 0.393 = 0.107 M
[I2]eq = 0.5 – 0.393 = 0.107 M
[HI]eq = 2x = 2(0.393) = 0.786 M


Q.10) The equilibrium mixture contains 1 mole of PCI5, 0.3 mole of PCI3 and 0.08 mole of Cl2 in a 10 L flask. Calculate Kc.

Answer:
Moles of PCl5 = 1 mole
Moles of PCl3 = 0.3 mole
Moles of Cl2 = 0.08 mole
Volume of flask = 10 L
Kc=?
As we know that concentration is the ratio of moles and volume, so converting the moles of given components into concentrations,
[PCl5] = 1/10 = 0.1 M
[PCl3] = 0.3/10 = 0.03 M
[Cl2] = 0.08/10 = 0.008 M


Q.11) The equilibrium concentration of N2 and O2 is 0.25M each while its Kc is 0.1 at 2000°C. Calculate the equilibrium concentration of NO for the following reaction. 
N2(g) + O2(g) ⇌  2NO.

Answer:
Equilibrium conc. of N2 = [N2]eq = 0.25 M
Equilibrium conc. of O2 = [O2]eq = 0.25 M
Kc = 0.1 at 2000oC
Equilibrium conc. of NO = [NO]eq = ?
N2(g)  +  O2(g)  ⇌  2NO
Using the formula for equilibrium constant,

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Q.12 i) a. State the Law of Mass Action and derive the equilibrium constant expression for the given equation.


                            A + B  → C + D

Answer:
Law of Mass Action:

“The rate at which a substance reacts is proportional to its active mass and the rate of a chemical reaction is proportional to the product of the active masses of the reacting substances”.
Derivation of Equilibrium Constant Expression
Consider a general reaction;

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Where A and B are the reactants while C and D are the products. The equilibrium concentration in mol dm-3 of A, B, C and D are represented in square brackets like [A], [B], [C], [D] respectively. According to the law of mass action, the rate of forwarding reaction is proportional to the product of molar concentrations of A and B.
Rate of forwarding reaction ∝ [A][B]
Rate of forwarding reaction ∝ Kf[A][B]
Kf is proportionality constant known as the rate constant for the forward reaction. Since C and D are the reactants for a backward reaction so the rate of the reverse reaction is given by
Rate of reverse reaction ∝ [C][D]
Rate of reverse reaction ∝ Kr[C][D]
Where Kr is the rate constant for the backward reaction.
At equilibrium,
Rate of forwarding reaction = Rate of the reverse reaction
Therefore, Kf[A][B]= Kr[C][D]
On rearranging we get,

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As,
Kf/Kr=Kc
So we can write,
Kc=[C][D]/[A][B]
Where, Kc is the ratio of the rate constants (Kf/Kr). It is known as equilibrium constant of the reaction. Kc is defined as the ratio of the product of the molar concentration of the products to that of the reactants.


Q.12 i) b. Given the equilibrium concentrations of N2, O2 and NO as 0.05M, 0.05M and 5.5 x 10-4 respectively. Calculate Kc for the decomposition of NO at 1500°C for the reaction.

Answer:
2NO → N2 + O2
[N2] = 0.05 M
[O2] = 0.05 M
[NO] = 5.5 x 10-4
Kc = ?
Calculations:
Equilibrium constant expression for this reaction is written as,
Kc=[N2][O2]/[NO]
Putting the values in the above expression

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Q.12 ii) Calculate the mass of CH3COOC2H5 formed from 100g of C2H5OH and 100g of CH3COOH, Kc = 4.0.

Answer:
C2H5OH + CH3COOH ⟶ CH3COOC2H5 + H2
Mass of C2H5OH = 100 g
Mass of CH3COOH = 100 g
Kc = 4.0
Mass of CH3COOC2H5 = ?
Calculations:
Molar mass of C2H5OH = 2(12)+5(1)+16+1 = 46 g/mol
Moles of C2H5OH = 100/46 = 2.174 moles
Molar mass of CH3COOH = 12+3(1)+12+16+16+1 = 60 g/mol
Moles of CH3COOH = 100/60 = 1.67 moles
Molar mass of CH3COOC2H5 = 12+3(1)+12+16+16+2(12)+5(1) = 88 g/mol
Kc expression for this reaction is written as,

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Balanced chemical equation shows that CH3COOC2H5 and H2O will have same number of moles or concentration, so we can write,
[CH3COOC2H5] = [H2O] = x
Rearranging for [CH3COOC2H5] and putting the values,

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Q.12 iii) a. State Le-Chatelier’s principle and discuss its application to Ammonia system.  

Answer:
Le-Chatelier’s principle:
“If a system at equilibrium is subjected to a stress by a change in temperature, pressure or concentration, the system tends to adjust itself so as to minimize the effect of that change”.
Application of Le-Chatelier’s principle to Ammonia System:
Ammonia is synthesized by Haber’s process. The equilibrium equation for this process is as follows:
N2 + 3H2 ⇌ 2NH3 △H = -92.384 kJ mol-1

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Le-Chatelier’s principle is applied to this process in order to obtain greater yield of ammonia. The effect of change in concentration, temperature and pressure is discussed below.
i) Effect of Concentration:
If hydrogen or nitrogen is added to the equilibrium mixture then more of ammonia is formed according to Le-Chatelier’s principle. This is due to the fact that an increase in the concentration of N2 or H2 will result in shifting the equilibrium towards right. If ammonia is removed from the reaction chamber, the forward reaction must be favored. Hence, efficient removal of ammonia results in a greater yield of ammonia.


ii) Effect of Pressure:
According to the principle, an increase in pressure on this system in an equilibrium state will change the system in a direction in which the volume is decreased. Since one volume of nitrogen and three volumes of hydrogen react to form two volumes of ammonia or a decrease in volume takes place in the forward direction. Hence, the formation of ammonia will be favored by high pressure.


iii) Effect of Temperature:
The formation of ammonia from its elements is an exothermic reaction, and its reverse reaction is endothermic. Thus, the increase in temperature results in an increased dissociation of ammonia, and a decrease in temperature results in the increased production of ammonia. In practice, at lower temperatures the reaction proceeds rather slowly and equilibrium is attained in a long time. Hence, the synthesis of ammonia is carried out at 450oC in the presence of finely divided iron which acts as a catalyst and a high pressure of 200-500 atmospheres is used.

Q.12 iii) b.What is the effect of raising temperature on each of the following equilibria.


i) N2 + 3H2 ⇌  2NH+ ΔH = -ve
ii) 2NO + O2 ⇌  2NO2 + ΔH = -ve
iii) 2SO2 +O2 ⇌  2SO3 + ΔH = -ve
iv) N2 + O2 ⇌  2NO; ΔH = +ve

Answer:
i) N2 + 3H2 ⇌  2NH3 + ΔH = -ve
As the reaction is exothermic in forward direction it’s reveres i.e. dissociation of NH3 will be an endothermic reaction. Hence an increase in temperature will favor the endothermic reaction i.e. reverse reaction and ammonia yield will decrease.
ii) 2NO + O2 ⇌  2NO2 + ΔH = -ve
As the reaction is exothermic in forward direction it’s reveres i.e. dissociation of NO2 will be an endothermic reaction. Hence an increase in temperature will favor the endothermic reaction i.e. reverse reaction and NO2 yield will decrease.
iii) 2SO2 +O2 ⇌  2SO3 + ΔH = -ve
As the reaction is exothermic in forward direction it’s reveres i.e. dissociation of SO3 will be an endothermic reaction. Hence an increase in temperature will favor the endothermic reaction i.e. reverse reaction and SO3 yield will decrease.
iv) N2 + O2 ⇌ 2NO; ΔH = +ve
This reaction is endothermic in the forward direction and hence an increase in temperature will favor the forward reaction and the yield of NO will increase.

Q.12 iv) a. With the help of equilibrium constant expression, how will you predict the direction and the extent of a chemical reaction?

Answer:
Direction of Chemical Reaction:

The direction of a reaction at any time can be predicted by means of two parameters of reaction system, equilibrium constant Kc and reaction quotient Qc. Kc is the ratio of equilibrium concentrations of reactants and products while Qc is the ratio of concentrations of reactants and products at any time, t. In order to predict the direction of reaction, the value of Kc is compared with the value of Qc. Therefore,
i) If Qc=Kc, the reaction is at equilibrium.
ii) If Qc>Kc, the system is not at equilibrium and the net reaction will occur in the reverse direction until equilibrium is reached.
iii) If Qc


The extent of a Chemical Reaction:
The extent of the reaction can be found from the value of the equilibrium constant.
i) If the equilibrium constant Kc is very large, this indicates that the reaction is almost complete. The concentration of products is much greater than those of reactants at equilibrium.
ii) If the value of Kc is very small, then the reaction proceeds a little in the forward direction. Rather than reverse reaction is favoured and a small amount of product is formed at equilibrium.
iii) If the value of Kc is neither very small nor very large, this shows that both reactants and products are in appreciable quantities at equilibrium.

Q.12 iv) b.The equilibrium constant for the reaction, 2O3 ⇌ 3O2 is 1.0 x 1050 at 25°C. Predict the extent of the formation of O2 at room temperature.

Answer:
The extent of Formation of O2:

2O3 ⇌ 3O2
Since the value of Kc is 1.0 x 1050, which is very large for this system. It means the reaction is almost complete and equilibrium mixture at 25oC mostly contains O3 gas and a very little amount of O2 gas.


Read more: Stoichiometry Chemistry Class 11 Notes Chapter 1 for kpk