11th Class Chemistry Chapter 8 Chemical Equilibrium Short Question Answers
11th Class Chemistry Chapter 8 Chemical Equilibrium Short Question Answers below
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1.What is the meant by stage of chemical equilibrium?
In reversible chemical reactions,two opposing reactions occue. A stage reaches for the reaction when the rates of two opposing reactions are equal. This stage is called stage of chemical equilibrium.
2. Justify that chemical equilibrium is dynamic in nature?
In reversible chemical reactions, the molecules of reactants collide and convert into products. At the same time the molecules of the products are converting to the reactants. When two opposing forces maintain the equal rates then equilibrium is there and that is dynamic equilibrium in nature.
3. The rates of chemical reaction depends upon the concentration of reactants. Why?
This is according to the law of mass action. The rates of chemical reactions are directly proportional to the product of the active masses of reactants. Greater the number of molecules greater the collisions and greater the rate.
4. Why is the equilibrium established, only when the reaction vessel is closed for a gaseous reaction?
If some of the products or the reactants in the open vessel leave the vessel during the progress of the reaction then the stage of equal rates is disturbed and equilibrium cannot be established. In such situation the reaction is pushed to the forward direction or backward direction.
5. Why the equilibrium constant value has its units for some of the reversible reactions, bur has no unit for some other reactions?
If the number of moles of reactants and products are equal in a reversible balanced equation, then the units are cancelled and the value of Kc has no units.
H2 + l2⇋ 2Hl Kc = [Hl]2/[H2][l2]
If the number of mole of reactants and products are unequal then Kc has net units
PCl3 ⇋ PCl3 + Cl2 Kc = [cons]+1
N2 + 3H2 ⇋ 2NH3 Kc = [conc]-2
H2 + l2⇋ 2Hl Kc = [Hl]2/[H2][l2]
If the number of mole of reactants and products are unequal then Kc has net units
PCl3 ⇋ PCl3 + Cl2 Kc = [cons]+1
N2 + 3H2 ⇋ 2NH3 Kc = [conc]-2
6. When four types of chemical equilibrium constants for a reaction become equal? How Kp and Kc are related?
When we have such reversible reaction, in which the total number of moles of reactants and products are equal, then Kp = Kx = Kn. In other words, Δn = 0, where Δn is the deference of number of moles of reactants and products.
In the following equation Kp and Kc are related by
Kp = Kc (RT)Δn
7. What happens to the directions of a reversible reaction, when the ratio of the concentration is less than actual Kc?
When the ratio of the concentration for a reversible reaction is less than Kc then it means that the reaction is not at equilibrium stage. It has to go to the forward direction to attain the actual value of Kc.
8. Why the factor of volume cancelled out in the final expression of Kc, when the number of moles of reactants and products are equal?
If the number of reactants and products are equal, then the factors of volume cancel each other and the factor of volume does not control the Kc value and also the equilibrium stage of the reaction.
H2 + l2 ⇋ 2Hl Kc = 4x2/(a-x)(b-x)
H2 + l2 ⇋ 2Hl Kc = 4x2/(a-x)(b-x)
9. Why the factor volume in Kc expression goes to the denominator, when the number of moles of products is greater than the reactants for a reversible reaction?
10. Those gaseous reactions which happen with the increase of volume go to the backward direction, when the volume is decreased. Why?
When such reactions proceed and attain the equilibrium position, then the volume of the gaseous mixtyre of reactants and products increases. Decrease of volume at equilibrium stage, make the reaction to go to that side where the volume is less, and that is side of the reactants.
11. How does the equilibrium constant of a chemical reaction tells us the direction of a chemical reaction?
When the concentrations of a reversible reaction are measured at any stage of its progress and equilibrium constant is calculated, then it is called calculated Q. If actual Kc is also known, then there are following three possibilities.
(i). Q < Kc. The reaction goes to the forward direction.
(ii). Q < Kc. The reaction goes to the backward direction.
(iii). Q < Kc
(i). Q < Kc. The reaction goes to the forward direction.
(ii). Q < Kc. The reaction goes to the backward direction.
(iii). Q < Kc
12. The change of volume of pressure disturbs the equilibrium position for some of the gaseous phase’s reactions but not the equilibrium constant. Why?
Those gaseous phase reversible reactions, which happen with changing number off the moles, are affected by the change of value at equilibrium stage. Their equilibrium position is disturbed, but equilibrium constant is not changed.
13. Solid ice at 00C can be melted by applying pressure without supply of heat from outside. Why?
When pressure is applied to the broken pieces of ice at 00C, then according to Le- chatelier’s principle, the ice moves to that direction where its volume should decrease i.e., towards liquid water. Actually ice occupies more value than liquid water.
14. Why are exothermoic reactions favoured to the forward direction by cooling, but reverse is true for endothermic reactions?
The account of heat which is being evolved by the exothermic reaction is take up by that body which has a cooling effect. So, the reactions move to that direction where is there is less energy. The endothermic reaction require extra energy to take place. If the system is cooled. It will go to that direction, where is a less energy and that is the backward direction of reaction.
15. The change of temperature disturbs the equilibrium position and the equilibrium constant of reaction. Justify?
All the reversible reaction are disturbed by changing their equilibrium position equilibrium constant by disturbing the temperature Kc. Actually, change of temperature changes the energy constant of reaction reactants and product.
16. What will be effect of change in pressure on NH3 and SO3 synthesis.
N2 + 3H2 ⇋ 2NH3
N2 + 3H2 ⇋ 2NH3
This is a gaseous reaction having less number of moles of products. So this reaction happens with the decrease of volume. The increase of pressure will shift the equilibrium position of reaction to the forward direction and greater amount of NH3,will be produced. Equilibrium constant does not change. the same principle of pressure is applied.
In 2O2 + O2 ⇋ 2SO3, the same principle of pressure is applicable.
In 2O2 + O2 ⇋ 2SO3, the same principle of pressure is applicable.
17 .What is the effect of rise in temperature on the solubility of Kl in water?
The heat of solution of Kl in water is positive. It means it is an endothermic process. When the temperature is decreased, it will absorb more heat and more dissolution will take place.
18. Why the solubility of gaseous in water is increased by increasing the temperature?
The solubility of glucose in water is a endothermic process. Increase of temperature pushes the system to that side where solubility increases.
19. The NH3 synthesis by Heber’s process is an exothermic reaction. It should be favored at low temperature, but the optimum temperature is 4000C. Why?
If low temperature is maintained for NH3 synthesis, then the number of collision per unit time decrease and the rate of reaction become slow. In order to increase the rate of reaction, temperature has to be increased.
20. How does a catalyst affect a reversible reaction?
A catalyst affects the rates of both steps equally, so the equilibrium position remains the same. It means that equilibrium constant should also be the same. Actually, a catalyst decreases the energy of activation of the chemical reaction by giving a new path to the reaction. In this way, a greater % age of reactant molecules is able to cross the energy barrier and the rate of reaction increases.
21. Why the ionic product of water (Kw) increases with the increase of temperature?
The value of Kw is 0.11 * 10-14 at 00C and is 250C. The reason is that the increase temperature increases the kinetic energy of water and possibility of bond breakage increases.
22. What is the concentration of hydroxide ion in a solution whose pH is 10?
pH = -log [H+]
-log [H+] = +pH
log [H+] = -pH = -10
[H+] = 10 -10
Since [H+] [OH] = 10-10 So [OH–] = 10-4
-log [H+] = +pH
log [H+] = -pH = -10
[H+] = 10 -10
Since [H+] [OH] = 10-10 So [OH–] = 10-4
23. Why the sum of pH and pOH of any aqueous solution is always equal to 14, i.e. pkw at 250C?
The [H+] = [OH–] = 10– 7 moles dm-3 for pure H2O at 250C. When certain quantity of an acid or a base is added to that, then the [H+]and [OH–] do not remain equal, but there is product is always 10-14 at 250C.
So pH + pOH 14= pKw
When temperature of the aqueous solution is above 250C then
pH + pOH < 14
pH + pOH = pKw
So pH + pOH 14= pKw
When temperature of the aqueous solution is above 250C then
pH + pOH < 14
pH + pOH = pKw
24. Calculate the pH of 10-4 mol dm-3 of Ba(OH)2?
Ba(OH)2 ⇋ Ba2+ + 2OH–
[OH–] = 2* 10-4 mol dm-3 pH -log2 * 10-4
[OH–] = 2* 10-4 mol dm-3 pH -log2 * 10-4
25.Why HCl acts as a weak acid in ethanoic acid as compared to, when dissolved in water?
When HCl is dissolved in water, it dissociates to a great extent, releases sufficient H+ ions and acts as a strong acid. HCl dissolved in ethanoic acid dissociates to a less extent. Ethanoic is itself acid and decreases the dissociation of HCl by common ion effect.
26. Define the effect of common ion on solubility? Give examples.
According to Le- Chatlier’s principle, if a common ion is added in a solution then the solubility of the electrolyte decreases and the solute is compelled to settle down. If Cl– are added in saturated solution of NaCl , then NaCl settle down, because its solubility decreases.
27. By diluting the solution of CH3COOH, the % age ionization changes, but the dissociation constant of the acid remains the same at a constant temperature. How?
CH3COOH is a very weak acid. Its dissociated power is small. When it is 0.1 M, it is 1.3 % dissociated, and 0.01 M is 4.17 % dissociated. It means the diluting the solution ten times, the % age dissociation increases three times. But the Ka value of the CH3COOH does not change by changing the morality. The reason is that when the concentration of CH3COOH in water is greater, then the ratio of dissociated to undisssociated molecules remains the same.
28. Why the strong acids have weak conjugate bases, and weak acids have strong conjugate bases?
According to protonic concept, acids are those species which are proton donors or have a tendency to donate a porton.
CH3COOH + H2O ⇋ CH3COOH– + H3O+
weak acid base conjugate conjugate
base acid
CH3COOH has a little tendency to break the oxygen hydrogen bond and to act as an acid. It means that CH3COOH– ion should have a strong tendency to react with a proton. So CH3COOH– should be a strong base. HCl is a very strong acid and so Cl– ion should have a least tendency to accept proton to act as a base.
29. How NaCl can be purified by common in effect?
The impure sample of NaCl is dissolved in H2O to prepare the saturated solution’
NaCl(s) ⇋ Na+ (aq) + Cl–(aq)
If HCl gas in passed in saturated solution of NaCl, the Cl– ions are generated in express in the solution
HCl(aq) ⇋ H+(aq) + Cl–(aq)
Due to excess or Cl–, the ionization of NaCl is suppressed and NaCl settles down in the form of precipitate.
Na+(aq)+ Cl–(aq) ⇋ NaCl(s) (ionization is suppressed)
Moreover the use of HCl in ll- group basic radicals is for common ion effect.
NaCl(s) ⇋ Na+ (aq) + Cl–(aq)
If HCl gas in passed in saturated solution of NaCl, the Cl– ions are generated in express in the solution
HCl(aq) ⇋ H+(aq) + Cl–(aq)
Due to excess or Cl–, the ionization of NaCl is suppressed and NaCl settles down in the form of precipitate.
Na+(aq)+ Cl–(aq) ⇋ NaCl(s) (ionization is suppressed)
Moreover the use of HCl in ll- group basic radicals is for common ion effect.
30. What is a solution?
A solution which resists the change of pH, when a small amount of an acid or a base is added in that, Buffers are prepared by mixing two components:
(i). Weak acid + salt with a strong base.
(ii). Weak base + salt with a strong acid.
(i). Weak acid + salt with a strong base.
(ii). Weak base + salt with a strong acid.
31. What is Henderson equation?
It is equation which is used for a preparation of a buffer solution of required pH, We have to adjust the ratio of the concentration of salt to acid or base. The acid of suitable pKa value has to be taken
pH = pKa + log [salt]/[Acid] and POH = pKb + log [Salt]/[Base]
pH = pKa + log [salt]/[Acid] and POH = pKb + log [Salt]/[Base]
32. What is buffer capacity?
Buffer capacity is a ability if a butter to resists the change of pH, when a few drops of an acid and base is added from outside. The best buffer is obtained (i) when pH of the butter is equal to pKa of the acid and the concentration of the salt and acid are equal (ii) The pOH of the butter is equal to pKb of the base and concentration of both components are equal.
33. By adjusting the ratio of salt and acid, we can adjust the pH of required buffer solution. How?
If we want to prepare an acidic buffer solution, the following equation is used
pH = pKa + log [salt]/[acidΔ
The pKa of the acid, which is being used for the preparation of this buffer solution is a constant quantity at a given temperature. By adjusting the ratio of salt and acid concentration, we can prepare a solution having pH of own choice.
pH = pKa + log [salt]/[acidΔ
The pKa of the acid, which is being used for the preparation of this buffer solution is a constant quantity at a given temperature. By adjusting the ratio of salt and acid concentration, we can prepare a solution having pH of own choice.
34. When the concentration of the salt is increased in an acidic buffer, then the pH of the solution increases. Why?
Henderson’s equation is
pH = pKa + log [Salt]/[acid]
According to this equation, when the [salt] is greater than [acid], then the long term becomes bigger and the pH becomes higher. In other words, the solution obtained is of lower acid strength and of higher pH.
pH = pKa + log [Salt]/[acid]
According to this equation, when the [salt] is greater than [acid], then the long term becomes bigger and the pH becomes higher. In other words, the solution obtained is of lower acid strength and of higher pH.
35. Why do we need buffer solution?
Buffer solution is used in much industrial process as electroplating, manufacture of leather, photographic materials. Buffer solution is used by analytical chemists and to cal liberate pH meter. They are also used in culture media.
36. Why HCl is added before passing H2S gas in second group basic radical analysis?
HCl is a strong acid, gives sufficient H+ ions in solution, suppresses the ionization of H2S and gives less S-2 ions is solution. These less ions are sufficient to give the ppt, of second group.
37. How does a buffer act? Give example if acidic buffer?
Let us consider the buffer solution consisted of CH3COOH and CH3COONa. They are dissociated in water. Sodium acetate being a very strong electrolyte as compared to acetic acid furnishes sufficient CH3COO–ion as compared to CH3COOH
CH3COOH ⇋ CH3COO–(aq) + H+(aq)
CH3COONa⇋ CH3COO–(aq) + Na+(aq)
When a few drops of an acid , say HCl are added I this solution, the H+ ions provided by HCl are taken up by CH3COO– (mostly obtained from CH3COONa), so incoming protons are consumed and pH is retained. When a few drops of a base say NaOH is added from outside, then the protons already present in the solution are consumed. To compensate those protons, there happens a further dissociation of CH3COOh and pH is retained.
CH3COOH ⇋ CH3COO–(aq) + H+(aq)
CH3COONa⇋ CH3COO–(aq) + Na+(aq)
When a few drops of an acid , say HCl are added I this solution, the H+ ions provided by HCl are taken up by CH3COO– (mostly obtained from CH3COONa), so incoming protons are consumed and pH is retained. When a few drops of a base say NaOH is added from outside, then the protons already present in the solution are consumed. To compensate those protons, there happens a further dissociation of CH3COOh and pH is retained.
38. Prove that pKa + pKb = 14.
pKa + pKb = 14
pKa + pKb = 14
Since Ka *Kb = kw
Take the log of equation
log Ka + Kb = log Kw
Log Ka + log Kb = log Kw
Multiply it with negative sign
-log Ka – log Kb = -logKw
pKa + pKb = pKw
Since pKw= 14, of 250C so the pKa and pKb of the conjugate acid base pair has a very simple relationship with each other.
Take the log of equation
log Ka + Kb = log Kw
Log Ka + log Kb = log Kw
Multiply it with negative sign
-log Ka – log Kb = -logKw
pKa + pKb = pKw
Since pKw= 14, of 250C so the pKa and pKb of the conjugate acid base pair has a very simple relationship with each other.
39. Calculate the pH of 10-4 moles dm-3 solution of HCl?
HCl ⇋ H+ + Cl–
10-4 mole dm-3 ⇋ 0+0 t = 0
0 ⇋ 10-4 mol dm-3 t= Eq]
So, [H+] = 10-4
-log[H+] = pH = 4
40. A mixture of NH4OH and NH4Cl give n=basic buffer. Justify?
The mixture of NH4ClOH and NH4Cl gives a buffer solution. The equation are;
NH4OH ⇋ NH4– + OH–
NH4Cl ⇋ NH4+ + Cl
Since OH– ions are produced in the solution pH will be more than 7, and solution will be basic.
NH4OH ⇋ NH4– + OH–
NH4Cl ⇋ NH4+ + Cl
Since OH– ions are produced in the solution pH will be more than 7, and solution will be basic.
41. Give application of solubility product?
(i). If [M+][X–] is less than Ksp, then solution is unsaturated.
(ii). If [M+][X–] = Ksp, solution is saturated.
(iii). If [M+][X–] > Ksp , solution is momentarily supersaturated.
(ii). If [M+][X–] = Ksp, solution is saturated.
(iii). If [M+][X–] > Ksp , solution is momentarily supersaturated.
42. Give units of Kc for NH3 synthesis?
N2(g) + 3H2(g) ⇋ 2NH3(g)
Kc = [NH3][N2][H2]3 = (mol dm-3)-2
(mol dm-3)(mol dm-3)3 (mol dm-3)-2
43. What are solubility product expressions of PbCl2, Ag2CrO4, AnXm?
Ksp = [Pb2+] [Cl–]3 Ksp = [Ag+][CrO42-] Ksp = [A+m]n[X+n]
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