Explore topic-wise InterviewSolutions in Current Affairs.

(a) Equilibrium constant at a given temperature is the ratio of product of molar concentrations of the products to that of the reactants, each concentration term being raised to the respective stoichiometric coefficients in the balanced chemical equation. 

(b) 1. The value of equilibrium constant is independent of the initial concentrations of the reactants and products. 

2. Equilibrium constant is temperature dependent having one unique value for a particular reaction represented by a balanced equation at a given temperature.

1. Equilibrium in a system having more than one phase is called heterogeneous equilibrium 

2. Equilibrium between solid Ca(OH)₂ and its saturated solution:

Ca(OH)₂(s) + (aq) ⇌ Ca₂ +(aq) + 2OH^-(aq)

At equilibrium the reaction does not stop. Both forward and backward reactions are taking place at equal rates. Thus, at equilibrium two exactly opposite changes occur at the same rate. Hence, chemical equilibrium is dynamic in nature.

heterogeneous

1. No.

2. Catalyst does not affect the equilibrium composition of a reaction mixture. It does not appear in the balanced chemical equation or in the equilibrium constant expression. It only helps to attain the equilibrium state in a faster rate.

1. K_p is the equilibrium constant in terms of the partial pressures of the reactants and products (Pressure should be expressed in bar as standard state is 1 bar). It is used for reactions involving gases. 

2. K_p = K_c(RT)^∆n 

where R = universal gas constant, T = absolute temperature and ∆n = number of moles of gaseous product(s) – number of moles of gasesous reactant(s).

1. The mass of a gas dissolved in a given mass of a solvent at any temperature is directly proportional to the pressure of the gas above the solvent.

2. Equilibrium between the CO₂ molecules in the. gaseous state and the CO₂ molecules dissolved in water under pressure,

CO₂(g) ⇌ CO₂ (in solution)

(c) A new reaction path

(d) May increase or decrease depending upon whether reaction is exothermic or endothermic

1. When the rate of forward reaction is equal to rate of backward reaction, the chemical reaction is said to be in equilibrium. 

2. Rate of chemical reaction is directly proportional to the product of molar concentration of the reactants.

3.  K = \(\frac{[NH_3]HCl]}{[NH_4Cl]}\)

1. Yes.

2. Here the total number of moles of the reactants is equal to that of the products. Hence pressure is having no influence in this equilibrium.

1. The following factors can influence the equilibrium state of a system: 

• Change in concentration of the reactants or products.
• Change in temperature.
• Change in pressure.
• Addition of inert gas.
• Presence of catalyst.

2. N₂(g) + 3H₂(g) ⇌ 2NH₃ (g); ∆_rH = -91.8 kJ mol^-1

When concentration of N or H is increased, a good yield of NH can be achieved. The rate of forward reaction can also be increased by removing NH from the reaction mixture.

When pressure is increased, the system will try to decrease pressure and for this system will proceed in that direction where there is minimum number of moles i.e., forward reaction. Thus, a good yield of NH₃ can be achieved by increasing pressure.

Since the formation of NH₃ is an exothermic reaction, a good yield of NH₃ can be achieved by decreasing the temperature. But if the temperature is decreased to very low value the reactant molecules do not have sufficient energy to interact. Hence, an optimum temperature of 500°C is used.

1. On decreasing pressure the reaction shifts in the direction in which there is increase in the number of moles. Thus, the rate of backward reaction increases on decreasing pressure. 

2. On increasing temperature, the rate of endothermic reaction increases. Here, backward reaction is endothermic. Hence, on increasing temperature the rate of backward reaction increases. 

3. Hydrogen, being a reactant increase in its partial pressure increases the rate of forward reaction.

(a) (i) When temperature is increased, the rate of forward reaction decreases since it is exothermic. 

(ii) When pressure is decreased the rate of forward reaction decreases since it is associated with decrease in number of moles. 

(iii) When a catalyst V₂O₅ is added the rate of both forward and backward reactions are increased by the same extent and equilibrium is reached earlier.

(b) [H⁺] = 0.01M x 2 = 0.02M = 2 x 10^-2M

pH = -log[H⁺] = -log(2 x 10^-2) = 1.6990

[OH^-] = \(\frac{10^{-14}}{[H+]}\) = \(\frac{10^{-14}}{2\times10^{-2}}\) = 5 x 10^-13

The hydroxyl ion concentration in a solution having pH = 4 will be 10^-14.

1. In a salt solution, if we increase the concentration of any one of the ions, according to Le Chatelier’s principle, it should combine with the ion of its opposite charge and some of the salt will be precipitated till K_sp = Q_sp . Similarly, if the concentration of one of the ions is decreased, more salt will dissolve to increase the concentration of both the ions till K_sp = Q_sp .

2. The common ion effect is used for almost complete precipitation of a particular ion as its sparingly soluble salt, with very low value of solubility product for gravimetric estimation.

1. If ∆G is negative, then the reaction is spontaneous and proceeds in the forward direction. 

If ∆G is positive, then reaction is considered non- spontaneous. instead, as reverse reaction would have a negative ∆G, the products of the forward reaction shall be converted to the reactants. If ∆G is 0, reaction has achieved equilibrium. At this point, there is no longer any free energy to drive the reaction. 

2. ∆G^⊖ = -2.303RT log K

= -2.303 x 8.134 x 300 x log K

= -2.303 x 8.134 x 300 x log 8

= 5187.5 mol^-1

(i) K^'_c = \(\frac{1}{K_c}=\frac{1}{57}\) = 0.0175

(ii) K^'''_c = K²_c = (57)² = 3249

(ii) K^'''_c = (K_c)^1/2 = (57)^1/2 = 7.55

1. Amount of CO decreases, because the system favours the reaction in which number of moles decreases with increase of pressure i.e., backward reaction. 

2. Increases. 

3. Equilibrium constant remains the same since temperature is constant.

1. Henry’s law 

2. K_p =K_c (RT)^∆n 

∆n = 3.2 = 1

K_p = 0.027 × (0.0831 × 1000)¹ = 2.2437

Na₂CO₃ will give a basic solution.

(a) -log[H⁺] = 5 Or  log[H⁺] = -5

∴ [H⁺] = antilog(-5) = 10^-5

(b) BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄^2-(aq)

K_sp = [Ba²⁺][SO₄^2-] = S² 

∴ S = \(\sqrt{K_{sp}}\) = \(\sqrt{1.5\times10^{-9}}\) = 3.873 x 10^-5

1. The applications of equilibrium constant are: 

• To predict the extent of a reaction on the basis of its magnitude. 

• To predict the direction of the reaction. 

• To calculate equilibrium concentrations.

2. Reaction quotient, Q_c at a given temperature is defined as the ratio of the product of concentrations of the reaction products to that of the reactants, each concentration term being raised to their individual stoichiometric coefficients in the balanced chemical equation, where the concentrations are not necessarily equilibrium values.

3. (i) When Q_c > K_c , the reaction will proceed in the direction of reactants (reverse reaction), i.e., net reaction goes from right to left.

(ii) When Q_c < K_c , the reaction will proceed in the direction of products (forward reaction), i.e., net reaction goes from left to right. 

(iii) When Q_c = K_c , the reaction mixture will be at equilibrium, i.e., no net reaction occurs.

(2) and (4) are correct.

The expression for ostwald dilution law is K_a = Cα² .

Both chlorine and oxygen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller size and as a result, a higher electron density per unit volume.

Uses of ClO_2:
(i) It is used for purifying water.
(ii) It is used as a bleaching agent.

[Hint : Due to bigger size of Cl than N.]

Concept:

Properties of iota:

i² = -1

\(\rm \frac 1 i = \frac i {i^2} = \frac i {-1} = -i\)

Calculation:

We have, 

\(\rm iz^3+z^2-z+i=0\)

On dividing by i, we get

\(⇒ \rm z^3+\frac {z^2} {i}-\frac z i+1=0\)

\(⇒ \rm z^3-iz^2+iz+1=0\)              (∵ \(\rm \frac 1 i \)= -i)

\(⇒ \rm z^3-iz^2+iz-i^2=0\)             (∵ i² = -1)

\(⇒ \rm z^2(z-i)+i(z-i)=0\)

\(⇒ \rm (z^2+i)(z-i)=0\)

So, z = i or z² = -i.

Now, z = i ⇒ |z| = |i| 

⇒  |z| = 1

And, z² = -i 

⇒ |z²| = |-i| = 1

⇒  |z| = 1

Moles of Ethane = 60/30 = 2 moles.
Now, we know that:

1 mole of gas occupies 22.4 litres volume at STP.
So, 2 moles of Ethane will occupy :

\(22.4 \times 2=44.8\) litres

Now, 1 litre = 1 dm³ 
So, volume of Ethane = 44.8 dm³.

Molarity is the ratio of the moles of a solute to the total liters of a solution. The solution includes both the solute and the solvent. Molality, on the other hand, is the ratio of the moles of a solute to the kilograms of a solvent.

The correct option is (c) NH₃.

NH₃ is better electron donor, and the lone pair is present in sp₃ hybrid orbital.

Micro-animals are animals so small that they can be visually observed only under a microscope.Microscopic arthropods, including dust mites, spider mites, and some crustaceans such as copepods and certain cladocera.

Tardigrades are one of nature's smallest animals. They are never more than 1.5 mm long, and can only be seen with a microscope. They are commonly known as "water bears".

Eusporangium: The sporangium develops from a GROUP of INITIAL cells and such a development is called development.
Leptosporangium: The sporangium develops from a SINGLE INITIAL cell and such a development is called Leptosporangiate development.