KPK 11th Class Chemistry Chapter 7 Chemical Equilibrium Short Questions Answers

Effect of temperature on the formation of No Consider the reaction

N2(g)+O2(g)+ heat ⇆ 2NO(g)

The reaction is endothermic in the forward direction and exothermic in the reverse direction. Thus increase in temperature will favour the formation of No.

Practically a high temperature of 3273K is used for the preparation of No on commercial scale.

SO3: Consider the reaction

2SO2(g)+O2g  ⇆ 2SO3(g)∆H=-94.58Kj/mol

The reaction is exothermic in the forward direction and endothermic in the reverse direction. Thus the decrease in temperature will favour formation of SO3. If we keep the temperature high. It will favour the reverse reaction and the value will fall due the decrease in the formation of SO3.

When a liquid is placed in a closed container. The liquid will evaporate with the increase in number of molecules in the gaseous state. They collide with the molecules on the liquid surface and return to it. After sometime the rates at which the molecules leave and return to the liquid phase become equal and equilibrium is established in the system. At equilibrium

The molecules in the vapour phase exert pressure in a similar manner as a gas world so. This pressure of vapours in equilibrium with its liquid phase is called vapour pressure of the liquid. As the evaporation and condensation process remain continued. Therefore, the equilibrium between evaporation of liquid and condensation of vapours is called dynamic equilibrium and non static.

A + B⇆C + D

[C]  [D]

________  = Kc
[A]   [B]

Let the reaction be completed 50%. It means 50% will be the concentration 0 the reactants and 50%  that of the product. Then the above equation will become

.                 [C]  [D]         (50%) (50%)
Kc = ______________ – ______________ = 1
.                 [A] [B]          (50%)  (50%)

When the concentration of one of the substance is changed, the system no longer remains in equilibrium. According to Le ___ Chatelier’s principle the system will shift so as to restore the constant value of kc. For example

aA + bB ⇆ cC + dD

.             [C]c [D]d
Kc = ______________
.              [A]a [B]b

If the concentration of A or B is increased, then their concentrations will increase. Thus the rate of forward will increase to decrease the concentration of A and B and to increase the concentration of C and D to keep the concentration ratio equal to kc

.           [C]c [D]d
Kc=  ______________
.           [A]a   [B]b

When a sparingly soluble salt such as silver chloride (AgCl) is placed in water. It dissolves until saturated solution is formed and equilibrium is established between the resulting saturated solution of its ions and un dissolved solid phase. The equilibrium condition is given by equation:

AgCl(s) ⇆ Ag+ (aq)     + Cl-(aq)

The equilibrium constant for this dissociation reaction is given as:
.        [Ag+] [Cl-]

Kc = ____________
.           [AgCl]

As the salt is sparingly soluble, the concentration of (AgCl(s)) does not change much. It is thus considered to be constant. Thus

[AgCl(s)]  = K

Substituting this “K” in the equilibrium constant expression gives

Kc.K = [Ag+]  [Cl-]

Kc.K =  Ksp =   [Ag+]  [Cl-]   =  1.8  × 10-10 at 25̊C

The product of the two constants can be expressed by a new constant called the solubility product. In the other words, the equilibrium constant for a solid electrolyte in equilibrium with its ions in solution is called the solubility product.