KPK Board 10th class Chemistry Ch 1 Chemical Equilibrium short questions answers

The state of reversible reaction during which the rate of forward reaction becomes equal to the rate of reverse reaction is called equilibrium state. Examples:

H2 + I2 -> 2HI

N2 + O2 -> 2NO

Dynamic equilibrium state of a chemical reaction can be identified by knowing the concentrations of reactants and products at regular intervals of time. When the concentration of reactants and production are observed constant, them the reaction is said to be in dynamic equilibrium state.

Forward Reaction	Reverse Reaction
It is a reaction in which reactants react to form products.	It is a reaction in which products react to form the reactants.
It takes place from left to right.	It taken place from right to left.
At the beginning the rate of reaction is very fast.	At the beginning the rate of reverse reaction is negligible [zero].
With the passage of time, it shown down gradually.	With the passage of time, it speeds up gradually.

To predict the direction of a chemical reaction ist we should calculate reaction quotient Qc, by putting the values of concentration of each reacting substance.

Qc= [products]/[Reactants]

Then the value of Qc is compared with the known value of Kc. Its Qc value is equal to Kc value, the system is in equilibrium. If Qc > Kc = The reaction will proceed in the reverse direction to reach equilibrium. If Qc < Kc; The reaction will proceed in the forward direction to reach equilibrium.

Active mass means molar concentration of reacting substances. According to the law of Mass action. “The rate of a chemical reaction is directly proportional to the active masses [ molar concentrations] of the reacting substances” it means that higher the active masses of the reacting substances. Faster will be the rate of a chemical reaction and smaller the active masses of the reacting substances, slower will be the rate of a chemical reaction.

Solution:

N2 + 3H2 2NH3

Kc = [NH3 ]2/[N2][H2 ]3

Kc = ((0.113)2)/((0.602) (0.420)3 )

Kc = (0.113)/(0.602)(0.0740) = 0.012769/0.044548 = ( 0.286 mol(-3)dm6)

Chemical for Equilibrium:

The following condition are necessary for the system to be in equilibrium state.

1.  Closed container.
2.  The concentration of reactants and products.
3.  The temperature of the system.
4. The pressure of the system.
5.  The volume of the system.
6. Catalyst, if used in the system, remain unchanged.

Ans.

1) N2 + 3H2 2NH3 Kc = [NH3 ]2 / [N2 ][H2 ]3

2) 2H2 + O2 2H2 O
Kc = [H2 O]2 /[H2 ]2 [O2]

3) 4NH(3 )+ 5O(2 ) 4NO + 6H(2 ) O
Kc = [NO]4 [H2 O]6/[NH3 ]4 [O2 ]5

Solution: 2SO2 + O(2 ) 2SO3
Kc = [So3]2 / [So2]2 [O2]
Kc = (3.50)2/(1.50)2 (1.25) = 7/ (3)(1.25) = 7 / 3.75 = ( 1.86 mol(-1).dm^(-3))

Effect of temperature on equilibrium state: According to le-chatelier’s principle; It temperature of the reaction increases the equilibrium shifts in the endothermic direction i.e. The direction in which heat is absorbed in an attempt to lower the temperature. An increase in temperature favors the endothermic reaction and decrease in temperature favors the exothermic reaction. Consider the reversible reaction; 2CO + O2 2CO2 = ∆H = – 41.84 Kj/mole. Since the reaction is exothermic in the forward direction and endothermic in the reverse direction so an increase in temperature will shift the equilibrium in reverse direction. At high tap endothermic reaction will proceed in the forward direction. At law tap exothermic reaction will proceed in the forward direction.