Le chatelier s principle , Factors affecting chemical equilibrium

Le chatelier’ s principle

The change in any condition of a system under equilibrium such as (concentration ,pressure or temperature ) the system is activated to direction which decrease or cancels the effect of change

Factors affecting chemical equilibrium

(reversible reaction)

1) Concentration

2) temperature

c)pressure

Equilibrium constant:

equilibrium constant

1)Effect of change in concentration on equilibrium system:

CO_(g) + 3H_2(g) ↔ CH_4(g) + H₂O_(v)

Increase the concentration of reactants(CO Or H2) the reaction direction is Forward reaction

Decreasing the concentration of reactants(CO or H2) the reaction direction is backward reaction

Increasing the concentration of of products(CH4 or H2O) the reaction direction is backward reaction

Decreasing the concentration of one of products(CH₄ or H₂O) the reaction direction is Forward reaction

Example :

Write the equilibrium constant equation of the decomposition of N₂O₅:

2N₂O_5(g) ↔ 4NO_2(g)+O_{2 (g)}

K_c =[NO₂][O₂] / [N₂O₅]

Notes:

-The concentration of pure water and the solid substance and the precipitated must not be written in the equilibrium constant equation GR?

-because their concentration remains constant

Example: write the equilibrium constant of:

Ca(HCO₃)_2(aq) ↔ CaCO_3(s) + CO_2(g) +H₂O_(L)

K_c= [CO₂] / [Ca(HCO₃)]

-The numerical value of the equilibrium constant K_c is not changes by changing the concentration of reactants and products at the same temperature.

-The value of K_c is changed by changing the temperature

Example: write the equilibrium constant equation of:

2H₂O_(v) ↔ 2H_2(g) + O_2(g)

NH_3(g)+HCl_(g) ↔NH₄Cl_(S)

Answer:

Kc= [H₂]² [O₂] / [H₂O]

Kc = 1/ [NH₃] [HCl]

Example: calculate N₂O₄ gas concentration in the following equation:

N₂O_4(g) ↔2NO_2(g)

K_c=4.81 x10^-5

Knowing that NO₂ gas concentration at equilibrium =0.0032 M

Answer:

K_c= [NO₂] ² / [N₂O₄]

4.81×10^-5= [0.0032]²/[N₂O₄]

= 0.213 M

Notes:

1)if K_c>1 the forward reaction is the predominant GR?

2)If K_c<1 The backward reaction is predominant

Example:

AgCl_{(s) ↔}Ag⁺_(aq) + Cl^–_(aq)

K_c=1.7 x10^-10

-It is difficult for AgCl to dissolve in water according this equation GR?

-because the value of K_c<1 so the reaction is backward (predominant).

Example:

H_2(g) + I_{2(g) ↔}2HI

1)calculate K_cknowing that at equilibrium the concentration of each Iodine and hydrogen =0.221 m and the concentration of hydrogen iodide = 1.563

2)To which direction is the reaction activated (forward or backward) GR?

Answer:

K_c= [HI]² / [H₂][I₂] = (1.563)² /0.221×0.221 = 50
The reaction is activated to the forward because K_c >1

Graph of Equilibrium Constant

Relation between concentration and time

The relation between rate of reaction and time

Example:

H_2(g)+ I_2(g) ↔ 2HI K_c=55.16 at 425c

If concentration of H2 =1×10^-3

I₂=1.5×10^-4 and HI=5×10^-3

Is this reaction at equilibrium ?

K_c= [HI]² /[H₂][I₂]

K_c=(5×10^-3)2 / 1×10^-3 x1.5 x10^-3

K_c=16.63

The reaction is not at equilibrium because the theoretical value Kc(16.67) not equal the actual value of Kc(55.16).

Example: at equilibrium

N_2(g)+ 3 H₂ ↔2NH_3(g)

The mixture volume =1 liter containing 0.3mol of N₂ and 0.2 mol of H₂ and 0.6 mol of ammonia gas NH₃

Calculate the K_c

Answer: concentration M=no. of moles /volume(l)

The concentration of gases =their number of moles

K_c = [NH₃]² /[N₂] [H₂]³

K_c =(0.6)² /0.3 x (0.2)³

K_c =150

2)Effect of changing in temperature on reversible system:

Reactions can be classified into:

1)Endothermic

Reactants+ heat ↔ products ΔH=-(negative value )

by increasing temperature the reaction moves( ⇒ ) forward so Kc increases

by decreasing temperature the reaction moves backward ( ⇐ ) so Kc increases

2)exothermic

reactants ↔ products +heat ΔH=+ (positive value)

by increasing temperature the reaction moves( ⇐ ) backward so K_c decreases

by decreasing temperature the reaction moves forward (⇒) so K_c increases

Example:

4HCl +O₂ ⇔2H₂O +2Cl₂+113 Kj

Colorless colorless colorless greenish yellow

What is the effect of the following on reaction color and K_c value :

1)increasing O₂ gas at the same temperature

2)decreasing the temperature

Answer:

1)increasing O₂ : the reaction moves forward and more of Cl₂ is formed and so more intense of greenish yellow color

-K_c is constant because the temperature not changed

2)decreasing temperature : in exothermic reaction moves forward so greenish yellow become more intense

-The equilibrium K_c will increase

Example:

In which of reaction does the rate of decomposition increase by raising temperature GR?

1)SO₃ ⇔ SO_2(g) + 1/2 O₂ H=⁺

2)N₂H_2(g) ⇔ N₂+2H₂ H= ^–

Answer:

-Because it is an endothermic that moves forward direction by increasing temperature

Example:

H₂ +I₂ ⇔ 2HI k_c=50 at 448 c

H₂ +I₂ ⇔ 2HI K_c=67 at 850 c

Is the reaction endothermic or exothermic GR?

The reaction is endothermic because by rising temperature the reaction moves forward

3)Effect of changing in pressure on equilibrium system:

When reactants and products are gases the equilibrium is measured by partial pressure of gases K_p

aA_(g) + bB_(g) ⇔ cC_(g) +dD_(g)

K_p= (Pc)^c (PD)^d / (PA)^a (PB)^b

Example:

N_2(g) +3H₂ ⇔2NH_3(g)

K_p=(P_NH3)² /(P_H2)(P_I2)

Example: write the balanced equation of the following K_p:

K_p= (P_HI)² /(P_H2) (P_I2)

H_2(g)+ I_2(g) ⇔ 2HI_(g)

Example: calculate the partial pressure of H₂ gas and Cl₂ :

2HCl_(g)⇔ H_2(g) +Cl_2(g)

Knowing that partial pressure of HCl is =7 atm

Answer:

K_p=(P_H2) (P_Cl2) / (P_HCl)

(P_H2) (P_CL2) = Kp(P_HCl)²

=2.45×10^-7 X (7)²

=1.2×10^-5 atm

The no. of moles of H₂= the no of moles of Cl₂

So their partial pressure is equal

P_H2 OR P_Cl2 = square root of 1.2×10^-5

Test yourself

[1]Write the equilibrium constant expression for each reaction.

$2 S O_{2 (g)} + O_{2 (g)} ⇌ 2 S O_{3 (g)}$

$N_{2} O_{(g)} + \frac{1}{2} O_{2 (g)} ⇌ 2 N O_{(g)}$

$C u_{(s)} + 2 A g_{(a q)}^{+} ⇌ C u_{(a q)}^{+ 2} + 2 A g_{(s)}$

$C a C O_{3 (g)} ⇌ C a C O_{(s)} + C O_{2 (g)}$

$2 N a H C O_{3 (s)} ⇌ N a_{2} C O_{3 (s)} + C O_{2 (g)} + H_{2} O_{(g)}$

[2] Calculate is the Kc of the following reaction?

with concentration

$S O_{2 (g)} = 0.2 M O_{2 (g)} = 0.5 M S O_{3 (g)} = 0.7 M$

answer :Kc: 24.5

[3]For the same reaction, the differing concentrations:

SO_2(g)=0.1M ,

O_2(g)=0.3 and MSO_3(g)=0.5M

What is the direction of the reaction (forward or backward)

Answers:Q1

$K_{c} = \frac{[S O_{3}]^{2}}{[O_{2}] [S O_{2}]^{2}}$

$K c = \frac{[N O]^{2}}{[O_{2}]^{0.5} [N_{2} O]}$

$K c = \frac{[C u^{+ 2}]}{[A g^{+}]^{2}}$

$K c = \frac{[C O_{2}]}{[C a C O_{3}]}$

$K_{c} = [H_{2} O] [C O_{2}]$

2 S O 2 (g) + O 2 (g) ⇌ 2 S O 3 (g) (21)

Chemistry For Third Secondary

Curriculum

Le chatelier s principle , Factors affecting chemical equilibrium