Chemical Equilibrium

By Nitesh Jain (B.tech, IIT Bombay)

Chemical Equilibrium and Equilibrium Constant

For a general reaction: 𝒂𝑨 + 𝒃𝑩 ⇌ 𝒄𝑪 + 𝒅𝑫 NOTE: Reactions are generally

reversible i.e. can proceed both ways

A reaction attains equilibrium when

𝑟𝑎𝑡𝑒 𝑜𝑓 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑟𝑓 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑏𝑎𝑐𝑘𝑤𝑎𝑟𝑑 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 (𝑟𝑏)

represents concentration

𝑘𝑓 𝑎𝑛𝑑 𝑘𝑏 are rate constants of forward and backward reactions respectively

From Law of Mass Action

𝒌𝒇[𝑨]𝒂[𝑩]𝒃= 𝒌𝒃[𝑪]𝒄[𝑫]𝒅 are rate constants of forward and backward reactions respectively

𝒌𝒇

𝒌𝒃

=[𝑪]𝒄[𝑫]𝒅

[𝑨]𝒂. [𝑩]𝒃

𝒌𝒇

𝒌𝒃

= 𝑲(𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)

𝑟𝑓 = 𝑘𝑓[𝐴]𝑎[𝐵]𝑏

𝑟𝑏 = 𝑘𝑏[𝐶]𝑐[𝐷]𝑑

Dynamic Nature of Chemical Equilibrium

NOTE: Chemical equilibrium is dynamic as, individual molecules are continually reacting,

even though overall composition of reaction mixture doesn’t change

For reaction: 𝑎𝐴 + 𝑏𝐵 ⇌ 𝑐𝐶 + 𝑑𝐷

𝐊 𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 concentration is called 𝑲𝑪 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝐾 =[𝐶]𝑐[𝐷]𝑑

[𝐴]𝑎[𝐵]𝑏

Equilibrium is established

𝑐𝑜𝑛

𝑐𝑒𝑛

𝑡𝑟𝑎

𝑡𝑖𝑜

𝑛𝑠

Time

[𝐶] and [𝐷]

𝐴 𝑎𝑛𝑑[𝐵]

Homogenous Gaseous Reaction

Let 𝒑𝒙 represent partial pressure of specie 𝑥 at equilibrium

𝑲𝒑 =𝒑𝑴

𝒎𝒑𝑵𝒏

𝒑𝑨𝒂𝒑𝑩

𝒃

Here, 𝐾𝑃 is known as pressure equilibrium constant

𝒑𝒙 =𝒏𝒙

𝒏𝒕𝒐𝒕𝒂𝒍

× 𝑷 𝑃 = total Pressure

𝑛𝑡𝑜𝑡𝑎𝑙 = 𝑡𝑜𝑡𝑎𝑙 #𝑜𝑓 𝑚𝑜𝑙𝑒𝑠

𝑎𝐴 + 𝑏𝐵 ⇌ 𝑚𝑀 + 𝑛𝑁 All 𝐴, 𝐵, 𝑀 𝑎𝑛𝑑 𝑁 are gases

then

Relation between Kp and KC

𝒂𝑨 + 𝒃𝑩 ⇌ 𝒎𝑴 + 𝒏𝑵

𝑲𝑪 can also be written for the reaction, but for the above reaction calculating. 𝑲𝒑 is easier

∆𝑛 = no. of moles of product − no. of moles of reactant 𝑲𝒑 = 𝑲𝑪 𝑹. 𝑻 ∆𝒏

Example: ∆𝑛 = 𝑚 + 𝑛 − 𝑎 + 𝑏 , 𝑓𝑜𝑟 𝑎𝑏𝑜𝑣𝑒 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛

Stoichiometric coefficients after balancing reaction

Both 𝐾𝑃 and 𝐾𝐶 given same information about the state of equilibrium

Equilibrium constant depend only on temperature

Units of 𝐾𝑝 and 𝐾𝑐 are not fixed and depend on stoichiometry of the reaction

Mole Fraction Equilibrium Constant (𝑲𝒙)

𝑓𝑜𝑟 𝑎 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛: 𝑎𝐴 + 𝑏𝐵 ⇌ 𝑚𝑀 + 𝑛𝑁

Mole fraction of specie 𝑍 =𝑛𝑧(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑍)

𝑛𝑡𝑜𝑡𝑎𝑙 (𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠)= 𝑥𝑧(𝑑𝑖𝑒𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑙𝑒𝑠𝑠)

𝐾𝑥 =𝒙𝑴

𝒎𝒙𝑵𝒏

𝒙𝑨𝒂𝒙𝑩

𝒃 𝐾𝑥 does not have any unit

∆𝑛 = no. of moles of product − no. of moles of reactant 𝑲𝒑 = 𝑲𝒙. 𝒑∆𝒏

= 𝑲𝑪(𝑹𝑻)∆𝒏

NOTE: 𝐾𝑝 = 𝐾𝐶 = 𝐾𝑥, 𝑓𝑜𝑟 ∆𝑛 = 0

𝐾𝑥 may depend 𝑜𝑛 𝑃 𝑜𝑟 𝑉, unlike 𝑲𝒑 𝒂𝒏𝒅 𝑲𝑪

Thermodynamic Definition of Equilibrium Constant

It involves activities of reactants and products, rather than their concentration, by definition

∴ Thermodynamic equilibrium constant has no unit

NOTE: For any pure solid or pure liquid, activity= 𝟏

Activity of a component (𝑿)

NOTE: It is a

dimensionless quantity

𝑋 (𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑋)

1𝑀

𝑝𝑥(𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝑋)

1 𝑎𝑡𝑚

Calculation of Degree of Dissociation

Consider a reaction: in a container of 𝑽𝒐𝒍𝒖𝒎𝒆 𝑽

𝑎𝐴 + 𝑏𝐵 ⇌ 𝑐𝐶 + 𝑑𝐷

initially

At equilibrium

Total number of moles at equilibrium = 𝑐𝛼 + 𝑑𝛼 + 𝑎0 − 𝑎0𝛼 + 𝑏0 − 𝑏0𝛼

= 𝑎0 + 𝑏0 + 𝛼(𝑐 + 𝑑 − 𝑎 − 𝑏)

= 𝑎0 + 𝑏0 + 𝛼(∆𝑛)

𝛼 is called degree of dissociation, which is defined per mole of limiting reactant

(assumed 𝐴 in example)

This expression derived for a general reaction would be helpful in calculating 𝜶 𝒐𝒓 𝒌

𝛼 is always less than 1

for equilibrium reactions 𝑎0 𝑏0

𝑐𝛼 𝑑𝛼 𝑎0(1 − 𝛼) 𝑏0(1 − 𝛼)

0 0

Degree of Dissociation from Density

If one mole of substance dissociates into 𝑛 moles, into container of Volume 𝑉

NOTE: The equation is not applicable to reactants in which

# 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔 = # 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔

NOTE: If all 𝑑’𝑠 are taken as vapor densities,

𝑑𝑡 =𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡

2

𝒙 =𝑴 − 𝒎

𝒏 − 𝟏 𝒎

𝑴 Initial molecular mass

𝒎 --Molecular mass of equilibrium

Reaction: 𝑨 ⇌ 𝒏. 𝑩

Initially moles 1 0

At equilibrium (1 − 𝑥) 𝑛. 𝑥

𝑑𝑡 ∝1

𝑉

𝑑0 ∝ 1

𝑉[1 + 𝑛 − 1 𝑥]

Degree of dissociation 𝑥 =𝑑𝑡 − 𝑑0

𝑛 − 1 𝑑0

𝑑𝑡 = theoretical density (assuming no dissociation)

𝑑0 = observed/ experimental density Both under same pressure

Pure Solids Involved in Chemical Reaction

Concentrations of pure solids involved in a chemical reaction, are assumed to remain constant

Example: 𝐶𝑎𝐶𝑂3 𝑠 ⇌ 𝐶𝑎𝑂 𝑠 + 𝐶𝑂2 (𝑔)

𝐾 =𝐶𝑎𝑂 [𝐶𝑂2]

[𝐶𝑎𝐶𝑂3]

𝐾 =𝑥1[𝐶𝑂2]

𝑥2

𝐾′ = [𝐶𝑂2] NOTE: 𝐾′ = 𝐾

𝑥2

𝑥1

becomes the

equilibrium constant of the reaction

𝐶𝑎𝑂 Constant [𝑥1]

[𝐶𝑎𝐶𝑂3] Constant [𝑥2]

Relation Between Equilibrium Constants of Forward & Backward Reactions

At equilibrium : Forward Reaction

For Backward Reaction:

𝒂𝑨 + 𝒃𝑩 → 𝒄𝑪 + 𝒅𝑫

For a reaction: 𝒂𝑨 + 𝒃𝑩 ⇌ 𝒄𝑪 + 𝒅𝑫

𝑲𝒇 =[𝑪]𝒄[𝑫]𝒅

[𝑨]𝒂. [𝑩]𝒃

𝒄𝑪 + 𝒅𝑫 → 𝒂𝑨 + 𝒃𝑩

𝑲𝒃 =[𝑨]𝒂[𝑩]𝒃

[𝑪]𝒄. [𝑫]𝒅

From 1 and 2

𝑲𝒇. 𝑲𝒃 = 𝟏 𝐾𝑓 =1

𝐾𝑏

i.e. 𝐾𝑓 and 𝐾𝑏 are reciprocal of each other

Rate of forward

reaction

Rate of

backward

reaction

rate

time

Equilibrium

state

1

2

Multiplication of Chemical Equation by Certain Factor

Multiplying the chemical equation by a factor n

New equilibrium constant will be old equilibrium constant

raised to a power equal to the multiplies factor 𝑲′ = 𝑲𝒏

Consider a reaction: 𝑨 + 𝟐𝑩 ⇌ 𝟑𝑪 + 𝟒𝑫

𝐾 =[𝑪]𝟑[𝑫]𝟒

[𝑨]𝟏[𝑩]𝟐

𝑛. 𝐴 + 2. 𝑛. 𝐵 ⇌ 3. 𝑛. 𝐶 + 4. 𝑛. 𝐷

𝐾′ =𝑪 𝟑𝒏 𝑫 𝟒𝒏

𝑨 𝒏. 𝑩 𝟐𝒏 = (𝑲)𝒏

Reactions Taking Place in more than One Step

Equilibrium constant of overall reaction is equal to the product

of equilibrium constant of each other step

This reaction takes place in steps as:

a) 𝐻2𝑂2 + 𝐼− ⇌ 𝑂𝐻− + 𝐻𝑂𝐼 ;

b) 𝐻+ + 𝑂𝐻− ⇌ 𝐻2𝑂;

c) 𝐻𝑂𝐼 + 𝐻+ + 𝐼− ⇌ 𝐼2 + 𝐻2𝑂 ;

𝑲𝟏. 𝑲𝟐. 𝑲𝟑 = 𝑲

Example: For reaction

𝐾 =𝐼2 [𝐻2𝑂]2

𝐻2𝑂2 [𝐻+]2[𝐼−]2 𝐻2𝑂2 + 2𝐻+ + 2𝐼− ⇋ 𝐼2 + 2𝐻2𝑂

𝐾1 =𝑂𝐻− [𝐻𝑂𝐼]

𝐻2𝑂2 [𝐼−]

𝐾2 =[𝐻20]

𝐻+ [𝑂𝐻−]

𝐾3 =𝐻20 [𝐼2]

[𝐻𝑂𝐼] 𝐻+ [𝐼−]

Reaction Quotient (𝑄)

For a reaction: 𝒂𝑨 + 𝒃𝑩 ⇌ 𝒎𝑴 + 𝒏𝑵

𝑄 > 𝐾: Reaction proceeds in backward direction

𝑄 =[𝑴]𝒎[𝑵]𝒏

[𝑨]𝒂. [𝑩]𝒃

Q tells Whether the reaction is at equilibrium or not

Direction of reaction when equilibrium constant of reaction is known

Not necessarily equilibrium concentration

𝐾 =[𝑴]𝒎[𝑵]𝒏

[𝑨]𝒂. [𝑩]𝒃,

𝑄 = 𝐾: Reaction is at equilibrium

𝑄 < 𝐾: Reaction proceeds in Forward Direction

Equilibrium concentrations

1

2

3

Le Chatelier’s Principle

Effect of change on a reaction at equilibrium, shifts the reaction in a

direction that minimizes that change

These changes can be change in temperature, pressure or concentration

These changes cannot change the equilibrium constant 𝐾𝑃 𝑜𝑟 𝐾𝐶 , except change

in temperature

Effects of Temperature on Equilibrium Constant

log 𝐾𝑝2 − log𝐾𝑝1 =∆𝐻°

2.303. 𝑅

𝑇2 − 𝑇1

𝑇1. 𝑇2

, 𝑇2 > 𝑇1

For an exothermic reaction, increase in temperature results in shifting reaction towards

endothermic path i.e. from right to left or towards reactants AND vice-versa for endothermic

reaction

NOTE: No shift in equilibrium for ∆𝐻 = 0

𝐾𝑝1 𝑎𝑛𝑑 𝐾𝑝2 are equilibrium constants at temperatures 𝑇1 𝑎𝑛𝑑 𝑇2 respectively

∆𝐻° is standard heat of reaction at constant pressure [assumed constant for small (𝑇2 − 𝑇1)]

𝐾𝑝2 > 𝐾𝑝1,𝑖𝑓 ∆𝐻° 𝒊𝒔 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 → endothermic reaction

𝐾𝑝2 < 𝐾𝑝1,𝑖𝑓 ∆𝐻° 𝒊𝒔 𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆 → exothermic reaction

Van’t Hoff equation:

Le Chatelier’s Principle: Effect of Pressure

At constant 𝑻, 𝑽 of reaction No. of moles ∝ 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒

𝒂𝑨 𝒈 + 𝒃𝑩 𝒈 ⇌ 𝒄𝑪 𝒈 + 𝒅𝑫(𝒈)

𝑲𝒑 =𝒑𝑪

𝒄. 𝒑𝑫𝒅

𝒑𝑨𝒂. 𝒑𝑩

𝒃 𝑃𝑥 = 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝑥

=𝒏𝑪

𝒄𝒏𝑫𝒅

𝒏𝑨𝒂𝒏𝑩

𝒃 ×

𝑷

𝒏𝒕𝒐𝒕𝒂𝒍

∆𝒏

From 1, where

𝑛𝑥 = # 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑋 𝑎𝑟 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚

1

∆𝑛 = 𝑐 + 𝑑 − (𝑎 + 𝑏) And from above, 𝐾𝑝 ∝ 𝑃 ∆𝑛

If 𝑃 is increased

NOTE: ∆𝑛 = 0 ⇒ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑖𝑠 𝑛𝑜𝑡 𝑎𝑓𝑓𝑒𝑐𝑡𝑒𝑑

∆𝑛 = +𝑣𝑒 ⇒ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑠ℎ𝑖𝑓𝑡𝑠 𝒕𝒐𝒘𝒂𝒓𝒅𝒔 𝒍𝒆𝒇𝒕 ( to increase 𝑎 + 𝑏)

∆𝑛 = −𝑣𝑒 ⇒ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑠ℎ𝑖𝑓𝑡𝑠 𝒕𝒐𝒘𝒂𝒓𝒅𝒔 𝒓𝒊𝒈𝒉𝒕 ( to increase 𝑐 + 𝑑)

For a reaction :

Le Chatelier’s Principle: Effect of Addition of Inert Gas

CASE 1: At constant Volume

Equilibrium is not affected at all

CASE 2: At constant Pressure

𝑎) ∆𝑛 = 0 ⇒ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑖𝑠 𝑛𝑜𝑡 𝑎𝑓𝑓𝑒𝑐𝑡𝑒𝑑

𝑏) ∆𝒏 = +𝒗𝒆 ⇒ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑖𝑠 𝒔𝒉𝒊𝒇𝒕𝒆𝒅 𝒕𝒐 𝒕𝒉𝒆 𝒓𝒊𝒈𝒉𝒕 (𝑡𝑜𝑤𝑎𝑟𝑑𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠)

𝑐) ∆𝒏 = −𝒗𝒆 ⇒ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑖𝑠 𝒔𝒉𝒊𝒇𝒕𝒆𝒅 𝒕𝒐 𝒍𝒆𝒇𝒕( 𝑡𝑜𝑤𝑎𝑟𝑑𝑠 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)

Le Chatelier’s Principle: Effect of Addition of Catalyst

Addition of catalyst changes the rate of reaction

Catalyst affects the activation energy (𝐸𝐴) of both forward and backward reactions equally

NOTE: Both reaction Quotient 𝑸 𝒂𝒏𝒅 𝑲 are not affected due to addition of catalyst.

But time requires to establish equilibrium is altered

𝑲𝒄 =𝑲𝒇

𝑲𝒃

=𝑲𝒇′

𝑲𝒃′

Both rate constants are affected equally

If 𝑲𝒇 𝒂𝒏𝒅 𝑲𝒃 are rate constants of forward and backward reactions respectively

Le Chatelier’s Principle: Effect of removal/addition of Reactants/Products

On addition or removal of reactants/products at equilibrium

Equilibrium shifts in direction that consumes or produces

that reactant(s)/product(s)

NOTE: This is valid only if volume of system is kept constant

If volume of system is not constant but at constant pressure

→ 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠ℎ𝑖𝑓𝑡𝑖𝑛𝑔 𝑜𝑓 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛 ∆𝑛

Relation between Free Energy Change and K

Standard conditions imply 1𝑀 or 1 𝑎𝑡𝑚 of reactants converting to 1𝑀 or 1 𝑎𝑡𝑚 of products

NOTE: In case all gaseous

reactants & products

𝐾 = 𝐾𝑝

In case of all solution

reactants & products,

𝐾 = 𝐾𝑐

In case of mixture of solution & gaseous

reactants and products , 𝐾 represents

thermodynamic equilibrium constant

∆𝐺 = ∆𝐺° + 𝑅. 𝑇. ln 𝑄 ∆𝐺° is standard free energy change for a

reaction and ∆𝐺 is free energy change

At equilibrium, ∆𝐺 = 0 and 𝑄 = 𝐾 ∴ 𝐹𝑟𝑜𝑚 1 ⇒ ∆𝐺° = −𝑅. 𝑇. ln 𝐾

= −2.303 𝑅. 𝑇. 𝑙𝑜𝑔𝐾

1

Feasibility of Reaction from Free Energy Change

∆𝐺° = −𝑅. 𝑇. ln 𝐾

∆𝐺° < 0 ⇒ 𝐾 > 1 𝑖. 𝑒. 𝑎 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑒𝑎𝑠𝑖𝑏𝑙𝑒

NOTE: For a spontaneous or feasible reaction ∆𝐺° < 0

∆𝐺° > 0 ⇒ 𝐾 < 1 𝑖. 𝑒. 𝑎 𝑟𝑒𝑣𝑒𝑟𝑠𝑒 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑒𝑎𝑠𝑖𝑏𝑙𝑒

∆𝐺° = 0 𝑜𝑟 𝐾 = 1 𝑖. 𝑒. 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑎𝑡 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚

1

2

3

Calculation of ∆𝐺° and ∆𝑆° from K

NOTE: Curve between log 𝐾 and log1

𝑇 gives a straight line

From Thermodynamics: ∆𝐺° = ∆𝐻° − T∆𝑆°

Knowing ∆𝐻° and ∆𝑆°, ∆𝐺° can be calculated from 1

∆𝑮° = −𝑹. 𝑻. ln 𝐾

From 1 and 2 𝐥𝐨𝐠 𝑲 =−∆𝑯°

𝟐. 𝟑𝟎𝟑𝑹.𝟏

𝑻+

∆𝑺°

𝟐. 𝟑𝟎𝟑𝑹

1

2

1

𝑇

log 𝐾

𝑠𝑙𝑜𝑝𝑒 =−∆𝐻°

2.303𝑅

intercept

=∆𝑆°

2.303𝑅

Calculation Kp and Kc

Consider a reaction: in a container of 𝑉𝑜𝑙𝑢𝑚𝑒 𝑉

𝒂𝑨 + 𝒃𝑩 ⇌ 𝒄𝑪 + 𝒅𝑫

initially

At equilibrium

𝑎0 𝑏0

𝑐𝛼 𝑑𝛼 𝑎0(1 − 𝛼) 𝑏0(1 − 𝛼)

0 0

General expression and can be used

directly for finding out 𝑲𝒄

NOTE: 𝐾𝑝 = 𝐾𝑐 𝑅. 𝑇 ∆𝑛, 𝐾𝑝 can also be calculated, if 𝛼 is known

𝛼 is degree of dissociation

𝑤. 𝑟. 𝑡 limiting reagent 𝐴

𝑲𝒄 =[𝑪]𝒆𝒒𝒖𝒊

𝒄[𝑫]𝒆𝒒𝒖𝒊𝒅

[𝑨]𝒆𝒒𝒖𝒊𝒂. 𝑩 𝒆𝒒𝒖𝒊

𝒃

𝑲 =

𝑪𝜶𝑽

𝒄

.𝒅𝜶𝑽

𝒅

𝒂𝟎 − 𝒂𝟎𝜶𝑽

𝒂.

𝒃𝟎 − 𝒃𝟎𝜶𝑽

𝒃

=𝐶𝑐. 𝑑𝑑

𝑎0𝑎 . 𝑏0

𝑏 ×𝛼𝑐+𝑑

(1 − 𝛼)𝑎+𝑏 ×1

𝑉∆𝑛 Where ∆𝑛 = 𝑐 + 𝑑 − (𝑎 + 𝑏)