Chemical equilibrium is briefly discussed with following topics:
Free energy change in a chemical reaction. Thermodynamic derivation of the law of chemical equilibrium.
Definition of ΔG and ΔG◦
Le Chatelier’s principle.
Relationships between Kp, Kc and Kx
HMCS Vancouver Pre-Deployment Brief - May 2024 (Web Version).pptx
F y b. sc. chemical equilibria
1. Syllabus of B.Sc. WITH CHEMISTRY
SEMESTER- II
CORE COURSE: DSC-2B
(6 credits: Theory-04, Practicals-02)
(Physical Chemistry & Organic Chemistry)
CHEMICAL EQUILIBRIA
1
2. Chemical Equilibrium
• Free energy change in a chemical reaction.
Thermodynamic derivation of the law of
chemical equilibrium.
• Definition of ΔG and ΔG◦
• Le Chatelier’s principle.
• Relationships between Kp, Kc and Kx
2
3. Free energy change in a chemical
reaction.
3
Reactant
Product
FreeEnergy
Reaction coordinates
Minimum free energy Equilibrium
-ve free energy change
For a spontaneous reaction
4. Concentration
Time
Equilibrium
Chemical equilibrium constant (K)
4
A + B C + D
𝐾 =
𝑘1
𝑘2
=
[𝑝𝑟𝑜𝑑𝑢𝑐𝑡]
[𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡]
𝐾 =
𝐶 [𝐷]
𝐴 [𝐵]
k1
k2
Chemical Equilibrium is a state of
reversible reaction when the two
opposing reactions occur at the same
rate and the concentrations of
reactants and product do not change
with time.
Product
Reactant
Catalyst do not change equilibrium
conditions, but it helps to attain it
faster.
6. Thermodynamic derivation of the law of chemical
equilibrium
6
aA + bB cC + dD
k1
k2
Consider a general reaction
Equilibrium constant (K) in terms of activity is given by 𝑲 =
𝒂 𝑪
𝒄
× 𝒂 𝑫
𝒅
𝒂 𝑨
𝒂
× 𝒂 𝑩
𝒃
The chemical potential (μ)of substance in a mixture is related to activity by the
following equation
𝝁 = 𝝁 ◦ + 𝑹𝑻 𝐥𝐧 𝒂
Where, μ ◦ is chemical potential of
pure substance, R is gas constant and
T is absolute temperature.
7. Thermodynamic derivation ….
7
𝑎μ 𝐴 = 𝑎(μ ◦ 𝐴 + 𝑅𝑇 ln 𝑎 𝐴)
For a moles of substance A equation can be written as
bμ 𝐵 = 𝑏(μ ◦ 𝐵 + 𝑅𝑇 ln 𝑎 𝐵)
cμ 𝐶 = 𝑐(μ ◦ 𝐶 + 𝑅𝑇 ln 𝑎 𝐶)
𝑑μ 𝐷 = 𝑑(μ ◦ 𝐷 + 𝑅𝑇 ln 𝑎 𝐷)
The change in free energy (ΔG) for the reaction is
∆𝐺 = ∆𝐺 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 − ∆𝐺 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡
∆𝐺 = 𝑑μ 𝐷 + cμ 𝐶 − 𝑎μ 𝐴 + bμ 𝐵
∆𝐺 = {𝑑(μ ◦ 𝐷 + 𝑅𝑇 ln 𝑎 𝐷)} + {𝑐(μ ◦ 𝐶 + 𝑅𝑇 ln 𝑎 𝐶)} − {𝑎(𝜇 ◦ 𝐴 + 𝑅𝑇 ln 𝑎 𝐴)} + {𝑏(𝜇 ◦ 𝐵 + 𝑅𝑇 ln 𝑎 𝐵)}
∆𝐺 = [ 𝑐μ ◦ 𝐶 + 𝑑μ ◦ 𝐷 − 𝑎μ ◦ 𝐴 + 𝑏μ ◦ 𝐵 ] + 𝑅𝑇𝑙𝑛
𝑎 𝐶
𝑐
× 𝑎 𝐷
𝑑
𝑎 𝐴
𝑎
× 𝑎 𝐵
𝑏
As ∆𝐺° = 𝑐μ ◦ 𝐶 + 𝑑μ ◦ 𝐷 − 𝑎μ ◦ 𝐴 + 𝑏μ ◦ 𝐵
∆𝐺 = ∆𝐺° + 𝑅𝑇𝑙𝑛
𝑎 𝐶
𝑐
× 𝑎 𝐷
𝑑
𝑎 𝐴
𝑎
× 𝑎 𝐵
𝑏
At equilibrium change in Gibbs free energy is zero
∆𝐺° = −𝑅𝑇𝑙𝑛
𝑎 𝐶
𝑐×𝑎 𝐷
𝑑
𝑎 𝐴
𝑎×𝑎 𝐵
𝑏
= −𝑅𝑇𝑙𝑛 K
Similarly for substance B, C, and D equations can be written as follows
8. Thermodynamic derivation ….
8
Thus change in free energy (ΔG) in terms of log for the reaction can be written as
∆𝐺° = −2.303𝑅𝑇𝑙𝑜𝑔 K
As R and T are constant we can infer 3 cases
If ∆𝑮° is negative log K must be positive and reaction proceeds spontaneously as
product of concentration terms is more than product of reactant concentration
terms.
If ∆𝑮° is positive log K must be negative and reaction does proceeds
spontaneously as product of concentration terms is less than product of reactant
concentration terms.
If ∆𝑮° is zero log K must be 0 and reaction has reached equilibrium as product of
concentration terms is equal to product of reactant concentration terms.
9. Definition of ΔG and ΔG◦
G = H – TS
H= enthalpy; T= temperature; S= entropy
Change Gibbs free energy in system at constant temperature is
ΔG = ΔH – TΔS
At standard state conditions Change standard Gibbs free energy is
ΔG◦ = ΔH◦ – TΔS◦
Criteria for spontaneity is ΔG<0, for that
Change in enthalpy should be lower
or
Change in entropy should be high 9
Gibbs free energy is a thermodynamic quantity that is a difference between
enthalpy and product of absolute temperature and entropy of the system.
Standard Gibbs free energy is a thermodynamic quantity that is a difference
between enthalpy and product of absolute temperature and entropy of the
system at standard state conditions.
10. Le Chatelier’s principle
Henry Le Chatelier (1884 )- French Chemist
“When a stress is applied on a system in
equilibrium, the system tends to adjust itself so
as to reduce the stress.”
Stress can be introduced in a reaction at
equilibrium by changing
A) Concentration B) Pressure C) Temperature
10
11. Le Chatelier’s principle- Concentration
11
2NH3(g)N2(g)+ 3H2(g)
Given equilibrium
conditions for
above reaction
Concentration of
N2 = 0.399 M
H2 = 1.197 M
NH3 = 0.202 M
New Equilibrium
conditions for above
reaction
Concentration of
N2 = 1.348 M
H2 = 1.044 M
NH3 = 0.304 M
1 mol/L N2 is
added
12. Le Chatelier’s principle- Pressure
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2NO2(g)N2O4(g)
Reactant side 4 moles and product side 2 moles
Increase in pressure will increase the amount of product and there will be shift in
equilibrium from left to right side.
2NO(g)N2(g)+ O2(g)
Reactant side 2 moles and product side 2 moles
Change in pressure will not result in change in equilibrium
2NO2(g)N2O4(g)
Reactant side 1 moles and product side 2 moles
Increase in pressure will decrease the amount of product and there will be shift in
equilibrium from right to left side.
13. Le Chatelier’s principle- Temperature
13
2NO2(g)N2O4(g) + 14 kcal
Endothermic reaction
Increase in temperature will increase the amount of product and there will be
shift in equilibrium from left to right side.
2HCl(g) + 44.2 kcalH2(g)+ Cl2(g)
Exothermic reaction
Increase in temperature will decrease the amount of product and there will be
shift in equilibrium from right to left side.
2NH3(g) + 22.2 kcalN2(g)+ 3H2(g)
450 °C
200 atm
Synthesis of ammonia is an exothermic process. Why high temperature
is employed in synthesis?
15. 𝐾 𝑐, 𝐾 𝑝 and 𝐾 𝑥
15
aA + bB cC + dD
k1
k2
𝐾 𝑐 =
C 𝑐 𝐷 𝑑
A 𝑎 B 𝑏
𝐾 𝑝 =
pCc
pD d
pA a
pBb =
C RT c
D RT d
A RT a
B RT b =
C c
D d
A a
B b
RT c
RT d
RT a
RT b = Kc
RT c+d
RT a−b
𝑲 𝒑 = 𝑲 𝒄 (RT)(Δn); (c+d)–(a+b) is difference in number of moles (Δn)
𝐾 𝑥 =
𝑥C 𝑐
𝑥𝐷 𝑑
𝑥A 𝑎
𝑥B 𝑏
p =(n/v) RT
Where, n/v is molar concentration
pA= A RT; pB=B RT; pC=C RT and pD=D RT
16. 𝐾 𝑐, 𝐾 𝑝 and 𝐾 𝑥
16
𝐾 𝑥 =
𝑥C 𝑐
𝑥𝐷 𝑑
𝑥A 𝑎
𝑥B 𝑏
𝐾 𝑝 =
𝑝C 𝑐
𝑝𝐷 𝑑
𝑝A 𝑎
𝑝B 𝑏 =
𝑥C P 𝑐
𝑥D P 𝑑
𝑥A P 𝑎
𝑥B P 𝑏 =
𝑥C 𝑐
𝑥𝐷 𝑑
𝑥A 𝑎
𝑥B 𝑏
𝑃 𝑐
P 𝑑
𝑃 𝑎
𝑃 𝑏
𝐾 𝑝 = 𝐾 𝑥
P 𝑐+ 𝑑
P 𝑎− 𝑏 = 𝐾 𝑥 P(Δn); (c+d)–(a+b) is difference in number of moles(Δn)
𝑲 𝒑 = 𝑲 𝒙 P(Δn)
Partial pressure = 𝑥 X Pressure
pA= 𝑥A P; pB= 𝑥B P; pC= 𝑥C P and pD= 𝑥D P
18. 𝑲 𝒄, 𝑲 𝒑 and 𝑲 𝒙 problem for understanding: At 500◦C the
reaction between nitrogen and hydrogen to form
ammonia has Kc= 6.0 X10-2. what is the numerical value of
Kp for the reaction.
18
2NH3(g)N2(g)+ 3H2(g)
𝑲 𝒑 = 𝑲 𝒄 (RT)(Δn)
Δn = number of moles of product – number of moles of reactants
Δn = 2 - 4= -2
T= 500 + 273 = 773 K
R = 0.0821 L atm mol-1 K-1
𝐾 𝑝 = 6.0 X10−2(0.0821 × 773)(−2)
𝑲 𝒑 = 𝟏. 𝟓 𝑿 10-5
