At any instant, the rate of a chemical reaction, at a given temperature is directly proportional to the product of the active masses of the reactants at that instant.
Rate of the reaction \(\alpha \) [Reactant]^x

\( \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})} \)
\(\mathrm{K}_{\mathrm{c}}=\frac{4 x^{2}}{(a-x)(b-x)}\)
This expression doesn't involve, V. So, increase or decrease of volume will not affect the equilibrium and hence the yield of the product.

This condition is not static and is dynamic, because both the forward and reverse reactions are still occurring with the same rate. No macroscopic change is observed.

(i) A system which can exchange both matter and energy with its surroundings is called an open system.
(ii) Hot water contained in an open beaker is an example for open system.
(iii) In this system, both water vapour and heat is transferred to the surroundings through the imaginary boundary.
(iv) All living things are open systems because they continuously exchange matter and energy with the surroundings.

(i) After freezing, the molecules attain an ordered state and therefore, entropy decreases.
(ii) At 0 K the constituent particles are in static form therefore, entropy is minimum. If the temperature.is raised to 115 K particles begin to move and entropy increases.
(iii) Reactant, NaHCO₃ is solid. Thus, its entropy is less in comparison to product which has high entropy.
(iv) Here, one molecule gives two atoms. Thus, number of particles increases and this leads to more disordered form.

A reaction that does occur under the given set of conditions is called a spontaneous reaction.
Example:
1. A waterfall runs downhill, but never up, spontaneously.
2. Heat flows from hotter object to a colder one.
3. Ageing process.

Zeroth law of thermodynamics states that 'If two systems at different temperatures are separately in thermal equilibrium with a third one, then they tend to be in thermal equilibrium with themselves'.

\(q_{rev}= (-\Delta_f H^{\ominus})=-286 kJ\ mol^{-1}=286000\ J\ mol^{-1}\)
\(\Delta{S}{_{(Surroundmgs)}}=\frac{q_{rev}}{T}=\frac{286000\ J\ mol^{-1}}{298\ K}\) = 959 J K^-1 mol^-1.

ΔG⁰ = - 2.303 RT log k_c
= - 2.303 x 8.314 x 298 log (1.8 x 10^-7)
= - 38484 J mol^-1
= - 38.484 kJ mol-1

State Function: Enthalpy, entropy, temperature, free energy
Path Function: Heat, work.

(i) The third law of thermodynamics states that the entropy of pure crystalline substance at absolute zero is zero.
(ii) It can also be stated as it is impossible to lower the temperature of an object to absolute zero in a finite number of steps
(iii) Mathematically, \(\lim _{ T\rightarrow 0 }{ S=0 } \) for a perfectly ordered crystalline state.

The heat of combustion of a substance is defined as "The change in enthalpy of a system when one mole of the substance is completely burnt in excess of air or oxygen". It is denoted by ΔH_C

Gibbs free energy is defined as G = H - TS

(i) non-spontaneous
(ii) non-spontaneous
(iii) spontaneous

(i) Entropy is a measure of the molecular disorderliness (randomness) of a system. dS = dq_rev/T
(ii) The entropy (S) is equal to heat energy exchanged (q) divided by the temperature (T) at which the exchange takes place. Therefore, The SI unit of entropy is JK^-1

The property that is independent of the mass or the size of the system is called an intensive property.
Examples: Refractive index, Surface tension, density, temperature, Boiling point, Freezing point, molar volume, etc.,

The enthalpy change of a reaction either at constant volume or constant pressure is the same whether it takes place in a single or multiple steps provided the initial and final states are same.

The first law of thermodynamics, also known as the law of conservation of energy, states that "The total energy of an isolated system remains constant though it may change from one form to another."
The mathematical statement of the First Law is: ΔU=q+w
Where q - the amount of heat supplied to the system; w - work done on the system

This equation gives the quantitative temperature dependence of equilibrium constant (K). The relation between standard free energy change (\(\triangle\)G^O) and equilibrium constant is
\(\Delta { G }^{ 0 }=-RTln\ K\)    ...(1)
We know that
\(\Delta { G }^{ 0 }=\Delta { H }^{ 0 }-T\Delta { S }^{ 0 }\)
Substituting (1) in equation (2)
\(-RTln\ K\ =\Delta { H }^{ 0 }-T\Delta { s }^{ 0 }\)
Rearranging
In \(K=\cfrac { -\Delta H^{ 0 } }{ RT } +\cfrac { { \Delta S }^{ 0 } }{ R } \)     ...(3)
Differentiating equation (3) with respect to temperature
\(\cfrac { d\left( In\quad K \right) }{ dT } =\cfrac { \Delta { H }^{ 0 } }{ { RT }^{ 2 } } \)       ...(4)
Equation 4 is known as differential form of Van't Hoff equation.
On integrating the equation 4, between T₁ and T₂ with their respective equilibrium constants K₁ and K_2.
\(\int _{ { k }_{ 1 } }^{ { K }_{ 2 } }{ d\left( In\ K \right) =\cfrac { \Delta { H }^{ 0 } }{ R } \int _{ { T }_{ 2 } }^{ { { T }_{ 2 } } }{ \cfrac { dT }{ { T }^{ 2 } } } } \)
\(\left[ In\quad K \right] _{ { K }_{ 1 } }^{ { K }_{ 2 } }=\cfrac { \Delta { H }^{ 0 } }{ R } \left[ -\cfrac { 1 }{ T } \right] ^{ { T }_{ 2 } }_{ { T }_{ 1 } }\)
\(In\quad { K }_{ 2 }-In\quad { K }_{ 1 }=\cfrac { \Delta { H }^{ 0 } }{ R } -\left[ \cfrac { 1 }{ { T }_{ 2 } } +\cfrac { 1 }{ { T }_{ 2 } } \right] \)
\(In\quad \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } =\cfrac { \Delta { H }^{ 0 } }{ R } \left[ \cfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 }{ T }_{ 1 } } \right] \)
\(log\quad \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } =\cfrac { \Delta { H }^{ 0 } }{ 2.303R } \left[ \cfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 }{ T }_{ 1 } } \right] \)  ...(5)
Equation (5) is known as integrated form of Van't Hoff equation.

If we know the value of k_c and Q, the reaction quotient, we can predict the direction of a reaction
If Q = Kc; the reaction is in equilibrium state.
If Q > Kc: the reaction will proceed in the reverse direction i.e., formation of reactants.
If Q < Kc: the reaction will proceed in the forward direction i.e., formation of products.

If Q > K_c; the reaction will proceed in the reverse direction (ie) formation of reactants to proceed to reach equilibrium.

\(\Delta n_{g}=\sum n p_{(g)}-\sum n R_{(g)}\)
As \(\Delta n_{p}(g)\) is greater \(\Delta n_{g}=+v e\)
\( \therefore K_{p}=K_{c}(R T)^{+v e} \)
\(\therefore K_{p}>K_{c} \)
So K is smaller than K_p.

Let us consider the general reaction in which all reactants and products are ideal gases
\(xA+yB\rightleftharpoons IC+mD\)
The equilibrium constant, K_c is
\({ K }_{ c }=\cfrac { \left[ C \right] ^{ I }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \)               .......(1)
and K_p is
\({ K }_{ p }=\cfrac { { p }_{ c }^{ I }\times { p }_{ D }^{ m } }{ { p }_{ a }^{ x }\times { p }_{ B }^{ y } } \)            ..........(2)
The ideal gas equation is
PY = nRT
or
\(P=\cfrac { n }{ V } RT\)
Since Active mass = molar concentration = n/V
p = active mass\(\times\)RT
Based on the above expression the partial pressure of the reactants and products can be expressed as,
\({ p }_{ A }^{ x }=\left[ A \right] ^{ x }\left[ RT \right] ^{ x }\)
\({ p }_{ B }^{ y }=\left[ B \right] ^{ y }\left[ RT \right] ^{ y }\)
\({ p }_{ C }^{ 1 }=\left[ C \right] ^{ I }\left[ RT \right] ^{ 1 }\)
\({ p }_{ D }^{ m }=\left[ D \right] ^{ m }\left[ RT \right] ^{ m }\)
On substitution in eqn. 2,
\({ K }_{ p }=\cfrac { { \left[ C \right] }^{ 1 }{ \left[ RT \right] }^{ 1 }{ \left[ D \right] }^{ m }{ \left[ RT \right] }^{ m } }{ { \left[ A \right] }^{ x }{ \left[ RT \right] }^{ x }{ \left[ B \right] }^{ y }{ \left[ RT \right] }^{ y } } \)            .........(3)
\({ K }_{ p }=\cfrac { { \left[ C \right] }^{ 1 }{ \left[ D \right] }^{ m }{ \left[ RT \right] }^{ I+m } }{ { \left[ A\quad \right] }^{ x }{ \left[ B \right] }^{ y }{ \left[ RT \right] }^{ x+y } } \)
\({ K }_{ p }=\cfrac { { \left[ C \right] }^{ I }{ \left[ D \right] }^{ m } }{ { \left[ A \right] }^{ x }{ \left[ B \right] }^{ y } } \left[ RT \right] ^{ \left( 1+m \right) -\left( x+y \right) }\)     ...........(4)
Sub (1) in (4)
\({ K }_{ p }={ K }_{ C }\left( RT \right) ^{ \left( \Delta { n }_{ g } \right) }\)
where,
\({ \Delta n }_{ g }\) is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase.

\(K_c={[NH_3]^4[O_2]^5\over [NO]^4[H_2O]^6}\)
\(4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} \rightleftharpoons 4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})}\)

Let us consider the formation of ammonia in which, 'a' moles nitrogen and 'b' moles hydrogen gas are allowed to react in a container of volume V. Let 'x' moles of nitrogen react with 3x moles of hydrogen to give 2x moles of ammonia.
\({ N }_{ 2 }\left( g \right) +3{ H }_{ 2 }\left( g \right) \rightleftharpoons { 2NH }_{ 3 }\left( g \right) \)

Applying law of mass action
\({ K }_{ C }=\cfrac { \left[ { NH }_{ 3 } \right] ^{ 2 } }{ \left[ { N }_{ 2 } \right] \left[ { H }_{ 2 } \right] ^{ 3 } } \)
= \(\cfrac { \left( \cfrac { 2x }{ V } \right) ^{ 2 } }{ \left( \cfrac { a-x }{ V } \right) \left( \cfrac { b-3x }{ V } \right) ^{ 3 } } \)
= \(\cfrac { \left( \cfrac { 4x }{ V } \right) ^{ 2 } }{ \left( \cfrac { a-x }{ V } \right) \left( \cfrac { b-3x }{ V } \right) ^{ 3 } } \)
\({ K }_{ C }=\cfrac { 4{ x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 2 } } \)
The equilibrium constant K_p can also be calculated as follows:
\({ K }_{ p }={ K }_{ C }\left( RT \right) ^{ \left( \Delta { n }_{ g } \right) }\)
\(\Delta \)n_g =n_p - n_r = 2 - 4 = -2
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \left( RT \right) ^{ -2 }\)
Total number of moles at equilibrium,
n = a - x + b - 3x + 2x = a + b - 2x
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \times \left[ \cfrac { PV }{ n } \right] ^{ -2 }\)
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \times \left[ \cfrac { n }{ PV } \right] ^{ 2 }\)
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \times \left[ \cfrac { a+b-2x }{ PV } \right] ^{ 2 }\)
\({ K }_{ p }=\cfrac { 4{ x }^{ 2 }\left( a+b\quad -2x \right) ^{ 2 } }{ { P }^{ 2 }\left( a-x \right) \left( b-3x \right) ^{ 3 } } \)