Balanced chemical equation for the reaction is 4 NO(g) + 6 H₂O(g) ⇌ 4 NH₃(g) + 5 .O₂(g)

Under non-equilibrium conditions, reaction quotient 'Q' is defined as the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants.

The modem periodic law states that, "the physical and chemical properties of the elements are periodic functions of their atomic numbers."

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

It is defined as the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.

The net nuclear charge experienced by valence electrons in the outermost shell is called the effective nuclear charge.
Z_eff=Z-S
Where Z is the atomic number and 'S' is the screening constant.

Ions of different elements having the same number of electrons are called isoelectronic ions.

Consider that 'a' moles of PCI₅ is taken in a container of volume V. Let 'x' moles of PCl₅ be dissociatedinto x moles of PC₁, and x moles of Cl₂ .
\(\mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}\)

Applying law of mass action
\({ K }_{ c }=\cfrac { \left[ { PCI }_{ 3 } \right] \left[ { CI }_{ 2 } \right] }{ { \left[ { PCI }_{ 5 } \right] } } =\cfrac { \left( \cfrac { x }{ V } \right) \left( \cfrac { x }{ V } \right) }{ \left( \cfrac { a-x }{ V } \right) } =\cfrac { { x }^{ 2 } }{ \left( a-x \right) V } \)
The equilibrium constant K_p can also be calculated as follows:
We know the relationship between the K_c and K _p
\(K_{p}=K_{c}(R T)\left(\Delta n_{g}\right)\)
Here the \(\Delta n_{g}=n_{p}-n_{r}=2-1=1\)
Hence K_p = K_c (RT)
We know that PV = nRT
\(RT=\cfrac { PV }{ n } \)
Where n is the total number of moles at equilibrium.
n = (a - x) + x + x = (a + x)
=  \({ K }_{ p }=\cfrac { { x }^{ 2 } }{ \left( a-x \right) V } \cfrac { PV }{ n } \)
= \({ K }_{ p }=\cfrac { { x }^{ 2 } }{ \left( a-x \right) V } \cfrac { PV }{ \left( a+x \right) } \)
= \({ K }_{ p }=\cfrac { { x }^{ 2 }P }{ \left( a-x \right) \left( a+x \right) } \)

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.

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
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \times \left[ \cfrac { PV }{ n } \right] ^{ -2 }\)
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \times \left[ \cfrac { n }{ PV } \right] ^{ 2 }\)
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \times \left[ \cfrac { a+b-2x }{ PV } \right] ^{ 2 }\)
\({ K }_{ p }=\cfrac { 4{ x }^{ 2 }\left( a+b\quad -2x \right) ^{ 2 } }{ { P }^{ 2 }\left( a-x \right) \left( b-3x \right) ^{ 3 } } \)

Variation along a period: Ionisation energy usually increases along a period. This is due to increase of nuclear charge and decrease in size as we move from left to right in a period.
Periodic variation in group: Ionisation energy decreases down a group. As we move down a group, the valence electron occupies new shells, the distance between the nucleus and the valence electron increases. So, the nuclear forces of attraction on valence electron decreases and hence ionisation energy also decreases down a group.

(i) Ionic radius of uni-univalent crystal can be calculated using Pauling's method from the inter ionic distance between the nuclei of the cation and anion.
(ii) Pauling assumed that ions present in a crystal lattice are perfect spheres, and they are in contact with each other therefore,
d=rC⁺ + r_A^-   ...(1)
Where d is the distance between the centre of the nucleus of cation C+ and anion A-and rC⁺, rA^- are the radius of the cation and anion respectively.
(iii) Pauling also assumed that the radius of the ion having noble gas electronic configuration is inversely proportional to. the effective nuclear charge.
\({ r }_{ C }^{ + }\alpha \frac { 1 }{ ({ Z }_{ eff }){ C }^{ + } } \) ....(2) and
\({ r }_{ A }^{ - }\alpha \frac { 1 }{ ({ Z }_{ eff }){ A }^{ - } } \)...(3)
Where Z_eff is the effective nuclear charge and Z_eff= Z - S
Dividing the equation 1 by 3
\(\frac { { r }_{ C }^{ + } }{ { r }_{ A }^{ - } } =\frac { ({ Z }_{ eff }){ A }^{ - } }{ ({ Z }_{ eff }){ C }^{ + } } \) ...(4)
On solving equation and (1) and (4) the values of r_C+ and r_A^- can be obtained.

i) When \(\Delta \mathrm{n}_{\mathrm{g}}=0\)
\(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{0}=\mathrm{K}_{\mathrm{C}}\)
Example:
\(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)
\(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\)
ii) When \(\Delta \mathrm{n}_{\mathrm{g}}=+ve\)
\(K_{p}=K_{c}(R T)^{+v e}\)
K_p>K_{c
\(2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\)
\(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\)}
iii) When _{\(\Delta \mathrm{n}_{\mathrm{g}}=-ve\)
^{\(\mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})-\mathrm{ve}\)}}
K_p<K_c
Example:
\(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) 
\(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\)

J. W. Dobereiner classified some elements such as chlorine, bromine and iodine with similar chemical properties into the group of three elements called as triads.
In triads, the atomic weight of the middle element nearly equal to the arithmetic mean of the atomic weights of the remaining two elements. However, only a limited number of elements can be grouped as triads.

This concept cannot be extended to some triads which have nearly same atomic masses such as [Fe, Co, Ni], [Ru, Rh, Pd] and [Os, Ir, Pt].

A pure solid always has the same concentration at a given temperature, as it does not expand to fill its container i.e., it has the same no. of moles of its volume. Therefore the concentration of pure solid is a constant. So the expression if K_C and K_P, is K_C = [CO₂], K, = P_CO2

The knowledge of equilibrium constant helps us to
1. Predict the direction in which the net reaction will take place
2. Predict the extent of the reaction and
3. Calculate the equilibrium concentrations of the reactants and products.
It is to be noted that these constants do not provide any information regarding the rates of the forward or reverse reactions.

The energy required to remove the most loosely held electron from an isolated gaseous atom is called ionization energy.
(i.e) Atom_(g) + Energy ⟶ Positive Ion_(g)+Electron
Li_(g) + 520Kj mol^-1 ⟶ Li_(g)⁺ + e^-

(i) F-1s² 2s² 2p_x² 2p_y² 2p_z¹
Cl- -1s² 2s² 2p⁶ 3s² 3p_x² 3p_y² 3p_z¹
(ii) Because of small size of fluorine, the 2p subshell becomes compact.
(iii) There occurs electron-electron repulsion
(iv) Due to its small size, large crowding of electrons occurs which reduces the nuclear charge. So the electron affinity of fluorine is less than chlorine.

(i) Noble gases have completely filled electronic configurations.
(ii) It has no tendency to attract the electrons towards itself
(iii) so the electron affinity of noble gases are almost zero.