Suppose at equilibrium, the molar concentration of both I₂(g) and Clig) is x mol L^-I

\({ K }_{ C }=\cfrac { \left[ { I }_{ 2 }\left( g \right) \right] \left[ { { Cl }_{ 2 }\left( g \right) } \right] }{ \left[ ICI(g) \right] ^{ 2 } } =\cfrac { \left( x \right) \times \left( x \right) }{ \left( 0.78-2x \right) ^{ 2 } } \) 
\(\cfrac { x }{ \left( 0.78-2x \right) } ={ (0.14) }^{ \frac { 1 }{ 2 } }=0.374\) or x = 0.374 (0.78 - 2x)
x = 0.292 - 0.748x or x = 1.748x = 0.292;\(x=\cfrac { 0.292 }{ 1.748 } =0.167\) 
[ICI] = (0.78 - 2 x 0.167) = (0.78 -0.334) = 0.446 M
[I₂]= 0.167 M; [CI2] = 0.167 M

Calculation of K_P

Total no. of moles in the equilibrium mixture = 1 - α + α + α = (1 + a) mol.
Let the total pressure of equilibrium mixture =p atm
Partial pressure of PCI_5,\({ PCl }_{ 3 }=\cfrac { \alpha }{ 1+\alpha } \times p\quad atm\) 
Partial pressure of \({ PCL }_{ 3 }=\cfrac { \alpha }{ 1+\alpha } \times p\quad atm\) 
Partial pressure of Cl₂=\({ PCl }_{ 2 }=\cfrac { \alpha }{ 1+\alpha } \)
\({ K }_{ P }=\cfrac { P_{ PCl_{ 3 } }\times { P }_{ cl_{ 2 } } }{ { P }_{ PC{ l }_{ 5 } } } =\cfrac { \left( \frac { \alpha }{ 1+\alpha } p\quad atm \right) \times \left( \cfrac { \alpha }{ 1+\alpha } p\quad atm \right) }{ \frac { 1-\alpha }{ 1+\alpha } p\quad atm } =\cfrac { { a }^{ 2 }p }{ 1-{ a }^{ 2 } } \)
P=4 atm and α.= 10%= \(\cfrac { 10 }{ 100 } =0.1\)
\({ K }_{ P }=\cfrac { \left( 0.1 \right) \times \left( 0.1 \right) \times \left( 4atm \right) }{ 1-\left( 0.1 \right) ^{ 2 } } =\cfrac { 0.04 }{ 0.99 } =0.004atm\)
= 0.96 atm

(i) \({ 2H }_{ 2(g) }+{ CO }_{ (g) }\rightleftharpoons { CH }_{ 3 }{ OH }_{ (g) }\)
According to Le-Chatelier's principle,
(a) addition of H₂ (increase in concentration of reactants) shifts the equilibrium in forward direction (more product is formed).
(b) addition of CH₃OH . (increase In concentration of preduct) shifts the equilibrium in backward direction.
(c) removal of CO also shifts the equilibrium in backward direction.
(d) removal ofCH₂OH shifts the equilibrium in forward direction.
(ii) When catalyst is added, the state of equilibrium is not disturbed but equilibrium is attained quickly, This is because the 'catalyst increases the rate offorward and backward reaction to the same extent.

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 formation of HI in which, 'a' moles of hydrogen and 'b' moles of iodine gas are allowed to react in a container of volume V. Let 'x' moles of each of H₂ and I₂react together to form 2x moles of HI.
\({ H }_{ 2\left( g \right) }+{ I }_{ 2\left( g \right) }\rightleftharpoons { 2HI }_{ \left( g \right) }\)

Applying law of mass action,
\({ K }_{ C }=\cfrac { { \left[ HI \right] }^{ 2 } }{ { \left[ H \right] }_{ 2 }\left[ { I }_{ 2 } \right] } \)
= \(\cfrac { \left( \cfrac { 2x }{ V } \right) ^{ 2 } }{ \left( \cfrac { a-x }{ V } \right) \left( \cfrac { b-x }{ v } \right) } =\cfrac { { 4x }^{ 2 } }{ \left( a-x \right) \left( b-x \right) } \)
The equilibrium constant Kp can also be calcu•lated as follows:
We know }he \rer~tionship between the K_c and K_p
Here the \(\Delta n_{ g }\)=n_p -n_r = 2 - 2 =0
Hence K = K_c ;\({ K }_{ p }=\cfrac { { 4x }^{ 2 } }{ \left( a-x \right) \left( b-x \right) } \)