The Concepts of Chemical Equilibrium for IIT-JEE

Illustrartion

0.0753 g of selenium vapours occupiedmol of 78ml at STP. Selenium is in state of equation.


Se₆(g)3Se₂(g)




Calculation deg. of dissociation






Factors affecting equation:-

(1) Chyange of conc. Of any reactant or product


(2) Change in temp. of system


(3) Change in pressure of system


(4) Tradition of catalyst


(5) Addition of inert gas.




Effect of addition of catalyst
:- Addition of catalyst does not disturb eqn. But helps in attainment of eqn. quickly.




(a)
If rxn takes place at constant vol.(closed vessel)
:- Addition of inert gas will not chang mole conc. Of reactants & products. So, state of eqn. Will remain unaffected.


(b) If rxn. Takes place at constant pressure, Hence dissociation of pcl3 increases.


Dumb quation:-[Q. How this happen?


Ans. Since p is cnstant, so addition of inert gas will Total volume. So, at equation, molec conc of each reactant old products will . So,kc is constant at constant temp.


So, to keep kc constant either [PCl₅]should or [PCl₃] & [Cl₂] should. This happen only if PCl₅dissociates.


Le chathelier’s principle:-
If a in eqn. Is subjected to a change of conc. Temp. of pressure, eqn. Shifts in direction that tends to undo the of the change imposes.




(1) Effect of conc.:-

A + BC + D


If rxn is in eqn. Mole of A or B is added, then ,according to Le chatelic’s principle effect will be to
conc. Of A & B react to frome C & D. I,e;eq m. shifts in forward direction. Shift in forward.




(2) Effect of temp.:-

Every rxn. is made up of two opposing rxns. If to rewards rxn. Is exothermic, then backwardrxn.




If temp. is

,I,e; heat is supplied to system; according to the chateliec princple. i,e; eqn. Will shift to side heat absorbs heat I,e; in backward direction & vice versa.


Note
:- Exothermic rxn are forward by low temp. whereas end. thermic rxn. are favoured by high temp.




(3) Effect of pressure:-
Significant role to play only in case of gaseons rxn.


N₂O₄(g)2NO₂(g)


1 mole    2 mole




Forward rxn. Occurs with

in no. of moles & viseversa. If pressure on system

, then according to Le chateliec’s principle , eqn. Will shift in direction in which pressure

or no of moles (b/c pressure is directly proportional to no. of moles). Since backward rxn. Takes place with in no of. moles. So,

in pressure will favovr. Combination of NO₂molecules to N₂O₄.




Note
:-


(1) Low pressure tavours those rxns which accompany by

in total no of moiles & vice versa.




(2) Pressure has no effect on eqn. Rxn which proceed which no change in total no. of moles. So,

in volume will shift eqn. In direction in which no of moles decreases.




I,e; eqn. Constant does not change with temp.


(ii)
log(k₂/k₁)= +velog k₂>log k₁ I,e; eqn. Constantwithin temp.


(iii)log(k₂/k₁) = -velog k₂< log K₁ I,e; eqn. Constantwithin temp.



Q.1 Calculate eqn. constant for rxn.


H₂(g) + CO₂(g)H₂
O(g) + CO(g) at 1539K, if eqn. constants at 1395K for following are










Q2.for reaction


A + B3C at 29⁰
C, 3L vessel contains 1,2,4 mole A, B & c resp. predict direction of rxn .


if (a)k_cfor rxn. is 10




(b) k_cfor rxn is 10.66




Ans.
:- A + B3C


Before rxn; [A] = ¹⁄₃


[B] = ²⁄₃


[C]= ⁴⁄₃


so,


and this [c] must

or [A] & [B] should so, rxn mole occurs in backward direction


(b) Q = K_c
, So, rxn. is in eqn.



Q.3 At temp, 2AB₂(g)2AB(g) + B₂(g)


with deg association ‘x’ which is small compared to unity. Device expression for ‘x’ in terms of eqns. Const. K_p& total pressure P.




Ans. 2AB₂(g)2AB(g) + B₂(g)




Mole after dissociation


Mole after dissociation


Total mole at eqn.






Q.4:- k_CFOR a₂(G) + b₂(g)2AB (g) at 100₀
C is 50. If one line flask containing 1 mole of A2 is connected with a 2L flask containing


2mole of B2 ; How many mole of AB will be formed at






Q.5     N₂(g) + O₂(g)2NO(g)


was established with an eqn. constain K
_c= 2.1 X 10^{- 3}. At eqn. mole % of No. was 1.8. estimates initial composition of air in mole fraction of N₂& O₂.


Ans: N
₂
(g) + O₂ 2NO


initial       a    (100 - a)    0


final     (a - x)     (100 - a - x)     2n


given


also


a = 79%, 100 - a = 21%




Q.6: k_pfor rxn. N₂+ 3H₂ 2NH₃is 1.6 X 10^-4atm^-2at 400⁰c what will be k_pat 500⁰
C ? Heat of rxn. in this temp. range is - 25.14 KCal.








Out K_pfor rxn. Also report K_p&G⁰forN₂+ 3H₂ 2NH₃at 25⁰C




Ans.       -
G⁰= 2.303 RT log₁₀K_p- ( - 16.5 X 10³) = 2.303 X 8.314 X 298 log₁₀K_p



K_p¹for N₂+ 3H₂ 2NH₃K_p¹= (K_p)²= (779.41)²= 6.07 X 10⁵atm^{- 1}Also
G⁰¹= 2.303 X 8.314 X 298 log₁₀(6.07 X 10⁵) J

=32.998 KJ


G⁰¹= -32.998 KJ








Q.9 At S4 Ok & 1 atom, N₂O₄is 66% dissociated into NO₂
what mol. of 10g N₂O₄occupy under these condition




Ans:-   N₂O₄2NO₂


initial 1             0


at eqn


total mole at eqn


but


Total mole of eqn. = 1 + 0.66 = 1.66


1 mole of N₂O₄
is taken, mole at eqn.=1.66



mole of n2o4 is taken ,mole at eqn. =

1 X v = 0.18X0.0821X340 = 5.02 L




Q.10 heat of rxn for rxn. At constant vol. is 1200 cal more than at constant pressure at 300k. calculate ratio of eqn. Constant k_p& k_c.




Ans.       
E =H = 1200 Cal. Whrere
E is heat of rxn. at constant volume &
H is heat of rxn. at constant pressure


H =E +nRT


nRT = -1200


n = -1200 / 2X300 = -2


K_p= K_c(RT)ⁿ= (0.0821 X 300)^-2= 1.648 X 10^-3



Q.1.   0.96g of HI sere heated to attom eqn. 2HIH₂+ I₂
rxn. mixture on titration required 15.7 ml of N/10 hypo sol. Calculate deg. of dissociation of HI.




Ans.             2HI  H₂+ I₂



mole at t = 0


= 7.5 X 10
^-3

at eqn.


meg. Of I₂
formed at eqn. = meg.of hypo used




Eqn. of I₂
=



Mole of I
₂
formed =




Deg. of dissociation of HI =


=

= 0.209 = 20.9 %




Q.2 An eqn. Mixture of


CO(g) + H₂O(g)CO₂(g) + H₂(g)




present in a vessel of 1 L capacity at 815c was found by analysis to contain 0.4 mole of Co, 0.3 mole of H₂O, 0.2 mole of Co₂and 0.6 mole of H₂
(a) Calculate kc

(b) Toconc. to 0.6 mole by adding co₂to vessel, how many mole must be added into eqn mix at mole at eqn 0.4 0.3 0.2 0.6





(b) Suppose ‘a’ mole of CO2 are forced in vessel at eqn.; by doing so rxn proceeds in backward direction


i.e                       Co₂+ H₂ CO + H₂O


at intial time (0.2 + a )   0.6     0.4    0.3


at eqn. (0.2+a-0.2) (0.6-0.2) (0.4+0.2) (0.3+0.2)


a           0.4            0.6       0.5






Q.3 Deg of dissociation is 0.4 at 400k and 1 atom for gaseous rxt PCl
_s PCL₃+ CL₂
Assume ideal gas calculate density of eqn mixture at 400k and 1 atm.


(Atomic mass of p=31& cl=35.5)




Ans:   PCl
_s PCL₃+ CL₂

intial mole 1    0        0


At eqn. (1-0.4)   0.4   0.4


Total mole at eqn. = 1 - 0.4 + 0.4 + 0.4 = 1.4






Q.5 Heat enthalpy standard free energy change for


eqn.   Z_n(S) + H₂O(g)Z_nO(s) + H₂(g)


eqn Are 224 kJ mol
^-1
and 33kjmol respectively at temp 920 calculate temp at which eqn constant becomes greater than


(1) Assume

remains constant.


Ans.    
G⁰= -2.303 RT logK


33000 = -2.303 X 8.314 X 920 log k


k_920k= 1.34 x 10^{- 2}







T₂= 1078.5 K.


SO, At any temp above this k will be greater than




Hard type



Q.1:- Solid nh4 on rapid healthy in aclosed vessel at 357 develop a constant pressure of 275 mmHg owing to partial decomposition of NH₄I into NH
₃& HI . calculate final pressure developed at eqn.k_pfor dissociation of HI is 0.015 at 357 C




Ans:- NH₄I(g)NH₃(g) + HI(g)—–(1)

for this eqn 2p=275 –> p=137.5


So k_pfor eq. (1)


K_p= P_NH3XP_HI= 137.5 X 137.5 ———(2)




For     2HIH₂+ I₂
1         0    0


(1 - x)   x/2    x/2






Dumb question

why we have taken kp (0.015)²?




Ans: K
_pfor division of HI = 0.015


HI

K
_p
= 0.015


2HIH₂+ I₂    K_p= (0.015)₂x = 0.2


Again since HI decomposes rxn proceeds in forward directioal (by Leehatetics Principal)




But a = (P + A) X 0.2


also k_p
for final eqn (3)




= 0.8(P + A)²But K_p
= 137.5 X 137.5


0.8(P + A)²
= 137.5 X 137.5


P + A = 153.73


A = 16.22


a = (p + A) X 0.2 = 153.73 X 0.2 = 30.75




Total pressure at eqn =PNH₃+PHI+PH₂+PI


=(P+A)+(P+A-a)+A/2+A/2


=2p+2A


=2(p+A)= 2x153.73


=307.46 mmHg




Q.2: when 1 - Pentyne (a)is treated with 4n atc. KOH at 175 it slowely get converted into an eqn mix of 1.3 % 1 pentyne 95.2% 2 pentyne(b) of 1,2 pentadiene ©eqn. was maintained at 175c calculate

for following eqn.


from calculated value of g1 and g2 indicate order of (a) (b)© write a reasonable rxn mechanism in showing all intermediates reading to (A)(B)©




= 16.178 KJ




Stability order for A & B is B > A


Similerly for B
C




So, stability order for B and Cis B > C


Total stability order is B><>A




Q3. N₂& 0₂
combine at a giventemp to produce no. at eqn the yield of no is x% by volume. If x =

where k is is eqn constant given rxn. at given temp and a and b volume % of n2 &o2 respectively in intial pure mixture what should be the intial composition of reacting mixture in order that max. yield of nois ensured? also repo max value of k at which x is maximum?




‘x’ is maximum only when condition of maximum are fulful




Note: this valid only when k<4 because if a = b Eq.(1)




Equilibrium


Physical Eqn.


Low of mass action


Law of chemical Eqn.


Eqn.constant


Revessible rxn.


Forward rxn


Backward rxn


Gibbs free energy and Spontaneity


Homogeneous Equilibrium


Heterogeneous Equilibrium Le chatelier’s principle

 

 

Source: goiit.com

      + CO(g) at 1539K, if eqn. constants at 1395K for following are<br />

Dumb question

why we have taken kp (0.015)²?




Ans: K
_pfor division of HI = 0.015


HI

K
_p
= 0.015


2HIH₂+ I₂    K_p= (0.015)₂x = 0.2


Again since HI decomposes rxn proceeds in forward directioal (by Leehatetics Principal)




But a = (P + A) X 0.2


also k_p
for final eqn (3)




= 0.8(P + A)²But K_p
= 137.5 X 137.5


0.8(P + A)²
= 137.5 X 137.5


P + A = 153.73


A = 16.22


a = (p + A) X 0.2 = 153.73 X 0.2 = 30.75




Total pressure at eqn =PNH₃+PHI+PH₂+PI


=(P+A)+(P+A-a)+A/2+A/2


=2p+2A


=2(p+A)= 2x153.73


=307.46 mmHg




Q.2: when 1 - Pentyne (a)is treated with 4n atc. KOH at 175 it slowely get converted into an eqn mix of 1.3 % 1 pentyne 95.2% 2 pentyne(b) of 1,2 pentadiene ©eqn. was maintained at 175c calculate

for following eqn.


from calculated value of g1 and g2 indicate order of (a) (b)© write a reasonable rxn mechanism in showing all intermediates reading to (A)(B)©




= 16.178 KJ




Stability order for A & B is B > A


Similerly for B
C




So, stability order for B and Cis B > C


Total stability order is B><>A




Q3. N₂& 0₂
combine at a giventemp to produce no. at eqn the yield of no is x% by volume. If x =

where k is is eqn constant given rxn. at given temp and a and b volume % of n2 &o2 respectively in intial pure mixture what should be the intial composition of reacting mixture in order that max. yield of nois ensured? also repo max value of k at which x is maximum?




‘x’ is maximum only when condition of maximum are fulful




Note: this valid only when k<4 because if a = b Eq.(1)




Equilibrium


Physical Eqn.


Low of mass action


Law of chemical Eqn.


Eqn.constant


Revessible rxn.


Forward rxn


Backward rxn


Gibbs free energy and Spontaneity


Homogeneous Equilibrium


Heterogeneous Equilibrium Le chatelier’s principle

 

 

Source: goiit.com

          temp. range is - 25.14 KCal.<br />

Dumb question

why we have taken kp (0.015)²?




Ans: K
_pfor division of HI = 0.015


HI

K
_p
= 0.015


2HIH₂+ I₂    K_p= (0.015)₂x = 0.2


Again since HI decomposes rxn proceeds in forward directioal (by Leehatetics Principal)




But a = (P + A) X 0.2


also k_p
for final eqn (3)




= 0.8(P + A)²But K_p
= 137.5 X 137.5


0.8(P + A)²
= 137.5 X 137.5


P + A = 153.73


A = 16.22


a = (p + A) X 0.2 = 153.73 X 0.2 = 30.75




Total pressure at eqn =PNH₃+PHI+PH₂+PI


=(P+A)+(P+A-a)+A/2+A/2


=2p+2A


=2(p+A)= 2x153.73


=307.46 mmHg




Q.2: when 1 - Pentyne (a)is treated with 4n atc. KOH at 175 it slowely get converted into an eqn mix of 1.3 % 1 pentyne 95.2% 2 pentyne(b) of 1,2 pentadiene ©eqn. was maintained at 175c calculate

for following eqn.


from calculated value of g1 and g2 indicate order of (a) (b)© write a reasonable rxn mechanism in showing all intermediates reading to (A)(B)©




= 16.178 KJ




Stability order for A & B is B > A


Similerly for B
C




So, stability order for B and Cis B > C


Total stability order is B><>A




Q3. N₂& 0₂
combine at a giventemp to produce no. at eqn the yield of no is x% by volume. If x =

where k is is eqn constant given rxn. at given temp and a and b volume % of n2 &o2 respectively in intial pure mixture what should be the intial composition of reacting mixture in order that max. yield of nois ensured? also repo max value of k at which x is maximum?




‘x’ is maximum only when condition of maximum are fulful




Note: this valid only when k<4 because if a = b Eq.(1)




Equilibrium


Physical Eqn.


Low of mass action


Law of chemical Eqn.


Eqn.constant


Revessible rxn.


Forward rxn


Backward rxn


Gibbs free energy and Spontaneity


Homogeneous Equilibrium


Heterogeneous Equilibrium Le chatelier’s principle

 

 

Source: goiit.com

Dumb question

why we have taken kp (0.015)²?




Ans: K
_pfor division of HI = 0.015


HI

K
_p
= 0.015


2HIH₂+ I₂    K_p= (0.015)₂x = 0.2


Again since HI decomposes rxn proceeds in forward directioal (by Leehatetics Principal)




But a = (P + A) X 0.2


also k_p
for final eqn (3)




= 0.8(P + A)²But K_p
= 137.5 X 137.5


0.8(P + A)²
= 137.5 X 137.5


P + A = 153.73


A = 16.22


a = (p + A) X 0.2 = 153.73 X 0.2 = 30.75




Total pressure at eqn =PNH₃+PHI+PH₂+PI


=(P+A)+(P+A-a)+A/2+A/2


=2p+2A


=2(p+A)= 2x153.73


=307.46 mmHg




Q.2: when 1 - Pentyne (a)is treated with 4n atc. KOH at 175 it slowely get converted into an eqn mix of 1.3 % 1 pentyne 95.2% 2 pentyne(b) of 1,2 pentadiene ©eqn. was maintained at 175c calculate

for following eqn.


from calculated value of g1 and g2 indicate order of (a) (b)© write a reasonable rxn mechanism in showing all intermediates reading to (A)(B)©




= 16.178 KJ




Stability order for A & B is B > A


Similerly for B
C




So, stability order for B and Cis B > C


Total stability order is B><>A




Q3. N₂& 0₂
combine at a giventemp to produce no. at eqn the yield of no is x% by volume. If x =

where k is is eqn constant given rxn. at given temp and a and b volume % of n2 &o2 respectively in intial pure mixture what should be the intial composition of reacting mixture in order that max. yield of nois ensured? also repo max value of k at which x is maximum?




‘x’ is maximum only when condition of maximum are fulful




Note: this valid only when k<4 because if a = b Eq.(1)




Equilibrium


Physical Eqn.


Low of mass action


Law of chemical Eqn.


Eqn.constant


Revessible rxn.


Forward rxn


Backward rxn


Gibbs free energy and Spontaneity


Homogeneous Equilibrium


Heterogeneous Equilibrium Le chatelier’s principle

 

 

Source: goiit.com

Dumb question

why we have taken kp (0.015)²?




Ans: K
_pfor division of HI = 0.015


HI

K
_p
= 0.015


2HIH₂+ I₂    K_p= (0.015)₂x = 0.2


Again since HI decomposes rxn proceeds in forward directioal (by Leehatetics Principal)




But a = (P + A) X 0.2


also k_p
for final eqn (3)




= 0.8(P + A)²But K_p
= 137.5 X 137.5


0.8(P + A)²
= 137.5 X 137.5


P + A = 153.73


A = 16.22


a = (p + A) X 0.2 = 153.73 X 0.2 = 30.75




Total pressure at eqn =PNH₃+PHI+PH₂+PI


=(P+A)+(P+A-a)+A/2+A/2


=2p+2A


=2(p+A)= 2x153.73


=307.46 mmHg




Q.2: when 1 - Pentyne (a)is treated with 4n atc. KOH at 175 it slowely get converted into an eqn mix of 1.3 % 1 pentyne 95.2% 2 pentyne(b) of 1,2 pentadiene ©eqn. was maintained at 175c calculate

for following eqn.


from calculated value of g1 and g2 indicate order of (a) (b)© write a reasonable rxn mechanism in showing all intermediates reading to (A)(B)©




= 16.178 KJ




Stability order for A & B is B > A


Similerly for B
C




So, stability order for B and Cis B > C


Total stability order is B><>A




Q3. N₂& 0₂
combine at a giventemp to produce no. at eqn the yield of no is x% by volume. If x =

where k is is eqn constant given rxn. at given temp and a and b volume % of n2 &o2 respectively in intial pure mixture what should be the intial composition of reacting mixture in order that max. yield of nois ensured? also repo max value of k at which x is maximum?




‘x’ is maximum only when condition of maximum are fulful




Note: this valid only when k<4 because if a = b Eq.(1)




Equilibrium


Physical Eqn.


Low of mass action


Law of chemical Eqn.


Eqn.constant


Revessible rxn.


Forward rxn


Backward rxn


Gibbs free energy and Spontaneity


Homogeneous Equilibrium


Heterogeneous Equilibrium Le chatelier’s principle

 

 

Source: goiit.com

br />

A = 16.22


a = (p + A) X 0.2 = 153.73 X 0.2 = 30.75




Total pressure at eqn =PNH₃+PHI+PH₂+PI


=(P+A)+(P+A-a)+A/2+A/2


=2p+2A


=2(p+A)= 2x153.73


=307.46 mmHg




Q.2: when 1 - Pentyne (a)is treated with 4n atc. KOH at 175 it slowely get converted into an eqn mix of 1.3 % 1 pentyne 95.2% 2 pentyne(b) of 1,2 pentadiene ©eqn. was maintained at 175c calculate

for following eqn.


from calculated value of g1 and g2 indicate order of (a) (b)© write a reasonable rxn mechanism in showing all intermediates reading to (A)(B)©




= 16.178 KJ




Stability order for A & B is B > A


Similerly for B
C




So, stability order for B and Cis B > C


Total stability order is B><>A




Q3. N₂& 0₂
combine at a giventemp to produce no. at eqn the yield of no is x% by volume. If x =

where k is is eqn constant given rxn. at given temp and a and b volume % of n2 &o2 respectively in intial pure mixture what should be the intial composition of reacting mixture in order that max. yield of nois ensured? also repo max value of k at which x is maximum?




‘x’ is maximum only when condition of maximum are fulful




Note: this valid only when k<4 because if a = b Eq.(1)




Equilibrium


Physical Eqn.


Low of mass action


Law of chemical Eqn.


Eqn.constant


Revessible rxn.


Forward rxn


Backward rxn


Gibbs free energy and Spontaneity


Homogeneous Equilibrium


Heterogeneous Equilibrium Le chatelier’s principle

 

 

Source: goiit.com