Equilibrium constant
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Equilibrium constant
Consider a reversible reacn
A +B <——> C+D
- Eq. Constt. Kc = [C] [D]/[A] [B]
“The equilibrium constant at given temp. is the ratio of product of active masses of products & active masses of reactants.
Kc =Kf /Kb
“The Eq. constt. at a given temp. is the ratio of the velocity constt. of forward & back ward reacn.
Factors affecting Equi. Constt.
1) Catalyst-
Presence of catalyst changes the value of rate constt. but doesn’t change the value of equi. constt. as it influences the values of Kf & Kb in the same proportion.
2) Temperature-
(a) The value of equi. constt. is higher at high temp. in the case of endothermic reaction, because Kf increases.
(b) The value of equi. constt. is lower at high temp in the case of exothermic reaction because the value of Kb increases.
Value of Equi. Constt. & Equation of Reacn
The value of equi. constt. depends upon the following factors-
- ) The mode representation of reacn–
Consider a reversible reacn–
A+B <—–> C+D
According to law of mass action,
Kc =[C][D] /[A][B]
Now, if reacn is reversed, then
C +D <——> A +B
Kc’ =[A][B]/[C][D]
Kc’ =1/Kc
2 — Stoichiometric representation of the Chemical eqn
(i) 2NO2 <——-> N2 +2O2
(ii) NO2 <——-> 1/2 N2 +O2
(iii) 1/2 N2 +O2 <——–> NO2
for eq. (i)
Kc =[N2] [O2]2/[NO2]2
for eq. (ii)
Kc’ = [N2] 1/2 [O2]/[NO2]
for eq. (iii)
Kc” =[NO2]/[N2]1/2 [ O2]
Then relation between Kc & Kc’
Kc’ = Kc
Relation between Kc & Kc”
Kc” =1/Kc
Kc” = 1/Kc’
Unit of Equilibrium Constt.
Unit of equi constt. depends upon the type of reacn
Unit of Kc=(mole/litre) Δn
here,
n=Total no of molecules of Products – Total No. of molecules of reactant
| Ex. H2 +I2 <——> 2HI
unit of Kc=(mole/l) Δn Dn=2–2=0 unit of Kc= (mole/l)0 =1 i.e. no unit Ans. Ex.2– N2 +3H2 <——–> 2NH3 unit of Kc=(mole/l) Δn Dn=2–4=-2 unit of Kc= (mole/l)-2 =litre2/mole2 Ans. Ex.3– PCl5 <——->PCl3 +Cl2 unit of Kc=(mole/l) Δn Δn=2–1= 1 unit of Kc= (mole/l)1 =mole /litre Ans. Ex.4–2 SO3 <——-> 2SO2 +O2 unit of Kc=(mole/l) Δn Δn=3–2=1 unit of Kc= (mole/l)1 =Mole /Litre Ans. Ex.5– N2 + O2 <——–> 2NO unit of Kc=(mole/l) Δn Δn=2–2= 0 unit of Kc= (mole/l)0 = 1 No unit Ans.
N2+2O2 <——–> 2NO2 at a particular temp. Equi. constt Kc= 100 Write down the equi. expression for the following reactions & determine the values of equi. constt.
Ans. For, N2+2O2 <——-> 2NO2 Kc =[NO2]2/[N2] [O2]2 =100 For eq.
Kc’ =[N2][O2]2 /[NO2]2 Kc’ =1/Kc Kc’= 1/100=0.01 Ans. For eq. NO2<——-> 1/2 N2 +O2 Kc” =[N2]1/2 [O2]/[NO2] Kc” =1/Kc =1/100 =0.1 Ans. Q— For the reaction, 2NH3<——-> N2 +3H2 ; Kc =25 Cal. Kc’ the following reaction, 1/2 N2 +3/2 H2<—–> NH3 Ans. Kc =[N2] [H2]3/[NH3]2 =25 Kc’ = [NH3]/[N2]1/2[H2]3/2 Kc’ =1/Kc =1/25 =0.2 Ans. Q —For the reacn, H2 +I2 <——>2HI At a given temp. Kc is 49. Find the Kc’ for the reverse reacn i.e. 2HI<——-> H2 +I2 Ans. Kc=[HI]2/[H2][I2] =49 Kc’ =[H2][I2]/[HI]2 Kc’ =1/Kc =1/49 =0.0204 Ans.
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