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The Low of Mass Action was first formulated by two Norwegian chemists, Guldberg and Waage in 1864. The basis of their formulation is the observation of huge deposit of Sodium Carbonate on Egyptian take shore. Large amount of NaCl in take water and CaCO3 on the take shore made the reverse reaction possible.
CaCl2(aq) + Na2CO3(aq)
CaCO3(s) + 2NaCl(aq)
The foreword reaction occurs spontaneously in the laboratory.
They States the Low as,
The rate of chemical reaction at a constant temperature, is directly proportional to the active mass of the reactants.
The active mass is thermodynamic quantity. We assumed active mass as,
  1. Molar Concentration (moles/lit) When the solution is dilute, that is when the system behaves ideally.
  2. Partial pressure in atmosphere unit for gaseous system and when the pressure of the system is very low.
  3. For pure solid and pure liquid, active mass is assumed to be unity since their mass does not effect the rate of reaction.

Application of Low of Mass Action to the Chemical Reaction:

Let us Consider a reaction,
A + B C + D
Let the reacting system contains reactants only and CA and CB are their Concentrations in molar units.
According to the mass action low, the rate of the foreword reaction,
RfCA × CB
Rf = Kf × CA × CB
Though the rate equation is an experimental quantity and powers of concentration terms are determined experimentally, here we consider a special type of rate equation where the stoichometric coefficients are raised in the of the conc. term(this is true for the elementary or one step reaction).
As the reaction proceeds in the foreword direction, conc. of A and B decreases and Rf also decreases. When the products are getting accumulated in the system, backwards reaction also starts and the rate of the backward reaction,
RbCC × CD
Rb = Kb × CC × CD
Here Kf and Kb are the rate constants of the foreword and backward reaction and they do not depends on Conc. at a given temperature.
As the reactions proceeds in the foreword reactions Rf is decreasing but Rb is increasing. A state is then attain when they are equal. This state is called the chemical equilibrium. There will be no further change in the composition of the system.
Thus at equilibrium,
Rf = Rb
or, Kf × CA × CB = Kb × CC × CD
∴ Kf/Kb = (CC × CD)/(CA × CB)
CA, CB, CC and CD are the equilibrium concentration of A, B, C and D.
Again, Kf/Kb = Kc, called concentration equilibrium constant of the reaction.
At a given temperature for a reaction, Kc is constant does not depend on the concentration of the reacting components.
Kc = (CC × CD)/(CA × CB)

Equilibrium Constant:

Concentration Equilibrium Constant:
If we write the equation as,
Ɣ1A1 + Ɣ2A2 Ɣ3A3 + Ɣ4A4
Where Ɣ1, Ɣ2, Ɣ3 and Ɣ4 are stoichiometric coefficient.
Kc = (C3Ɣ3 × C4Ɣ4)/(C1Ɣ1 × C2Ɣ2)
Where Kc is called concentration equilibrium constant of the reaction and C1C2C3 and C4 are the equilibrium concentration of A, B, C and D.
However the values of equilibrium constant of a chemical reaction depends on the mode of writing its Stoichiometric (balanced) equation.
Thus, for the reaction of formation of NH3 from N2 and H2, we can write the equation as,
N2 + 3 H2 2 NH3
The equilibrium constant can be written as,
Kc = (CNH3)2/(CN2) (CH2)2
But if the equation written as,
1/2 N2 + 3/2 H2 NH3
Then, c = (CNH3)/(CN2)3/2(CH2)1/2
It is clear then Kc and c are not Equal in magnitude.
Thus, Kc = (c)1/2
The general rule is that if the equation is multiplied by n then the relation of Equilibrium concentration n will be raised to the power.
That is, Kc = (c)n
Pressure Equilibrium Constant:
When all the reactants and products are gases (that is, gas - phase reacting system), the expression of equilibrium constant for the equation,
Ɣ1A1 + Ɣ2A2 Ɣ3A3 + Ɣ4A4
Where Ɣ1, Ɣ2, Ɣ3 and Ɣ4 are stoichiometric coefficient.
Kp = (P3Ɣ3 × P4Ɣ4)/(P1Ɣ1 × P2Ɣ2)
Where Kp is called Pressure equilibrium constant of the reaction and P1P2P3 and P4 are the equilibrium partial pressure of reacting components.
For the dissociation N2O4 2 NO2, obtain an expression for the fraction of original N2O4 dissociated at equilibrium in terms of Kp and total pressure.
The reaction is, N2O4 2 NO2
Let a mole of N2O4 is taken initially and x moles of N2O4 is dissociated at equilibrium then mole number of N2O4 and NO2 at equilibrium is (a - x) and 2x.
Total moles number at equilibrium = (a -x + 2x) = (a + x).
Equilibrium partial pressure of the components is {(a - x)/(a + x)}P and {2x/(a + x)}P where P = Total pressure of the reacting system.
The expression of the equilibrium constant,
Kp = (PNO2)2/PN2O4
= {2x/(a + x)}P/{(a - x)/(a + x)}P
Kp = (4x2P)/(a2 - x2)
The fraction of the original N2O4 dissociated at equilibrium ɑ = x/a.
Replacing (x/a) by ɑ, we have,
Kp = (4ɑ2P)/(1 - ɑ2)
Mole Fraction Equilibrium Constant:
If we write the equation as,
Ɣ1A1 + Ɣ2A2 Ɣ3A3 + Ɣ4A4
Where Ɣ1, Ɣ2, Ɣ3 and Ɣ4 are stoichiometric coefficient.
Kx = (x3Ɣ3 × x4Ɣ4)/(x1Ɣ1 × x2Ɣ2)
Where Kx is called Mole fraction Equilibrium Constant of the reaction and x1x2x3 and x4 are the equilibrium mole fraction of reacting components.
Mathematical expression of law of mass action.
Equilibrium Constant.

Relation Between Kp and Kc:

The interrelations of these Equilibrium constants are as follows,
Kp = (P3Ɣ3 × P4Ɣ4)/(P1Ɣ1 × P2Ɣ2)
The ideal gas Equation, PV = nRT, may be written as,
P = (n/V)RT = CRT
Where C is the concentration of gas expressed as amount per unit volume.
= {(C3RT)Ɣ3 × (C4RT)Ɣ4}/{(C1RT)Ɣ1 × (C2RT)Ɣ2}
Kp = Kc(RT)ΔƔ
ΔƔ = (Ɣ3 + Ɣ4) - (Ɣ1 + Ɣ2)
  1. For the reaction in which total number of reactant molecules and of resultant molecules are same:
  2. H2 (g) + I2 (g) 2HI
    KP = (PHI)2/{(PH2)(PI2)}
    = (CHIRT)2/(CH2RT) (CI2RT)
    = [(CHI)2/{(CH2) (CI2)}] × [(RT)2/(RT)(RT)]
    = Kc
    Thus when (Ɣ3 + Ɣ4) = (Ɣ1 + Ɣ2),
    Kp = Kc
  3. For the reactions in which the number of molecules of reactants differ from that of the resultant:
  4. 2SO2(g) + O2(g) SO3(g)
    Here, Kp = Kc×(RT){1 - (2+1)}
    = Kc×(RT)-2
    Thus when, (Ɣ3 + Ɣ4) (Ɣ1 + Ɣ2),
    Kp  Kc
Reaction
(Gaseous System)
Relation between
Kp and Kc
H2 + Cl2  2HCl Kp = Kc
CO H2O   2HCO2 Kp = Kc
CO + NO2  NO CO2 Kp = Kc
PCl5  PCl3 Cl2 Kp = Kc RT
2H2 + O2  2H2O Kp = Kc RT -1
2CO + O2  2CO2 Kp = Kc RT -1
N2 + 3H2  2NH3 Kp = Kc RT -2
Calculate the Kc value of the reaction N2 + 3H2 ⇆ 2NH3 at 4000C, Given Kp at the same temperature 1.64 × 10-4.
Kp = 0.5.
For Solution see,
Chemical Equilibrium Questions and Answers
At 1000C the vapour density of N2O4 is 25 at 1 atm. Show that Kp = 9.6.
N2O4 (g) 2NO2 (g)
Let 1 moles of N2O4 is taken initially (t = 0) and x mole of N2O4 has reacted at equilibrium.
So the mole number of each component is (1-x) and 2x and total moles at equilibrium,
(1-x+2x) = (1+x).
So total moles has increased from 1 to (1+x).
Let Volume is increases from V1 to V2
So, (1+x) = V2/V1
As density and hence vapour density is inversely proportional to volume so vapour density will decrease from d1 to d2.
Hence, (1+x) = V2/V1= d1/d2.
Molecular weight of N2O4 is 92 and vapour density,
= 92/2
= 46
Due to dissociation it is = 25.
∴ 1+x = 46/25
or, x = 0.84
Now partial pressure are,
PNO2 = {2x/(1+x)}P
= (2×0.84)/1.84
= 0.913 atm
PN2O4 = {(1-x)/(1+x)}P
= 0.16/1.84
= 0.087
Kp = (PNO2)2/PN2O4
= (0.913)2/0.087
≃ 9.6

Relation Between Kp and Kx:

Kp = Kx(P)ΔƔ
ΔƔ = (Ɣ3 + Ɣ4) - (Ɣ1 + Ɣ2)
Show that the equilibrium condition for any chemical reaction is given by ΔG = 0.
Vant Hoff reaction isotherm is ΔG = - RT lnKa + RT lnQa
But when the reaction attain equilibrium, Qa = Ka
Thus, ΔG = 0.

Van't Hoff Equation - Effect on temperature on Equilibrium Constant:

Equilibrium Constant KP of a reaction is constant at a given temperature but when temperature is changed, the value of equilibrium constant also changed. The quantitative relation, relation, Known as Van't Hoff equation is derived by using Gibbs - Helmholtz equation.
The Gibbs - Helmholtz equation is,
ΔG0 = ΔH0 + T[d(ΔG0)/dT]P
Zero superscript is indicating the stranded values.
or, - (ΔH0/T2 )= -(ΔG0/T2 )+(1/T)[d(ΔG0)/dT]P
or, - (ΔH0/T2 ) = [d/dT(ΔG0/T)]P
Again Van't Hoff isotherm is,
- RT lnKP = ΔG0
or, - R lnKP = ΔG0/T
Differentiating with respect to T at constant P,
- R [dlnKP/dT]P = [d/dT(ΔG0/T)]P
Comparing the above two equation we have,
dlnKP/dT = ΔH0/T2
This is the differential form of Van't Hoff reaction equation.
Greater the value of ΔH0, the faster the equilibrium constant (KP) changes with temperature (T).
Separating the variables and integrating,
dlnKP = (ΔH0/R)(dT/T2)
Assuming that ΔH0 is independent of temperature.
or, lnKP = - (ΔH0/R)(1/T)+C(integrating constant)
The integration constant can be evaluated and identified as ΔS0/R using the relation,
ΔG0 = ΔH0 - TΔS0
Thus the Van't hoff Equation is,
lnKP = - (ΔH0/R)(1/T) + ΔS0/R
For a reaction 2A + B ⇆ 2C, ΔG0(500 K) = 2 KJ mol-1 Find the KP at 500 K for the reaction A + ½B ⇆ C.
ΔG0(500K) for the reaction A + ½B C is,
2 KJ mol-1/2 = (1 KJ mol-1)/2. 
The relation is ΔG0 = - RT lnKP
or, 1  = - (8.31 × 10-3 ) × (500) lnKP 
 or, lnKP = 1/(8.31 × 0.5) 
or, lnKP = 0.2406
KP = 1.27
Plot of lnKP vs 1/T:
For exothermic reaction, ΔH0 = (-) ve.
Examples are the formation of ammonia from H2 and N2.
N2 + H2 2NH3 ΔH0 = (-) ve
For endothermic reaction, ΔH0 = (+) ve.
Examples are the dissociation of HI into H2 and I2.
HI H2 + I2 ΔH0 = (+) ve
For the reaction, ΔH0 = 0
lnKP is independent of T. Provided ΔS0 does not change much with T.
Van't Hoff Equation - Effect on temperature on Equilibrium Constant and Expression of Le -Chatelier Principle from Van't Hoff Equation
Plot of lnKP vs 1/T
Since for ideal system, H is not a function of P and ΔH0 = ΔH and Van't Hoff equation is,
dlnKP/dT = ΔH/RT2
and the integrated equation is,
lnKP = (ΔH/R)(1/T) + ΔS/R
However S of an ideal gas depends strongly on P, so ΔS and ΔG per mole of reaction in the mixture differ quite substantially from ΔS0 and ΔG0.
If we consider the integrated Van't Hoff equation at two temperature then, it becomes,
ln(KP2/KP1) = (ΔH/R){(T2 -T1)/T1T2)}
Where KP1 and KP2 are the equilibrium constants of the reaction at two different temperature T1 and T2 respectively.
Thus determination of KP1 and KP2 at two temperature helps to calculate the value of ΔH of the reaction.
The above relation is called Van't Hoff reaction isobar since P remains constant during the change of temperature.
Use Gibbs - Helmholtz equation to derive the Vant Hoff reaction isochore. Under What condition do you expect a linear relationship between logk and 1/T ?
Again KP = KC (RT)ΔƔ
or, lnKP = lnKC + ΔƔ lnR + ΔƔlnT
Differentiating with respect to T,
dlnKP/dT = dlnKC/dT + ΔƔ/T
But dlnKP/dT = ΔH0/RT2
Hence, dlnKC/dT = ΔH0/RT2 - ΔƔ/T
or, dlnKC/dT = (ΔH0 - ΔƔRT)/RT2
∴ dlnKC/dT = ΔU0/RT2
ΔU0 is standers heat of reaction at constant volume. This is really the Van't Hoff reaction isochore in differential form.
The integrated form of the equation at two temperature is,
ln(KC2/KC1) = (ΔU0/R){(T2 -T1)/T1T2)}
For ideal system ΔU0 = ΔU. Since the reaction occurs in an equilibrium box at constant Volume this is called Van't Hoff reaction isochore.
Two important assumptions are:
  1. The reacting system is assumed to behave ideally.
  2. ΔH is taken independent of temperature for small range of temperature change. 
Due to assumption involved ΔH and ΔU do not produce precise value of these reactions. As ΔG is independent of pressure for ideal system ΔH and ΔU also independent of pressure.
Hence, ΔG0 = - RT lnKP
or, [dlnKP/dT]T = 0
Show that the equilibrium condition for any chemical reaction is given by ΔG = 0.
Vant Hoff reaction isotherm is, 
ΔG = - RT lnKa + RT lnQa
But when the reaction attain equilibrium Qa = Ka
∴ ΔG = 0
Expression of Le -Chatelier Principle from Van't Hoff Equation:
Van't Hoff equations gives quantitative expression of Le-Chatelier Principle.
From the equation,
lnKP = -(ΔH/R)(1/T) + C
It is evident that for endothermic reaction (ΔH〉0), increase of T increases the value of a of the reaction.
But for exothermic reaction (ΔHㄑ 0), with rise in temperature, a is decreased.
This change of KP also provides the calculation of quantitative change of equilibrium yield of products.
This above statement is in accordance with Le - Chatelier Principle which states that whenever stress is placed on any system in a state of equilibrium, the system is always react in a direction to reduce the applied stress.
Thus when T is increased, the system in equilibrium will try to move in a direction in which heat is absorbed that is endothermic reaction is favors.

Given,
(i) H2 (g) + 1/2 S2 (g) = H2S (g), KP1 
(ii) 2H2 (g) + S2 (g) = 2H2S (g), KP2 
Show that, KP2 = (KP1)2
For the first reaction, ΔG10 = - RT lnKP1
For the second reaction, ΔG20 = - RT lnKP2
Since, ΔG20 = 2ΔG10, therefore, it follows that,
- RT lnKP2 = - 2RT lnKP1 
KP2 = (KP1)2
Chemical Equilibrium Questions and Answers
Equilibrium Constant
Use Gibbs - Helmholtz equation to derive the Vant Hoff reaction isochore. Under What condition do you expect a linear relationship between logk and 1/T ?
The Gibbs - Helmholtz equation is,
ΔG0 = ΔH0 + T[d(ΔG0)/dT]P 
Zero superscript is indicating the stranded values.
or, - (ΔH0/T2 ) = -(ΔG0/T2 ) + 1/T[d(ΔG0)/dT]P
or, - (ΔH0/T2 ) = [d/dT(ΔG0/T)]P
Again Vant Hoff isotherm is,
- RT lnKP = ΔG0
or, - R lnKP = ΔG0/T
Differentiating with respect to T at constant P,
- R [dlnKP/dT]P = [d/dT(ΔG0/T)]P
Comparing the above two equation we have,
dlnKP/dT = (ΔH0/T2
This is the Van't Hoff Equation isochore.
Greater the value of ΔH0, the faster the equilibrium constant(KP)changes with temperature(T), ΔH0 should remains constant for the linear plot of logKP vs 1/T.
How does the equilibrium constant for a reaction, 2A + 3B 4C + Heat, Change when (i) pressure is increases (ii) Temperature is decreases (iii) a catalyst added?
  1. Equilibrium constant remains same when P is increases .
  2. The reaction is exothermic, hence ΔH = (-)ve so the equilibrium constant is increased with decreases of temperature.
  3. Equilibrium constant remains same though a catalyst is added.
  4. ΔG0 = - RT lnKa but ΔG0 is not changed due to addition of catalyst as the latter does not participate in the reaction. The initial state and the final state of a chemical reaction remains same weather the catalyst is added or not.
    Hence the equilibrium constant (Ka) remains unchanged.
For a reaction,2A + B 2C, ΔG0(500 K) = 2 KJ mol-1. Find the KP at 500 K for the reaction A + ½B C.
ΔG0 (500 K) for the reaction,
A + ½B C 
= (2 KJ mol-1)/2
= 1 KJ mol-1
The relation is, ΔG0 = - RT lnKP
or, 1 = - (8.31 × 10-3) × (500) lnKP
or, lnKP = 1/(8.31 × 0.5)
= 0.2406
KP = 1.27
Justify or criticize the following: "The Equilibrium yield of products can not change if the equilibrium constant is kept fixed."
This statement is not correct.
Equilibrium yield of the product is changed if pressure is changed (Δγ ≠ 0), if inert gas is added at constant P (Δγ ≠ 0), and any of the reacting component is added a depleted.
For example, N2 + 3 H2 2NH3
Equilibrium yield of NH3 is increased if P is increased though equilibrium constant kept fixed. 
Justify or criticize the following: Heat of reaction is the same weather a catalyst is used or not.
H is a state function hence ΔH (heat of a reaction) does no change if initial state and final state of a Process is same.
A catalyst can not change the initial and final state of a chemical reaction, hence ΔH remains same weather a catalyst is used or not.
Therefore the statement is correct.

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