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### 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,
 Î”G0 = Î”H0 + T[d(Î”G0)/dT]P
Zero superscript is indicating the stranded values.
or, - (Î”H0/T2 )= -(Î”G0/T2 )+(1/T)[d(Î”G0)/dT]P
or, - (Î”H0/T2 ) = [d/dT(Î”G0/T)]P
Again Van't Hoff isotherm is,
 - RT lnKP = Î”G0
or, - R lnKP = Î”G0/T
Differentiating with respect to T at constant P,
- R [dlnKP/dT]P = [d/dT(Î”G0/T)]P
Comparing the above two equation we have,
 dlnKP/dT = Î”H0/T2
This is the differential form of Van't Hoff reaction equation.
Greater the value of Î”H0, the faster the equilibrium constant (KP) changes with temperature (T).
Separating the variables and integrating,
dlnKP = (Î”H0/R)(dT/T2)
Assuming that Î”H0 is independent of temperature.
or, lnKP = - (Î”H0/R)(1/T)+C(integrating constant)
The integration constant can be evaluated and identified as Î”S0/R using the relation,
Î”G0 = Î”H0 - TÎ”S0
Thus the Van't hoff Equation is,
 lnKP = - (Î”H0/R)(1/T) + Î”S0/R
For a reaction 2A + B ⇆ 2C, Î”G0(500 K) = 2 KJ mol-1 Find the KP at 500 K for the reaction A + ½B ⇆ C.
Î”G0(500K) for the reaction A + ½B C is,
2 KJ mol-1/2 = (1 KJ mol-1)/2.
The relation is Î”G0 = - 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, Î”H0 = (-) ve.
Examples are the formation of ammonia from H2 and N2.
 N2 + H2 ⇆ 2NH3 Î”H0 = (-) ve
For endothermic reaction, Î”H0 = (+) ve.
Examples are the dissociation of HI into H2 and I2.
 HI ⇆ H2 + I2 Î”H0 = (+) ve
For the reaction, Î”H0 = 0
lnKP is independent of T. Provided Î”S0 does not change much with T.
 Plot of lnKP vs 1/T
Since for ideal system, H is not a function of P and Î”H0 = Î”H and Van't Hoff equation is,
dlnKP/dT = Î”H/RT2
and the integrated equation is,
lnKP = (Î”H/R)(1/T) + Î”S/R
However S of an ideal gas depends strongly on P, so Î”S and Î”G per mole of reaction in the mixture differ quite substantially from Î”S0 and Î”G0.
If we consider the integrated Van't Hoff equation at two temperature then, it becomes,
 ln(KP2/KP1) = (Î”H/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 Î”H 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)Î”Æ”
or, lnKP = lnKC + Î”Æ” lnR + Î”Æ”lnT
Differentiating with respect to T,
dlnKP/dT = dlnKC/dT + Î”Æ”/T
But dlnKP/dT = Î”H0/RT2
Hence, dlnKC/dT = Î”H0/RT2 - Î”Æ”/T
or, dlnKC/dT = (Î”H0 - Î”Æ”RT)/RT2
 ∴ dlnKC/dT = Î”U0/RT2
Î”U0 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) = (Î”U0/R){(T2 -T1)/T1T2)}
For ideal system Î”U0 = Î”U. 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. Î”H is taken independent of temperature for small range of temperature change.
Due to assumption involved Î”H and Î”U do not produce precise value of these reactions. As Î”G is independent of pressure for ideal system Î”H and Î”U also independent of pressure.
Hence, Î”G0 = - RT lnKP
or, [dlnKP/dT]T = 0
Show that the equilibrium condition for any chemical reaction is given by Î”G = 0.
Vant Hoff reaction isotherm is,
Î”G = - RT lnKa + RT lnQa
But when the reaction attain equilibrium Qa = Ka
 ∴ Î”G = 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 = -(Î”H/R)(1/T) + C
It is evident that for endothermic reaction (Î”H〉0), increase of T increases the value of a of the reaction.
But for exothermic reaction (Î”Hã„‘ 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.

### Ideal Gas Equation

A gas at equilibrium has definite value of Pressure(P), Volume(V), Temperature(T) and Composition(n). These are called state Variables and are determined experimentally. The state of the gas can be defined by these variables. Boyle's(1662), Charles's(1787) and Avogadro laws gives the birth of an equation of state for Ideal Gas.
 Boyle's law,  V ∝ 1/P When n and T are constant for a gas. Charl's Low,  V ∝ T When n and P are constant for a gas. Avogadro's Low,  V ∝ n When P and T are constant for a gas.
When all the variables are taken into account, The variation rule states that,
V ∝ (1/P) × T × n
or, V = R × (1/P) × T × n
 ∴ PV = nRT
 Ideal Gas Equation.
Where R is the Universal Gas Constant. This is called ideal gas equation of state for ideal gas.
This equation is found to hold most satisfactory when P tense to zero. At ordinary temperature and pressure, the equation is found to deviated about 5%.
Value Of Universal Gas constant (R) at NTP:
At NTP 1 mole gas at 1 atm Pressure occupied 22.4 lit of gas.
Thus, R = (PV)/(nT)
Putting the values above equation,
We have, R = ( 1 atm × 22.4 lit)/(1 mol × 273 K)
 = 0.082 lit atm mol-1 K-1
Value of R in C.G.S. and S.I. system:
P = 1 atm = 76 cm Hg
= 76 cm × 13.6 gm cm-2 × 981 cm sec-2
= 76 × 13.6 × 981 dyne cm-2
Thus, R = (76 × 13.6× 981 dyne cm-2 ×22.4 × 103 cm3)/(1 mol × 273 K)
 = 8.314 × 107 dyne cm2 mol-1 K-1
Again, Work (W) = Force(F) × Displacement(d),
So, erg = dyne cm2.
 Thus, R = 8.314 × 107 erg mol-1 K-1
We Know That, 1 J = 107 erg,
Thus the vale of R in S.I. Unit,
 = 8.314 J mol-1 K-1
Again, 4.18 J = 1 Cal,
hence, R = 8.314 / 4.18 Cal mol-1 K-1
= 1.987 Cal mol-1 K-1
 ≃ 2 Cal mol-1 K-1
Physical Significance of Gas Constant R:
The universal gas constant R = PV/nT
Thus, it has the units of (Pressure × Volume)/(amount of gas × temperature).
Now the dimension of pressure and volume are,
Pressure = (force/area)
= (force/ length2)
= force × length-2
and Volume = length3
R = (force×length-2×length3)/(amount of gas×Kelvin)
= (force × length)/(amount of gas × kelvin)
= (Work or Energy)/(amount of gas × kelvin)
Thus, the dimensions of R are energy per mole per kelvin and hence it represents the amount of work or energy) that can be obtained from one mole of a gas when its temperature is raised by one kelvin.
Determine the value of gas constant R when pressure is expressed in Torr and Volume in dm3.
61.54 Torr dm3 mol-1 K-1
For Solution See Problem 6
Derive the value of R when, (a) pressure is expressed in atom, and volume in cm3and (b) Pressure in dyne m-2 and volume mm3.
(a) 82.05 atm cm3 mol-1K-1
(b) 8.314 × 1014 dyne m-2 mm3 mol-1 K-1
For Solution See Problem 7
Determination of Molar mass from Ideal Gas Equation:
The Ideal Gas Equation is,
PV = nRT
or, PV= (g/M)RT
Where g = weight of the gas in gm and M = Molar mass of the gas.
Again, P = ( g/V) (RT/M)
We know that, Density (d) = Weight (g)/Volume (V).
 ∴ P = dRT/M
Find the Molar mass of ammonia at 5 atm pressure and 300C temperature (Density of ammonia = 3.42 gm lit-1).
17 gm mol-1
For Solution See Problem 8
What is the molecular weight of a gas, 12.8 gms of which occupy 10 liters at a pressure of 750 mm and at 27°C ?
31.91 gm mol-1
For Solution See Problem 9
Determination of Number of Molecule Present in Ideal Gas From Ideal Gas Equation:
The Ideal gas equation for n mole gas is,
PV = nRT
Again, PV= (N/N0) RT
Where N = Number of molecules present in the gas and N0 = Avogadro Number.
Thus, = (N/V) × (R/N0) × T
 ∴ P = N′ k T
Where N′ = number of molecules present per unit Volume and
k = Boltzmann Constant = R/N0
= 1.38 × 10-16 erg molecule-1 K-1
Calculate the number of molecules present per ml of an ideal gas maintained at pressure of 7.6 × 10-3 mm of Hg at 0°C.
We have given that, V = 1ml = 10-6 dm3
P = 7.6 × 10-3 mmHg
= (7.6 × 10-3 mmHg) (101.235 kPa/760 mmHg)
= 1.01235 × 10-3 kPa
Amount of the gas, n = PV/RT
= (1.01235 × 10-3 kPa)(10-6 dm3)/(8.314 kPa dm3 mol-1 K-1)(273 K)
= 4.46 × 10-13 mol
Hence the number of molecules, N = n N0
= (4.46 × 10-13 mol)(6.023 × 1023 mol-1)
= 2.68 × 10-11

### Formulation of Kinetic Gas Equation

After knowing the experimental gas lows, it is of interest to develop a theoretical model based on the structure of gases, which can correlated to the experimental facts. Fortunately such theory has been developed and known as the Kinetic Theory of Gases based upon the certain postulates which are supposed to be applicable to an Ideal Gas.

### Postulates of Kinetic Theory:

1. The gas is composed very small discrete particles, now called molecules. For a gas the mass and size of the molecules are same and different for different gases.
2. The molecules are moving in all directions with verity of speeds. Some are very fast while other are slow.
3. Due to random motion, the molecules are executing collision with the walls of the container (wall collision) and with themselves (inter molecular collision). These collision are perfectly elastic and so there occurs no loss of kinetic energy or momentum of the molecules by this collision.
4. The gas molecules are assumed to be point masses, that is their size is very small in comparison to the distance they travel.
5. There exist no inter molecular attraction specially at law pressure, that is one molecule can exert pressure independent of the influence of other molecules.
6. The pressure exerted by a gas is due to the uniform wall collisions. Higher the frequency of the wall collision greater will be the pressure of the gas.
7. This explains Boyle's law since when volume is reduced, wall collision becomes more frequent and pressure is increased.
8. Through the molecular velocity are constantly changing due to inter molecular collision, average kinetic energy(Ñ”) of the molecules remains fixed at a given temperature.
9. This explain the Charl's law that when T is increased , velocity are increased, wall collision become more frequent and pressure(P) is increased when T kept constant or Volume(V) is increases when P kept constant.
Root Mean Square Speed:
Root mean square(RMS) speed is defined as the square root of the average of the squares of speeds.
 CRMS2 = (N1C12 + N2C22 + N3C32 + ..)/N

### Derivation of the Kinetic Gas Equation:

Let us take a cube of edge length l containing N molecules of a gas of molecular mass m and RMS speed is CRMS at temperature T and Pressure P.
 Molecular Velocity and its Components.
Let in a gas molecules,
N1 have velocity C1
N2 have velocity C2
N3 have velocity C3, and so on.
Let us concentrate our discussion to a single molecule among N1 that have resultant velocity C1 and the component velocities are Cx, Cy and Cz.
 So that, C12 = Cx2 + Cy2 + Cz2
The Molecule will collide walls A and B with the Component Velocity Cx and other opposite faces by Cy and Cz.
Change of momentum along X-direction for a single collision,
= mCx - (- mCx) = 2 mCx
Rate of change of momentum of the above type of collision,
= 2 mCx × (Cx/l)
= 2 mCx2/l
Similarly along Y and Z directions, the rate of change of momentum of the molecule are 2 mCy2/l and 2 mCz2/l respectively.
Total rate change of momentum for the molecule,
 = 2 mCx2/l + 2 mCx2/l +2 mCz2/l  = 2 (m/l) (Cx2 + Cy2 + Cz2) = 2 mC12/l
For similar N1 molecules, it is 2 mN1C12/l
Taking all the molecules of the gas, the total rate of change of momentum,
 = (2 mN1C12/l)+(2 mN2C22/l)+(2 mN3C32/l)+ .. = 2 mN {(N1C12 + N2C22 + N3C13 ..)/N} = 2 mNCRMS2
Where CRMS2 = Root Mean Square Velocity of the Gases.
According to the Newton's 2nd Low of Motion, rate of change of momentum due to wall collision is equal to force developed within the gas molecules.
That is, P × 6l2 = 2 mNCRMS2/l
or, P × l3 = 1/3 mNCRMS2
 ∴ PV = 1/3 mNCRMS2
Here, l3 = Volume of the Cube Contain Gas Molecules.
The other form of the equation is,
PV = 1/3 mN × mN/V × CRMS2
 ∴ PV = 1/3 d CRMS2
Where mN/V is the density(d) of the gas molecules.
 Kinetic Gas Equation
This equation are also valid for any shape of the gas container.
Calculate the pressure exerted by 1023 gas particles each of mass 10-22 gm in a container of volume 1 dm3. The root mean square speed is 105 cm sec-1.
We have, N = 1023
m = 10-22 gm = 10-25 Kg,
V = 1 dm3 = 10-3 m3
and CRMS2 = 105 cm sec-1 = 103 m sec-1
Therefore, from Kinetic Gas Equation,
PV = 1/3 mNCRMS2
or, P = 1/3 × mN/V × CRMS2
Putting the value we have,
P=(1/3)(10-25 Kg×1023/10-3 m3)×(103 m sec-1)2
∴ P = 0.333 × 107 Pa
Expression of Root Mean Square Velocity(CRMS2) from Kinetic Gas Equation:
Let us apply the kinetic equation for 1 mole Ideal Gas.
In that case mN = mN0 = M and PV = RT.
Hence from Kinetic Gas Equation,
PV = 1/3 mNCRMS2
or, RT = 1/3 mNCRMS2
 or, CRMS2 = 3RT/M ∴ CRMS2 = √3RT/M
Thus root mean square velocity depends on the molar mass(M) and temperature(T) of the gas.
Calculate the root mean square speed of oxygen gas at 270C.
We know that, CRMS2 = √3RT/M .
Here, M = 32 gm mol-1 ,and T = 270C = (273+27)K = 300 K.
CRMS2 = √(3 × 8.314 × 107 erg mol-1K-1 × 300 K)/(32 gm mol-1)
= 48356 cm sec-1
Calculate the RMS speed of NH3 at N.T.P.
At N.T.PV = 22.4 dm3 mol-1 = 22.4 × 10-3 m3 mol-1
P = 1 atm = 101325 Pa
and M = 17 × 10-3 Kg mol-1
Thus, CRMS2 = √3RT/M
= √(3 × 101325 × 22.4 × 10-3) /(17 × 10-3)
= 632 m sec-1
Expression of Average Kinetic Energy(Ä’):
The average kinetic energy(Ä’) is defined as,
Ä’ = 1/2 m CRMS2.
Again from Kinetic Gas Equation,
PV = 1/3 mNCRMS2
or, PV = (2/3)  N × (1/2) m CRMS2
or, PV = 2/3 N Ä’
For 1 mole ideal gas,
PV = RT and N = N0
Thus RT = 2/3 N0 Ä’
 or, Ä’ = (3/2)(R/N0)T  ∴ Ä’  = 3/2 kT
Where k = R/N0 and is known as the Boltzmann Constant. Its value is 1.38 × 10-23 JK-1.
The total kinetic energy for 1 mole of the gas is,
 ETotal = N0 (Ä’) = 3/2RT
Thus Average Kinetic Energy is dependent of T only and it is not dependent on the nature of the gas.
Calculate the kinetic energy of translation of 8.5 gm NH3 at 270C.
We know that total kinetic energy for 1 mole of the gas is,
ETotal = (3/2)RT
= (3/2)(2 cal mol-1 K-1 × 300 K)
= 900 cal mol-1
Again 8.5 gm NH3 = (8.5/17) mol = 0.5 mol
Thus Kinetic Energy of 8.5 gm NH3 at 270C is (0.5 × 900) cal
= 450 cal
Calculate the RMS velocity of oxygen molecules having a kinetic energy of 2 K.cal mol-1. At what temperature the molecules have this value of KE ?
T = 673.9K or 400.90C

### Chemical Equilibrium Questions and Answers

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, Î”G10 = - RT lnKP1
For the second reaction, Î”G20 = - RT lnKP2
Since, Î”G20 = 2Î”G10, therefore, it follows that,
- RT lnKP2 = - 2RT lnKP1
 ∴ KP2 = (KP1)2
 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,
Î”G0 = Î”H0 + T[d(Î”G0)/dT]P
Zero superscript is indicating the stranded values.
or, - (Î”H0/T2 ) = -(Î”G0/T2 ) + 1/T[d(Î”G0)/dT]P
or, - (Î”H0/T2 ) = [d/dT(Î”G0/T)]P
Again Vant Hoff isotherm is,
- RT lnKP = Î”G0
or, - R lnKP = Î”G0/T
Differentiating with respect to T at constant P,
- R [dlnKP/dT]P = [d/dT(Î”G0/T)]P
Comparing the above two equation we have,
 dlnKP/dT = (Î”H0/T2
This is the Van't Hoff Equation isochore.
Greater the value of Î”H0, the faster the equilibrium constant(KP)changes with temperature(T), Î”H0 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 Î”H = (-)ve so the equilibrium constant is increased with decreases of temperature.
3. Equilibrium constant remains same though a catalyst is added.
4. Î”G0 = - RT lnKa but Î”G0 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, Î”G0(500 K) = 2 KJ mol-1. Find the KP at 500 K for the reaction A + ½B C.
Î”G0 (500 K) for the reaction,
A + ½B C
= (2 KJ mol-1)/2
= 1 KJ mol-1
The relation is, Î”G0 = - 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 (Î”Î³ ≠ 0), if inert gas is added at constant P (Î”Î³ ≠ 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 Î”H (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 Î”H remains same weather a catalyst is used or not.
Therefore the statement is correct.

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