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Showing posts with label CHEMICAL EQULIBRIUM. Show all posts
Showing posts with label CHEMICAL EQULIBRIUM. Show all posts

Dec 23, 2018

Van't Hoff Equation

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/dTG0/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/dTG0/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/




For a reaction 2A + B ⇆ 2C, ΔG0(500K) = 2 KJ mol-1 
Find the KP at 500K 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, KJ mol-1  = - 8.31 × 10-3 KJ mol-1 K-1 × 500K lnKP 
or, lnKP  = 1/(8.31 × 0.5) = 0.2406

∴ KP = 1.27

Plot of lnKP vs 1/T :

For exothermic reaction, ΔH0 = (-) ve.
Examples are the formation of ammonia from Hand N2.
Van't Hoff Equation - Effect on temperature on Equilibrium Constant
Plot of lnKP vs 1/T
For endothermic reaction, ΔH0 = (+) ve.
Examples are the dissociation of HI into Hand I2.
For the reaction, ΔH0 = 0, lnKP is independent of T. Provided ΔS0 does not change much with T.
Since for idel system, H is not a function of P and ΔH0 = ΔH
and Van't Hoff equation is dlnKP/dT = ΔH/RTand 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 reaction 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 = (ΔH0 - ΔƔRT)/RT2
or, dlnKC/dT = ΔU0/RT2
ΔU0 is standerd 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:
(i) The reacting system is assumed to behave ideally and 
(ii) 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 a and b 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
Thus, 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.   




Chemical Equilibrium Questions and Answers

Chemical Equilibrium Questions and Answers
Equilibrium-Constant

For the dissociation N2O4  2NO2, obtain an expression for the fraction of original N2O4 dissociated at equilibrium in terms of KP and total pressure.

The reaction is N2O4  2NO2 
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 
or, KP = (4x2P)/(a2 - x2
The fraction of the original N2O4  dissociated at equilibrium ɑ = x/a.
Replacing (x/a) by ɑ, we have
KP = (2P)/(1 - ɑ2)

At 1000C the vapour density of N2O4 is 25 at 1 atm. Show that KP = 9.6.

Let 1 moles of a 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/V1d1/d2.
Molecular weight of a 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 
and PN2O4  = {(1-x)/(1+x)}P = 0.16/1.84 = 0.087
∴ KP = (PNO2)2/PN2O4 = (0.913)2/0.087 ≃ 9.6



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.

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/dTG0/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/dTG0/T)]P
Comparing the above two equation we have,
(dlnKP/dT) = (ΔH0/T2)
This is the Vant Hoff reaction 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 logK 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 ?

(i) Equilibrium constant remains same when P is increases .
(ii) The reaction is exothermic, hence ΔH = (-)ve so the equilibrium constant is increased with decreases of temperature.
(iii) Equilibrium constant remains same though a catalyst is added.
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(500K) = 2 KJ mol-1 
Find the KP at 500K 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, KJ mol-1  = - 8.31 × 10-3 KJ mol-1 K-1 × 500K 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, in N2 + 3 H2 ⇆ 2NH3
Equilibrium yield of NHis increased if P is increased though equilibrium constant kept fixed.
Examples of other cases can also be cited to illustrate.

Justify or criticise 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.