Equilibrium-Constant |

For the dissociation N

_{2}O_{4}⇆ 2NO_{2}, obtain an expression for the fraction of original N_{2}O_{4}dissociated at equilibrium in terms of K_{P}and total pressure.
The reaction is

**N**_{2}O_{4}⇆ 2NO_{2}
Let a mole of N

_{2}O_{4}is taken initially and x moles of N_{2}O_{4}is dissociated at equilibrium then mole number of N_{2}O_{4}and NO_{2}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,

K

_{P}= (P_{N}_{O2})^{2}/P_{N2O4}= {2x/(a + x)}P**/**{(a - x)/(a + x)}P
or, K

_{P}= (4x^{2}P)/(a^{2}- x^{2})
The fraction of the original N

_{2}O_{4}dissociated at equilibrium ɑ = x/a.
Replacing (x/a) by ɑ, we have

**K**

_{P}= (4ɑ^{2}P)/(1 - ɑ^{2})
At 100

^{0}C the vapour density of N_{2}O_{4}is 25 at 1 atm. Show that**K**= 9.6._{P}
Let 1 moles of a is taken initially (t = 0) and x mole of N

_{2}O_{4}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 V

_{1}to V_{2}.
So, (1+x) =

_{ }V_{2}/V_{1}.
As density and hence vapour density is inversely proportional to volume so vapour density will decrease from d

_{1}to d_{2}.
Hence (1+x) =

_{ }V_{2}/V_{1}_{= }d_{1}/d_{2}.
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

or, x = 0.84

Now partial pressure are, P

_{NO2 }
= {2x/(1+x)}P = (2×0.84)/1.84 = 0.913 atm

and P

_{N2}_{O4 }= {(1-x)/(1+x)}P = 0.16/1.84 = 0.087**∴ K**

_{P}= (P_{N}_{O2})^{2}/P_{N2O4}= (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 lnK

_{a}+ RT lnQ_{a}
But when the reaction attain equilibrium Q

_{a}= K_{a}**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 ΔG

^{0}= ΔH^{0}+ T[d(ΔG^{0})/dT]_{P}
Zero superscript is indicating the stranded values.

or, - (ΔH

^{0}/T^{2}) = -(ΔG^{0}/T^{2}) + 1/_{T}[d(ΔG^{0})/dT]_{P}
or, - (ΔH

^{0}/T^{2}) = [d/dT(ΔG^{0}/T)]_{P}
Again Vant Hoff isotherm is, - RT lnK

_{P}= ΔG^{0}
or, - R lnK

_{P}= ΔG^{0}/T
Differentiating with respect to T at constant P,

- R [dlnK

_{P}/dT]_{P}= [d/dT(ΔG^{0}/T)]_{P}
Comparing the above two equation we have,

**(dlnK**

_{P}/dT) = (ΔH^{0}/T^{2})
This is the Vant Hoff reaction isochore.

Greater the value of ΔH

^{0}, the faster the equilibrium constant(K_{P})changes with temperature(T)
ΔH

^{0}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.

ΔG

^{0}= - RT lnK_{a}but ΔG^{0}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 (K

_{a}) remains unchanged.
For a reaction 2A + B ⇆ 2C, ΔG

^{0}(500K) = 2 KJ mol^{-1}
Find the K

_{P}at 500K for the reaction A + ½B ⇆ C .
ΔG

^{0}(500K) for the reaction A + ½B ⇆ C is 2 KJ mol^{-1}/2 = 1 KJ mol^{-1}/2
The relation is ΔG

^{0}= - RT lnK_{P}
or, 1 KJ mol

^{-1}= - 8.31 × 10^{-3}KJ mol^{-1}K^{-1}× 500K lnK_{P}
or, lnK

_{P}= 1/(8.31 × 0.5) = 0.2406**∴ K**

_{P}= 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 N

_{2}+ 3 H_{2}⇆ 2NH_{3}
Equilibrium yield of NH

Examples of other cases can also be cited to illustrate._{3 }is increased if P is increased though equilibrium constant kept fixed.
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.