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Dec 23, 2018

Chemical Equilibrium Questions and Answers

Chemical Equilibrium Questions and Answers

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.