Q 11 :

The reaction A(g)⇌B(g)+C(g) was initiated with the amount 'a' of A(g). At equilibrium it is found that the amount of A(g) remaining is (a-x) at a total pressure of p. The equilibrium constant Kp of the reaction can be calculated from the expression:            [2026]

  • x2a2+x2×p

     

  • x2a2-x2×p

     

  • a+x2x2×p

     

  • a2-x2x2×p

     

(2)

A(g)⇌B(g)+C(g)t=0 (mol)a00t=∞ (mol)a-xxxtotal=a+xt=∞ (mol fraction)a-xa+xxa+xxa+xt=∞ (partial pressure)a-xa+xpxa+xpxa+xp

Kp=pB×pCpA=(xa+xp)×(xa+xp)(a-xa+xp)

      =x2p(a+x)(a-x)=x2pa2-x2



Q 12 :

One mole each of He and A(g) are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium.

A(g)⇌B(g)

Kc for this reaction at 400 K is 4.0. The partial pressures (in atm) of He and B(g) are respectively (at equilibrium)

(Assume He, A(g) and B(g) behave as ideal gases)

(Given: R = 0.082 L atm K-1 mol-1)                            [2026]

  • 3.28, 2.624  

     

  • 2.624, 3.28

     

  • 3.28, 0.656

     

  • 0.656, 6.56

     

(1)

A(g)⇌B(g)t=0 (mol)10t=∞ (mol)1-xxt=∞ (conc)1-x10x10

KC=[B][A]=x/10(1-x)/10=x1-x=4

x=0.8

PHe=nHeRTV=1RTV=1×0.082×40010 L=3.28 atm

PB=nBRTV=xRTV=0.8×0.082×40010 L=2.624 atm



Q 13 :

Given is a concentrated solution of a weak electrolyte AxBy of concentration 'c' and dissociation constant 'K'. The degree of dissociation is given by:          [2026]

  • [K×cx+y-1xxyy]x+y

     

  • (Kcx+y-1xxyy)1x+y

     

  • (cx+y-1xxyyK)x+y

     

  • (cx+y-1xxyyK)1x+y

     

(2)

AxBy⇌xAy++yBx-t=0 (conc.)c00t=∞ (conc.)c(1-α)xcαycα

K=[Ay+]x[Bx-]y[AxBy]=(xcα)x(ycα)yc(1-α)=xxyycx+y-1αx+y(1-α)

For a concentrated solution α≪1, so 1-α≃1

K=xxyycx+y-1αx+y

α=(Kxxyycx+y-1)1x+y



Q 14 :

Consider the following gas phase reaction being carried out in a closed vessel at 25°C.

2A(g)→4B(g)+C(g)

time(min) Total pressure of the system(mm Hg)
30 300
∞ 600

 

The pressure of C(g) at 30 minutes time interval would be ___ mm Hg. (nearest integer)             [2026]



(20)

2A(g)→4B(g)+C(g)t=0 (pressure)P°0    0t=30 min (pressure)P°-2p4p    pP°+3p=300  …(i)t=∞ (pressure)02P°  P°/22.5P°=600  …(ii)

From (i) and (ii)

P°=240 and p=20 

Pressure of C at t = 30 min is p = 20 mmHg.