# Chemical Equilibrium

Chemical Equilibrium

## 40% of a mixture of 0.2 mol of N2 and 0.6 mol of H2 react to give NH3 according to the equation: N2(g) + 3H2(g) ⇔ 2NH3(g) at constant  temperature  and pressure. Then the ratio of the final volume to the initial volume of gases are

Q: 40% of a mixture of 0.2 mol of N2 and 0.6 mol of H2 react to give NH3 according to the equation: N2(g) + 3H2(g) ⇔ 2NH3(g) at constant  temperature  and pressure. Then the ratio of the final volume to the initial volume of gases are

(a) 4:5

(b) 5:4

(c) 7:10

(d) 8:5

Sol: N2  +  3H2   ⇔  2NH3

t = 0,  0.2        0.6                  0

at = t  ,    0.2–n   0.6–3x            2x

40% of N2 = 0.2 ×0.4 = 0.08

40% of H2= 0.6 ×0.4 = 0.24

∴ number of moles of N2 remaining = 0.2 – 0.08 = 0.12

number of moles of H2 remaining  = 0.6–0.24 = 0.36

number of moles of NH3 formed = 0.16

Total number of moles = 0.12 + 0.36 + 0.16 = 0.64

$\frac{Final \; Volume}{Initial\; Volume} = \frac{Final \; moles}{Initial\; moles}$

=0.64/0.80 = 4/5

Ans: (A)

## One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when N2O4 (g) decomposes to NO2(g). If the resultant pressure is 2.4 atm, the percentage dissociation by mass of N2O4 (g) is

Q: One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when N2O4 (g) decomposes to NO2(g). If the resultant pressure is 2.4 atm, the percentage dissociation by mass of N2O4 (g) is

(A) 10%

(B) 20%

(C) 30%

(D) 40%

Sol: N2O4(g)   ⇔     2NO2(g)

Equilibrium      1–α        2α

Where α is the degree of dissociation

$\frac{(P_2) Pressure \; of \;N_2 O_4 \;at \;600 K}{(P_1)Pressure \; of \;N_2 O_4 \;at \; 300 K} = \frac{600}{300}$ ( Q V is constant )

∴ P2 = 2 atm

After dissociation of N2O4 at 600K,

PN2O4 = 2 (1–α) = 2–2a

= 2 × 2α = 4 α

Total pressure = 2 – 2α + 4α = 2 + 2α

2 + 2α = 2.4      (Given)

α = 0.2

∴ Percentage dissociation  = 20%

Ans: (B)

## K for the synthesis of HI (g) is 50. The degree of dissociation of HI is

Q:  K for the synthesis of HI (g) is 50. The degree of dissociation of HI is

(A) 0.10

(B) 0.14

(C) 0.18

(D) 0.22

Sol: 2HI(g)  ⇔  H2(g) + I2(g)

1–α            α/2            α/2

Where α is the degree  of dissociations

Kdiss = 1/KSynthesis = 1/50

$\frac{1}{50} = \frac{\alpha/2 \times \alpha/2}{(1-\alpha)^2}$

5√2 α = 2 – 2α

2 α + 5 √2 α = 2

α  = 0.22

Ans: (D)

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## KP for a reaction at 25°C is 10 atm. The activation energy for forward and reverse reactions are 12 and 20 kJ / mol respectively. The KC for the reaction at 40°C will be

Q:  KP for a reaction at 25°C is 10 atm. The activation energy for forward and reverse reactions are 12 and 20 kJ / mol respectively. The KC for the reaction at 40°C will be

(A) 4.33 ×10–1 M

(B ) 3.33 ×10–2 M

(C) 3.33 × 10–1 M

(D) 4.33 × 10–2 M

Sol: Enthalpy changed of a reaction is given by

ΔH = (Ea)f – (Ea)b

Where (Ea)f and (Ea)b are energies of activation for the forward and backward reactions.

ΔH = 12–20 = –8 kJ / mol

KP for the reaction at 25°C = 10 atm. Since KP is expressed in
atmosphere, Δn = +1

QKP = KC (RT)Δn,    KC = 10/(0.0821 ×298) = 0.4 M

KC at 40°C is given by

$log\frac{(K_c)_{40}}{(K_c)_{25}} = \frac{\Delta H}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

$= \frac{-8 \times 1000}{2.303 \times 8.314} \times \frac{15}{298 \times 313}$

= –0.06719

(KC)40/(KC)25 = 0.85

(KC)40 = 0.85 × 0.4 = 0.34 M

Ans: (C)

## In a closed container at 1 atm pressure 2 moles of SO2(g) and 1 mole of O2(g) were allowed to react to form SO3(g) under the influence of acetalyst. Reaction 2SO2(g) + O2(g) ⇔ 2SO3(g) occurred . At equilibrium it was found that 50% of SO2(g) was converted to SO3(g). The formal pressure of O­2 (g) at equilibrium will be

Q:  In a closed container at 1 atm pressure 2 moles of SO2(g) and 1 mole of O2(g) were allowed to react to form SO3(g) under the influence of acetalyst. Reaction 2SO2(g) + O2(g) ⇔ 2SO3(g) occurred . At equilibrium it was found that 50% of SO2(g) was converted to SO3(g). The formal pressure of O­2 (g) at equilibrium will be

(A) 0.66 atm

(B) 0.493 atm

(C) 0.33 atm

(D)  0.20 atm

Sol:  For  ,  H2(g)   +   O2(g) ⇔     2SO3(g)

t = 0         2 moles     1 moles                0

t = equilibrium ,    2(1 – 0.5)  1(1– 0.5)       1 mole

2–1 + 0.5 +1  = 2.5

P’ = P × mF

P2‘ = (P × 0.5)/2.5   = 1 ×(1/5)  = 0.20 atm

Ans:  (D)