JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Equilibrium

A homogeneous ideal gaseous reaction

AB_{2 (g)} ⇌ A_{(g)} + 2 B_{(g)}

is carried out in a 25 litre flask at

2 7^{\circ} C

. The initial amount of

A B_{2}

was 1 mole and the equilibrium pressure was 1.9 atm. The value of

K_{p}

x \times 1 0^{- 2}

. The value of x is ______ (Integer answer).

[R = 0.08206 dm^{3} atm K^{- 1} mol^{- 1}]

Answer: 73

Solution

For the reaction

AB_{2} ⇌ A + 2 B

, let

α

be the degree of dissociation.

Initial moles:

n_{AB_{2}} = 1

n_{A} = 0

n_{B} = 0

At equilibrium:

n_{AB_{2}} = 1 - α

n_{A} = α

n_{B} = 2 α

Total moles at equilibrium:

n_{total} = (1 - α) + α + 2 α = 1 + 2 α

Using

P V = n R T

1.9 \times 25 = (1 + 2 α) \times 0.08206 \times 300

47.5 = (1 + 2 α) \times 24.618

1 + 2 α = 1.93

α = 0.465

Partial pressures:

P_{AB_{2}} = \frac{1 - α}{1 + 2 α} \times 1.9 = \frac{0.535}{1.93} \times 1.9 = 0.527 atm

P_{A} = \frac{α}{1 + 2 α} \times 1.9 = \frac{0.465}{1.93} \times 1.9 = 0.458 atm

P_{B} = \frac{2 α}{1 + 2 α} \times 1.9 = \frac{0.93}{1.93} \times 1.9 = 0.916 atm

K_{p} = \frac{P _{A} \times P _{B}^{2}}{P _{AB_{2}}} = \frac{0.458 \times ( 0.916 ) ^{2}}{0.527} = 0.73 = 73 \times 1 0^{- 2}

Therefore,

x = 73

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Integer answer