Consider the following gaseous equilibrium in a closed container of volume ' V ' at T(K) . P_ 2 (~g)+Q_ 2 (~g) leftharpoons 2 PQ(g) 2 moles each of P_ 2 (~g), Q_ 2 (~g) and PQ(g) are present at equilibrium. Now one mole

JEE Main 202624 JanuaryEvening Shift

JEE Main 2026 · 24 January · Evening Shift · Chemistry · Chemical Equilibrium

Consider the following gaseous equilibrium in a closed container of...

Consider the following gaseous equilibrium in a closed container of volume ' V ' at

\mathrm{T}(\mathrm{K})

\mathrm{P}_{2}(\mathrm{~g})+\mathrm{Q}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{PQ}(\mathrm{g})

2 moles each of

\mathrm{P}_{2}(\mathrm{~g}), \mathrm{Q}_{2}(\mathrm{~g})

and

\mathrm{PQ}(\mathrm{g})

are present at equilibrium. Now one mole each of '

\mathrm{P}_{2}

' and '

\mathrm{Q}_{2}

' are added to the equilibrium keeping the temperature at

\mathrm{T}(\mathrm{K})

. The number of moles of

P_{2}, Q_{2}

and

P Q

at the new equilibrium, respectively, are

C. $2.67,2.67,2.67$

Initial equilibrium: 2 moles each of

P_2

Q_2

PQ

with

K = \frac{[PQ]^2}{[P_2][Q_2]} = \frac{4}{4} = 1

.
After adding 1 mole each of

P_2

and

Q_2

: initial composition is

(3, 3, 2)

. Reaction quotient

Q = \frac{4}{9} < K

, so reaction shifts right.
Let x moles of reactants combine:

\frac{(2+2x)^2}{(3-x)^2} = 1

2+2x = 3-x

3x = 1

→

x = \frac{1}{3}

Final moles:

P_2 = 3 - \frac{1}{3} = 2.67

Q_2 = 2.67

PQ = 2 + \frac{2}{3} = 2.67