'a’ moles of PCl5, undergoes, thermal dissociation as: PCl5 leftharpoons PCl3 + Cl2, the mole fraction of PCl3 at equilibrium is 0.25 and the total pressure is 2.0 atmosphere. The partial pressure of Cl2 at equilibrium is:

Question

'a’ moles of PCl5, undergoes, thermal dissociation as: PCl5 leftharpoons PCl3 + Cl2, the mole fraction of PCl3 at equilibrium is 0.25 and the total pressure is 2.0 atmosphere. The partial pressure of Cl2 at equilibrium is: (A) 2.5 (B) 1.0 (C) 0.5 (D) None of these

Tardigrade Admin · Accepted Answer

underset oversetat t=0t=tq underset oversetaa-xPCl5 leftharpoons underset overset0xPCl3 + underset overset0xCl2 Ppcl3 =XPcl3 × PT= 0.25 × 2=0.5 atm PPCl3 =PCl2 =0.5 atm

Q. $^{'} a ’$ moles of $PC l_{5}$ , undergoes, thermal dissociation as: $PC l_{5} ⇌ PC l_{3} + C l_{2},$ the mole fraction of $PC l_{3}$ at equilibrium is $0.25$ and the total pressure is $2.0$ atmosphere. The partial pressure of $C l_{2}$ at equilibrium is:

$2.5$

$1.0$

$0.5$

None of these

$t = tq a tt = 0 a - x a PC l_{5} ⇌ x 0 PC l_{3} + x 0 C l_{2}$
$P_{p c l_{3}} = X_{P c l_{3}} \times P_{T} = 0.25 \times 2 = 0.5$ atm
$P_{PC l_{3}} = P_{C l_{2}} = 0.5$ atm

Q. ′a’ moles of PCl5​, undergoes, thermal dissociation as: PCl5​⇌PCl3​+Cl2​, the mole fraction of PCl3​ at equilibrium is 0.25 and the total pressure is 2.0 atmosphere. The partial pressure of Cl2​ at equilibrium is:

2.5

1.0

0.5