Question

The value of kp for the reaction at 27°C Br2(l) + Cl2(g) ⇌⇌ 2BrCl(g) is '1 atom'. At equilibrium in a closed container partial pressure of BrCI gas is 0.1 atm and at this temperature the vapour pressure of Br2(l) is also 0.1 atm. Then what will be minimum moles of Br2(l) to be added to 1 mole of Cl2 , initially, to get above equilibrium situation please provide answer with accurate explaination​

Answer 1

Br

2

(l)+Cl

2

(g)⇌2BrCl(g) at 27

0

C K

P

=1 Assuming sufficient quantity is there

At t=0 C 10moles 0moles initially to maintain saturated vapour

↓ pressure equilibrium

pressures P

0

Br

2

1.5atm

at t=0

(P)(164)=10×0.082×300

P=1.5atm

At ↱ (doesn't change because of gas liquid physical equilibrium)

Eq'b P

0

Br

0

1.5(1−α) 2(1.5)(α)

Total pressure ⇒

P

0

Br

2

+1.5(1+α)=2.25

0.25+(1+α)1.5=2.25

1+α=

1.5

2

=4/3⇒α=1/3

So 10/3 moles of Cl

2

used to form BrCl

2

↪ So 10/3 moles Br

2

also used.

Next = Moles required to maintain 0.25 atm saturated vapour pressure :

(0.25)(164)=n×(0.082)×(300)

n=5/3moles/:

So minimum initial requirement ⇒10/3+5/3=5moles