Question

P +Q ⇒ R + S. If initially the concentration of PandQ are equal but at equilibrium
concentration of S will be twice that of P, then what will be the value of Kc

Answer 1

Answer:

Open_in_app

P(g)+2Q(g)⇌R(g)

At equilibrium 3 M 4 M x M k

c

=

3×4

2

x

just After doubling the volume of the vessel

[P(g)]=1.5 M (

2

3 M

)

[Q(g)]=2 M

P(g)+2Q(g)⇌R(g)

t

2

=0 1.5 M 2 M x

at new eqlb 1.5−y 2−2y x+y

given conc of Q at new eqlb =3 M

⇒2−2y=3

⇒t=

2

−1

M=−0.5 M

So conc at equilibrium are

(p(g))=1.5+0.5=2 M

[Q(g)]=3 M

[R(g)]=x−0.5

k

c

=

2×3

2

x−0.5

equating k

c

3×16

x

=

2×9

x−0.5

3x=8x−4

4=5x

x=

5

4

=0.8

So, k

c

=

3×16

0.8

=1.67×10

−2

conc of R at first eqlb stage =0.8 M

conc of R at second eqlb stage =(0.8−0.5)M

=0.3 M