The vapour pressure of mercury is 0.002 mm Hg at 27° C Hg(.l.)leftharpoons Hg(.g.) The value of equilibrium constant KC is

Question

The vapour pressure of mercury is 0.002 mm Hg at 27° C Hg(.l.)leftharpoons Hg(.g.) The value of equilibrium constant KC is (A) 1.068× 10- 7 M (B) 0.002 M (C) 8.12× 10- 5 M (D) 3.9× 10- 5 M

Tardigrade Admin · Accepted Answer

Hg(.l.)leftharpoons Hg(.g.) Δ ng=1 Kp=(0.002/760)=2.63× 10- 6 atm K p = K c ( RT )Δ n K c =( K p / RT )=(2.63 × 10-6/0.0821 × 300) =1.068 × 10-7 M

Q. The vapour pressure of mercury is 0.002 mm Hg at $2 7^{\circ} C$

$H g (l) ⇌ H g (g)$

The value of equilibrium constant $K_{C}$ is

$1.068 \times 1 0^{- 7}$ M

$0.002$ M

$8.12 \times 1 0^{- 5}$ M

$3.9 \times 1 0^{- 5}$ M

$H g (l) ⇌ H g (g)$

$Δ n_{g} = 1$

$K_{p} = \frac{0.002}{760} = 2.63 \times 1 0^{- 6}$ atm

$K_{p} = K_{c} (RT)^{Δ n}$

$K_{c} = \frac{K _{p}}{RT} = \frac{2.63 \times 1 0 ^{- 6}}{0.0821 \times 300}$

$= 1.068 \times 1 0^{- 7} M$

Q. The vapour pressure of mercury is 0.002 mm Hg at 27∘C Hg(l)⇌Hg(g) The value of equilibrium constant KC​ is

1.068×10−7 M

0.002 M

8.12×10−5 M

3.9×10−5 M