You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equilibrium Constant and Standard Free Energy Change

Find Kp for...

Question

Find $K_{p}$ for the following reactio,
$S O_{2} {C l}_{2} (g) ⇌ S O_{2} (g) + {C l}_{2} (g)$
If sulphuryl chloride decomposes to the extent of $91.2$ % at $102^{o} C$ and total pressure $1$ atmosphere

Open in App

Solution

Let us assume that initially, 100 moles of $S O_{2} C l_{2}$ are present.
91.2 moles of $S O_{2} C l_{2}$ decompose to form 91.2 moles of $S O_{2}$ and 91.2 moles of $C l_{2}$ .
$100 - 91.2 = 8.8$ moles of $S O_{2} C l_{2}$ remains.
The mole fraction of $S O_{2} C l_{2}$ at equilibrium $= \frac{8.8}{100} = 0.088$
The partial pressure of $S O_{2} C l_{2}$ at equilibrium $= 0.088 \times 1 a t m = 0.088 a t m$ .
The mole fraction of $S O_{2}$ at equilibrium $= \frac{91.2}{100} = 0.912$
The partial pressure of $S O_{2}$ at equilibrium $= 0.912 \times 1 a t m = 0.912 a t m$ .
The mole fraction of $C l_{2}$ at equilibrium $= \frac{91.2}{100} = 0.912$
The partial pressure of $C l_{2}$ at equilibrium $= 0.912 \times 1 a t m = 0.912 a t m$ .
$K_{P} = \frac{P_{S O_{2}} P_{C l_{2}}}{P_{S O_{2} C l_{2}}}$
$K_{P} = \frac{0.912 a t m \times 0.912 a t m}{0.088 a t m}$
$K_{P} = 9.45 a t m$

Suggest Corrections

Similar questions

{S O}_{2}

{C l}_{2}

3 L

{S O}_{2} {C l}_{2}

{C l}_{2}

1

2

K_{p}

10

{S O}_{2} {C l}_{2} (g) ⇌ {S O}_{2} (g) + {C l}_{2} (g)

S O_{2} C l_{2} (g) ⇌ S O_{2} (g) + C l_{2} (g)

2000

10

C l_{2} =

S O_{2} = 40

K_{p}

\frac{K_{p}}{K_{c}} \frac{i n a t m}{i n M}

S O_{2} {C l}_{2} (g) ⇌ {S O}_{2} (g) + {C l}_{2} (g)

3 \times 10^{a}

a

S O_{2} C l_{2} (g) ⇌ S O_{2} (g) + C l_{2} (g), K_{p} > K_{x}

Join BYJU'S Learning Program