Question

A sample of S8(g) is placed in an otherwise empty rigid container at 1325 K at an initial pressure of 1.00 atm, where it decomposes to S2(g) by the following reaction. S8(g) equilibrium reaction arrow 4 S2(g) At equilibrium, the partial pressure of S8 is 0.25 atm. Calculate Kp for this reaction at 1325 K.

Answer 1

Answer:

324

Explanation:

S₈(g) ⇄ 4S₂(g)

K(P)(equilibrium expesssion) = (P)S₂⁴(g) /(P)S₈(g)

S₂ = 3.00atm

S₈ = 0.25atm⇄

=(3.00)⁴ / 0.25

K = 324

Answer 2

Answer:

Kp for the reaction is 324

Explanation:

Equation of reaction:

One mole of S8 reversibly decomposes to 4 moles of S2

Kp = (PS2)^4/(PS8)

Initial pressure of S8 = 1 atm

Equilibrium pressure of S8 = 0.25 atm

Let the equilibrium pressure of S2 be y

From the equation of reaction, mole ratio of S8 to S2 is 1:4, therefore equilibrium pressure of S8 in terms of y is (1 - 0.25y)

Equilibrium pressure of S8 = (1 - 0.25y) = 0.25

1 - 0.25y = 0.25

1 - 0.25 = 0.25y

0.25y = 0.75

y = 0.75/0.25 = 3

Equilibrium pressure of S2 is 3 atm

Kp = 3^4/0.25 = 81/0.25 = 324