S ₈ on heating at a temperature above 1000 K, changes to S ₂(g) . When S ₈ is heated above 1000 K at an initial pressure of 1 atm, the partial pressure of S ₈ falls by 32% at equilibrium. The equilibrium constant for the conversion is:

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 3.94 ~atm ^3

S₈(g) 4S₂(g) Initial pressure of S₈ = 1 atm. Pressure falls by 32%: P_ S₈ = 1 - 0.32 = 0.68 atm P_ S₂ = 4 0.32 = 1.28 atm K_p = (P_ S₂ )^4 P_ S₈ = (1.28)^4 0.68 = 2.684 0.68 3.94 atm ^3

S ₈ on heating at a temperature above 1000 K, changes to S ₂(g) . When S ₈ is heated above 1000 K at an initial pressure of 1 atm, the partial pressure of S ₈ falls by 32% at equilibrium. The equilibrium constant for the conversion is:

Options

Correct answer

Step-by-step solution

More from Chemical Equilibrium