Question

Phosphorus tribromide decomposes to form phosphorus and bromide, like this:

4PBr3(g)-->P4(g)+6Br2(g)

Also, a chemist finds that a certain temperature at the equilibrium mixture of phosphorus tribromide, phosphorus, and bromine has the following composition:

Compound pressure at equiibrium
PBr3 97.4 atm
P4 99.2 atm
Br2 97.2 atm

Calculate the value of the equilibrium constant Kp for this reaction. Round your answer to 2 significant digits.

Answer 1

Kp = 9.3x10⁴

Step-by-step explanation:

The reaction is this:

4PBr₃ (g) ⇄ P₄ (g) + 6Br₂ (g)

We define Kp from the partial pressures in equilibrium

PBr₃ → 97.4 atm

P₄ → 99.2 atm

Br₂ → 97.2 atm

Kp = (Partial pressure Br₂)⁶ . (Partial pressure P₄) / (Partial pressure PBr₃)⁴

Kp = 97.2⁶ . 99.2 / 97.4⁴

Kp = 929551.4533

929551.4533 → 9.3x10⁴