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

4pbr3(> 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.

Kp = 9.3x10⁴

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

radioactive decay occurs in unstable atomic nuclei – that is, ones that don't have enough binding energy to hold

the nucleus together due to an excess of either protons or neutrons.