Phosphorus tribromide decomposes to form phosphorus and bromine, 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.

Kp = 929551.4

Explanation:

First of all, we state the equilibrium:

4PBr₃ (g) ⇄ P₄(g) + 6Br₂(g)

In order to determine Kp, we need the partial pressure of each gas at equilibrium. Expression for Kp is:

{(Parial Pressure P₄) . (Partial Pressure Br₂)⁶} / (Partial Pressure PBr₃)⁴

Kp = 99.2 . 97.2⁶ / 97.4⁴

Kp = 929551.4

Take account that Kp can be also calculated from Kc.

Kp = Kc (RT)^Δn where Δn is the value for (final moles - initial moles) of any gas

the potential energy of the reactants is 200j.

from the energy diagram, the energy of the product formed is 350j; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.

the answer is 1 i believe