Answered By Leonardo F

We need first to determine the amount of moles reacted and formed for each component of the reaction. In this case, we have originally 4.0 mol of our reagent, and nothing initially for the products. If, in the chemical equilibrium, there is 1.8 mol of  $Br_2$Br₂ , we know that the amount formed of  $Br_2$Br₂ is 1.8 mol as well. According to the stoichiometric proportion of the reaction (2 mol of  $PO_2$PO₂ for 1 mol of  $Br_2$Br₂), we will form double of this amount for  $PO_2$PO₂ ( $1.8\times2=3.6$1.8×2=3.6 mol) and, consequentially, the amount of  $PO_2Br$PO₂Br reacted is also 3.6 mol (2:2 proportion). 

Now, we can calculate the amount of  $PO_2Br$PO₂Br in the chemical equilibrium, by subtracting the amount reacted from the initial amount:  $4-3.6=0.4$4−3.6=0.4 mol. So, in the chemical equilibrium, we will have 0.4 mol of  $PO_2Br$PO₂Br , 3.6 mol of  $PO_2$PO₂ and 1.8 mol of  $Br_2$Br₂ .

The expression for the equilibrium constant (Kc) for this reaction is:

 $Kc=\frac{\left[PO_2\right]^2\left[Br_2\right]}{\left[PO_2Br\right]^2}$Kc=[P‍O₂]²[Br₂][PO₂Br]²  

If we divide the amount of moles of each component in the chemical equilibrium by the total volume of the container (2.0 L), we will have their concentrations:

  $\left[PO_2\right]=\frac{3.6mol}{2.0L}$[PO₂]=3.6mol2.0L  = 1.8 mol/L

 $\left[Br_2\right]=\frac{1.8mol}{2.0L}$[Br₂]=1.8mol2.0L  = 0.9 mol/L

 $\left[PO_2Br\right]=\frac{0.4mol}{2.0L}$[PO₂Br]=0.4mol2.0L  = 0.2 mol/L

Putting these values in the expression for Kc:

  $Kc=\frac{\left[PO_2\right]^2\left[Br_2\right]}{\left[PO_2Br\right]^2}=\frac{\left(1.8\right)^2\left(0.9\right)}{\left(0.2\right)^2}\approx73$Kc=[PO₂]²[Br₂][PO₂Br]² =(1.8)²(0.9)(0.2)² ≈73 

The value of Kc is approximately 73.