A 1.500 mol sample of PCL5 dissociates at 160C and 1.00 atm to give 0.203 mol of PCL3 at equilibrium. What is the composition of the final reaction mixture?
To determine the composition of the final reaction mixture, we first need to write and balance the chemical equation for the dissociation of PCl5:
PCl5(g) ⇌ PCl3(g) + Cl2(g)
We are given the initial moles of PCl5 (1.500 mol) and the moles of PCl3 at equilibrium (0.203 mol). Since PCl3 is a product, its moles can be considered as the amount that has "reacted" or been consumed.
To find the moles of Cl2 at equilibrium, we can use the stoichiometry of the balanced equation. From the equation, we can see that for every 1 mole of PCl5 that reacts, 1 mole of PCl3 and 1 mole of Cl2 are formed.
Since we know that 0.203 mol of PCl3 has formed, the same amount of Cl2 will have formed as well.
Now, let's calculate the moles of each component at equilibrium:
Moles of PCl5 at equilibrium = Initial moles - Moles of PCl3 formed
= 1.500 mol - 0.203 mol
= 1.297 mol
Moles of PCl3 at equilibrium = Moles of PCl3 formed
= 0.203 mol
Moles of Cl2 at equilibrium = Moles of PCl3 formed
= 0.203 mol
Therefore, the composition of the final reaction mixture is:
PCl5: 1.297 mol
PCl3: 0.203 mol
Cl2: 0.203 mol