the equilibrium contant kc=49 at 230^0C, for the reaction PCl3 + Cl2= PCL5. If 0.50mol of each of phosphorus trichloride and chlorine are added to a 5.0L reaction vessel, what is the equilibrium composition of the mixture at 230 degrees celcias?
chemistry i mean sory
To determine the equilibrium composition of the mixture at 230 degrees Celsius, we'll need to use the equilibrium constant (Kc) and the given initial amounts of the reactants.
First, let's write the balanced equation for the reaction:
PCl3 + Cl2 → PCl5
From the equation, we can see that the stoichiometric ratio (mole ratio) between PCl3 and PCl5 is 1:1, and between Cl2 and PCl5 is also 1:1.
Given that 0.50 mol of PCl3 and 0.50 mol of Cl2 are added to a 5.0 L reaction vessel, we can calculate their initial concentrations (in mol/L):
[PCl3] = (0.50 mol) / (5.0 L) = 0.10 M
[Cl2] = (0.50 mol) / (5.0 L) = 0.10 M
Now, using the equilibrium constant equation:
Kc = [PCl5] / ([PCl3] * [Cl2])
We know that Kc = 49, so we can rearrange the equation to solve for [PCl5]:
[PCl5] = Kc * ([PCl3] * [Cl2])
Substituting the given values:
[PCl5] = (49) * (0.10 M) * (0.10 M)
≈ 0.49 M
Therefore, the equilibrium composition of the mixture at 230 degrees Celsius will have approximately 0.49 M of PCl5.