phsophorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to
PCl3 (g) +Cl2 (g)->PCl5 (g) Kc = 2.01 at 500k
If a 1.000L reaction vessel is charged with .300 mol of PCl5(g) and allowed to equilibrate at this temperature, what would be the partical pressure of PCl5(g)
To determine the partial pressure of PCl5(g) in the reaction vessel at equilibrium, we need to use the given equilibrium constant (Kc) and the stoichiometry of the balanced equation.
The equilibrium constant expression for the given reaction is:
Kc = [PCl5(g)] / ([PCl3(g)] * [Cl2(g)])
Here, [PCl5(g)], [PCl3(g)], and [Cl2(g)] represent the molar concentrations (in mol/L) of PCl5(g), PCl3(g), and Cl2(g) at equilibrium, respectively.
From the balanced equation, we can see that 1 mole of PCl3 reacts with 1 mole of Cl2 to produce 1 mole of PCl5. So, at equilibrium, the number of moles of PCl3 and Cl2 would be equal to the number of moles of PCl5.
Given that the reaction vessel is initially charged with 0.300 mol of PCl5(g), at equilibrium, the number of moles of PCl3 and Cl2 would also be 0.300 mol each.
Since the reaction vessel has a volume of 1.000 L, the molar concentration of PCl5 at equilibrium can be calculated as:
[PCl5(g)] = (moles of PCl5) / (volume of vessel)
= 0.300 mol / 1.000 L
= 0.300 M
Now, using the equilibrium constant expression:
Kc = [PCl5(g)] / ([PCl3(g)] * [Cl2(g)])
We can rearrange the equation to find [PCl5(g)]:
[PCl5(g)] = Kc * ([PCl3(g)] * [Cl2(g)])
Substituting the values:
[PCl5(g)] = 2.01 * (0.300 M * 0.300 M)
= 0.1809 M
Finally, we need to convert the molar concentration to partial pressure using the ideal gas law:
PV = nRT
Since we are dealing with partial pressure (P) and volume (V), we can rearrange the equation as:
P = (nRT) / V
Substituting the values:
P = (0.1809 M * 0.0821 L·atm/(mol·K) * 500 K) / 1.000 L
= 7.485 atm
Therefore, the partial pressure of PCl5(g) in the reaction vessel at equilibrium is 7.485 atm.