Pure PCL5 is introduced into an evacuated chamber and comes to equilibrium at 247C and 2 atm. The equilibrium gaseous mixture contain 40% chlorine by volume. Calculate Kp for the reaction.
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To calculate Kp for the reaction, we first need to write the balanced chemical equation for the reaction:
PCl5(g) ⇌ PCl3(g) + Cl2(g)
The equilibrium constant expression for this reaction in terms of partial pressures is:
Kp = (P[PCl3] * P[Cl2]) / P[PCl5]
Given that the equilibrium gaseous mixture contains 40% chlorine by volume, we can assume that the total volume of the mixture is 100 volumes. Therefore, the volume of PCl5 is 100 volumes.
To convert the volume of each gas to partial pressure, we can use the ideal gas law:
PV = nRT
Since we have the same volume and temperature for all gases, we can ignore the constants and write:
P[PCl5] = P[PCl3] + P[Cl2]
Since the equilibrium gaseous mixture contains 40% chlorine by volume, the volume of Cl2 is 40 volumes. Therefore, the volume of PCl3 is (100 - 40) = 60 volumes.
Now we can calculate Kp:
Kp = (P[PCl3] * P[Cl2]) / P[PCl5]
= (60/100 * 40/100) / (100/100)
= (0.6 * 0.4) / 1
= 0.24
Therefore, Kp for the reaction is 0.24.
To calculate the equilibrium constant (Kp) for the reaction, we first need to write the balanced equation for the reaction involving pure PCL5:
PCL5(g) ⇌ PCL3(g) + Cl2(g)
From the given information, we know that the equilibrium gaseous mixture contains 40% chlorine by volume. This implies that the volume ratio of Cl2 to the total gaseous mixture is 40:100, or simply 0.4.
Now, let's assume that the initial volume of pure PCL5 in the chamber is V. At equilibrium, the volume of Cl2 will be 0.4V, since it accounts for 40% of the total volume.
First, we need to determine the partial pressure of Cl2 at equilibrium. We know that the total pressure is 2 atm. Therefore, the partial pressure of Cl2 is 0.4 times the total pressure:
P(Cl2) = 0.4 * 2 atm = 0.8 atm
Next, we can write the expression for Kp using the partial pressures of the gases:
Kp = (P(PCL3) * P(Cl2)) / (P(PCL5))
However, the partial pressure of PCL3 is not provided. But we can use the fact that the system is at equilibrium to assume that PCL3 and PCL5 concentrations are equal at equilibrium. Therefore, we can write:
P(PCL3) = P(PCL5)
Substituting back into the Kp expression, we get:
Kp = (P(PCL5) * P(Cl2)) / (P(PCL5))
Since P(PCL5) cancels out, we are left with:
Kp = P(Cl2)
Therefore, Kp in this case is equal to the partial pressure of Cl2 at equilibrium, which is 0.8 atm.
........PCl5 ==> PCl3 +Cl2
XCl2 = 0.40
Ptotal = 2 atm
pCl2 = XCl2*Ptotal = 0.4*2
PPCl5 = Ptotal - PCl2 = ?
Substitute into Kp expression and solve for Kp.