Answer:
Equilibrium concentration of PCl5 is 0.004 M
Equilibrium concentration of PCl3 is 0.079 M
Explanation:
Kc = [Cl2][PCl3]/[PCl5]
Let the equilibrium concentration of PCl3 be y M
Initial concentration of PCl5 = 0.285 mol/3.45 L = 0.083 M
From the equation of reaction, the mole ratio of PCl5 to PCl3 is 1:1, therefore equilibrium concentration of PCl5 is (0.083 - y) M
Also, from the equation of reaction, mole ratio of PCl3 to Cl2 is 1:1, therefore equilibrium concentration of Cl2 is also y M
Kc = 1.8
1.8 = y×y/(0.083 - y)
1.8(0.083 -y) = y^2
y^2 = 0.1494 - 1.8y
y^2 + 1.8y - 0.1494 = 0
The value of y must be positive and is obtained using the quadratic formula.
y = [(-1.8 + sqrt(1.8^2 - 4×1×0.1494)) ÷ 2(1)] = [(-1.8 + sqrt(3.8376)) ÷ 2] = [0.158 ÷ 2] = 0.079
Equilibrium concentration of PCl5 = 0.083 - y = 0.083 - 0.079 = 0.004 M
Equilibrium concentration of PCl3 = y = 0.079 M
Answer:
[PCl₅] = 0.00348 M
[PCl₃] = 0.07912 M
Explanation:
Step 1: Data given
Kc = 1.80 at 250 °C
Number of moles PCl5 = 0.285 moles
Volume = 3.45 L
Temperature = 250 °C
Step 2: The balanced equation
PCl5(g)⇔PCl3(g)+Cl2(g)
Step 3: Calculate molarity of PCl5
Molarity PCl5 = moles PCl 5 / volume
Molarity PCl5 = 0.285 moles / 3.45 L
Molarity PCl5 = 0.0826 M
Step 4: The initial concentration
[PCl5] = 0.0826 M
[PCl3]= 0M
[Cl2] = 0M
Step 5: The concentration at the equilibrium
[PCl5] = (0.0826 - x)M
[PCl3] = xM
[Cl2] = xM
Step 6: Calculate [PCl5] and [PCl3]
Kc = [ (PCl₃) ] [ Cl₂ ] / [PCl₅]
1.80 = x² / (0.0826 - x)
x = 0.07912 M
[PCl₅] = (0.0826 - 0.07912)M = 0.00348 M
[PCl₃] = 0.07912 M
