Answer:
the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M
Explanation:
for the reaction
PCl₅(g) → PCl₃(g) + Cl₂(g)
where
Kc= [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M
and [A] denote concentrations of A
if initially the mixture is pure PCl₅ , then it will dissociate according to the reaction and since always one mole of PCl₃(g) is generated with one mole of Cl₂(g) , the total number of moles of both at the end is the same → they have the same concentration → [PCl₃(g)] = [Cl₂]=0.27 M
therefore
Kc= [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M* 0.27 M /[PCl₅] = 20 M
[PCl₅] = 0.27 M* 0.27 M / 20 M = 3.64*10⁻³ M
[PCl₅] = 3.64*10⁻³ M
the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M
The equilibrium concentration of PCl5 is 0.0036 M.
The equation of the reaction is; PCl5(g) ⇄ PCl3(g) + Cl2(g)
The number of moles of PCl3 at equilibrium is = 0.27 M × 1.00-L = 0.27 moles
Now;
Kc = [PCl3] [Cl2]/[PCl5]
Kc = 2.0 × 10^1 or 20
We can see that;
Number of moles of PCl3 = Number of moles of Cl2 = 0.27 moles
Let the equilibrium concentration of PCl5 be x
20 = (0.27)^2/x
x = (0.27)^2/20
x = 0.0036 moles
Since the volume does not change;
equilibrium concentration of PCl5 = 0.0036 moles/1.00-L = 0.0036 M
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