Respuesta :
Answer : The equilibrium concentration of [tex]PCl_5[/tex] is, 0.50 M
Explanation : Given,
Initial moles of [tex]PCl_5[/tex] = 0.65 mole
Volume of solution = 1.0 L
Moles of [tex]PCl_3[/tex] at equilibrium = 0.15 mole
The balanced equilibrium reaction will be,
[tex]PCl_5\rightleftharpoons PCl_3+Cl_2[/tex]
Initial moles 0.65 0 0
At eqm. (0.65-x) x x
Moles of [tex]PCl_3[/tex] at equilibrium = x = 0.15 mole
Moles of [tex]Cl_2[/tex] at equilibrium = x = 0.15 mole
Moles of [tex]PCl_5[/tex] at equilibrium = (0.65-x) = (0.65-0.15) = 0.50 mole
Now we have to calculate the concentration of [tex]PCl_3,Cl_2\text{ and }PCl_5[/tex] at equilibrium.
Formula used : [tex]Concentration=\frac{Moles}{Volume}[/tex]
[tex]\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}=\frac{0.15mole}{1.0L}=0.15M[/tex]
[tex]\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}=\frac{0.15mole}{1.0L}=0.15M[/tex]
[tex]\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of solution}}=\frac{0.50mole}{1.0L}=0.50M[/tex]
Therefore, the equilibrium concentration of [tex]PCl_5[/tex] is, 0.50 M
Initially, 0.65 mol of PCl₅ is placed in a 1.0 L flask at equilibrium, there is 0.15 mol of PCl₃ in the flask then the concentration of PCl₅ at the equilibrium is 0.50 M.
How do we calculate concentration?
Concentration in terms of molarity of any substance present in solution will be calculated as:
M = n/V, where
n = moles of solute
V = volume of solution
Concentration of PCl₅ = 0.65/1 = 0.65M
Concentration of PCl₃ = 0.15/1 = 0.15M
Given chemical reaction with ICE table will be represent as:
PCl₅ → PCl₃ + Cl₂
Initial: 0.65 0
Change: -0.15 +0.15
Equilibrium: 0.65-0.15 0.15
So, equilibrium concentration of PCl₅ = 0.65-0.15 = 0.50 M
Hence required concentration of PCl₅ is 0.50 M.
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