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Answer: The final concentration of [tex]Cl_2[/tex] at equilibrium is 0.36 M
Explanation:
Moles of [tex]PCl_5[/tex] = 0.64 mole
Volume of solution = 1.0 L
Initial concentration of [tex]PCl_5[/tex] = [tex]\frac{moles}{Volume}=\frac{0.64mol}{1.0L}=0.64 M[/tex]
The given balanced equilibrium reaction is,
[tex]PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)[/tex]
Initial conc. 0.64 M 0M 0M
At eqm. conc. (0.64-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[Cl_2]\times [PCl_3]}{[PCl_5]}[/tex]
Now put all the given values in this expression, we get
[tex]0.47=\frac{(x)\times (x)}{(0.64-x)}[/tex]
By solving the term 'x', we get :
x = 0.36
Thus, the final concentration of [tex]Cl_2[/tex] at equilibrium is x = 0.36 M
The concentration of Cl2 is 0.36M.
Number of moles of PCl5 = 0.64 mol
Volume = 1.0 L
Concentration of PCl5 = 0.64 mol/1.0 L = 0.64 M
The equation of the reaction is;
PCl5 (g)→ PCl3 (aq) + Cl2 (g)
I 0.64 0 0
C -x +x +x
E 0.64 - x x x
Kc = [PCl3] [Cl2]/[PCl5]
0.47 = x^2/0.64 - x
0.47(0.64 - x) = x^2
0.3 - 0.47x =x^2
x^2 + 0.47x - 0.3 = 0
x = 0.36 M
The concentration of Cl2 is 0.36M.
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