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The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol of Cl2 in a 1.00-L reaction vessel at 230°C, what is the concentration of PCl3 when equilibrium has been established?A) 0.049 MB) 0.11 MC) 0.59 MD) 0.30 ME) 0.83 M

Respuesta :

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration [tex]PCl_3[/tex] and [tex]

Cl_2[/tex].

[tex]\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}[/tex]

[tex]\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M[/tex]

[tex]\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}[/tex]

[tex]\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M[/tex]

The given equilibrium reaction is,

                            [tex]PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)[/tex]

Initially                 0.70        0.70              0

At equilibrium    (0.70-x)   (0.70-x)           x

The expression of [tex]K_c[/tex] will be,

[tex]K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}[/tex]

[tex]K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}[/tex]

Now put all the given values in the above expression, we get:

[tex]49=\frac{(x)}{(0.70-x)\times (0.70-x)}[/tex]

By solving the term x, we get

[tex]x=0.59\text{ and }0.83[/tex]

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of [tex]PCl_3[/tex] at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of [tex]Cl_2[/tex] at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of [tex]PCl_5[/tex] at equilibrium = x = 0.59 M

Therefore, the concentration of [tex]PCl_3[/tex] at equilibrium is 0.11 M

okpalawalter8

The concentration of PCl_3 is mathematically given as

CPCl_3= 0.11 M, Option B is correct

What is the concentration of PCl3 when equilibrium has been established?

Question Parameters:

Generally, the equation for the Chemical Reaction  is mathematically given as

[tex]PCl_3(g)+Cl_2(g) ---- > PCl_5(g)[/tex]

Therefore

[tex]K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}[/tex]

[tex]K_c=\frac{(x)}{(0.70-x)* (0.70-x)}[/tex]

[tex]49=\frac{(x)}{(0.70-x) * (0.70-x)}[/tex]

x=0.59 and 0.83

In conclusion, The resultant  concentration of PCl_3 at equilibrium

CPCl_3 = (0.70-x)

CPCl_3= (0.70-0.59)

CPCl_3= 0.11 M

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