Answer : The correct option is, (B) 0.11 M
Solution :
First we have to calculate the concentration
and
.
![\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7DPCl_3%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DPCl_3%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D)
![\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7DPCl_3%3D%5Cfrac%7B0.70moles%7D%7B1.0L%7D%3D0.70M)
![\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7DCl_2%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DCl_2%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D)
![\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7DCl_2%3D%5Cfrac%7B0.70moles%7D%7B1.0L%7D%3D0.70M)
The given equilibrium reaction is,
![PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)](https://tex.z-dn.net/?f=PCl_3%28g%29%2BCl_2%28g%29%5Crightleftharpoons%20PCl_5%28g%29)
Initially 0.70 0.70 0
At equilibrium (0.70-x) (0.70-x) x
The expression of
will be,
![K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BPCl_3%5D%5BCl_2%5D%7D)
![K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%28x%29%7D%7B%280.70-x%29%5Ctimes%20%280.70-x%29%7D)
Now put all the given values in the above expression, we get:
![49=\frac{(x)}{(0.70-x)\times (0.70-x)}](https://tex.z-dn.net/?f=49%3D%5Cfrac%7B%28x%29%7D%7B%280.70-x%29%5Ctimes%20%280.70-x%29%7D)
By solving the term x, we get
![x=0.59\text{ and }0.83](https://tex.z-dn.net/?f=x%3D0.59%5Ctext%7B%20and%20%7D0.83)
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
at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M
The concentration of
at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M
The concentration of
at equilibrium = x = 0.59 M
Therefore, the concentration of
at equilibrium is 0.11 M