The reaction, `PC1_(5)hArrPC1_(3) + C1_(2)` is started in a 5 litre container by taking one mole of `PC1_(5)`. If `0.3` mole of `PC1_(5)` is there at equilibrium, concentration of `PC1_(3)` and `K_(c)` will respectively be:

Options

A. `0.14, 0.326`
B. `0.12, 0.120`
C. `0.07, 0.150`
D. `20, 0.010`

Answer

A

Explanation

`{:(PCl_(5),hArr,PCl_(3),+,Cl_(2)),(1,,0,,0 " " "Initial mole"),((1-0.7)/5,,(0.7)/5,,(0.7)/5" " Conc.at equilibrium"):}`
Total mole of `PCl_(3)=0.7`
Concentration=0.14
`K_(c) = (x^(2))/((1-x)V) = (0.7xx0.7)/(0.3xx5) = (49)/(150) = 0.326`