Question

`PCI_(5), PCI_(3)` and `CI_(2)` are in equilibrium at 500 K in a closed container and their concentration are `0.8 gt 10^(-3) " mol" L^(-1) " and " 1.2 xx 10^(-3) "mol" L^(-1) " and " 1.2 xx 10^(-3) "mol" L^(-1)` respectively. The value of `K_(c)` for the reaction `PCI_(5) (g) hArr PCI_(3) (g) + CI_(2) (g)` will be
A. `1.8 xx 10^(3) mol L^(-1)`
B. `1.8 xx 10^(-3)`
C. `1.8 xx 10^(-3) mol L^(-1)`
D. `0.55 xx 10^(4)`

Answer 1

`PCI_(5) (g) hArr PCI_(3) (g) +CI_(2)(g)`
`K_(c) =[[PCI_(3)(g)][CI_(2)9(g)]]/[[PCI_(5)(g)]] ("mol " L^(-1))`
`=((1.2 xx 10^(-3) " mol "L^(-1))xx(1.2xx 10^(-3) " mol " L^(-1)))/((0.8 xx 10^(-3) "mol " L^(-1)))`
`= 1.8 xx 10^(-3) " mol " L^(-1)`