Home
Class 11
CHEMISTRY
PCl(5), PCl(3) and Cl(2) are at equilibr...

`PCl_(5), PCl_(3) and Cl_(2)` are at equilibrium at 500 K in a closed container and their concentrations are `0.8xx10^(-3) ` mol `L^(-1), 1.2xx10^(-3) ` mol `L^(-1) and 1.2xx10^(-3) ` mol `L^(-1)` respectively. The value of `K_(c)` for the reaction `PCl_(5)(g) hArrPCl_(3)(g)+Cl_(2)(g)` will be

A

`1.8 xx10^(3)"mol L"^(-1)`

B

`1.8 xx10^(-3)`

C

`1.8 xx10^(-3)"mol"^(1)L`

D

`0.55 xx10^(4)`

Text Solution

Verified by Experts

The correct Answer is:
B
For the reaction, `PCl_5 hArr PCl_3 + Cl_2`
At 500K in a closed container,
`[PCl_5] = 0.8 xx10^(-3) " mol L"^(-1)`
`[PCl_3] = 1.2 xx10^(-3) " mol L"^(-1)`
`[Cl_2] = 1.2 xx10^(-3) " mol L"^(-1)`
`K_c=([PCl_3][Cl_2])/([PCl_5])=((1.2 xx10^(-3))xx(1.2 xx10^(-3)))/((0.8 xx10^(-3)))`
`=1.8 xx10^(-3)`
Promotional Banner

Similar Questions

Explore conceptually related problems

For the reaction, PCl_(5)(g)to PCl_(3)(g)+Cl_(2)(g)

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

For PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g), write the expression of K_(c)

Initially, 0.8 mole of PCl_(5) and 0.2 mol of PCl_(30 are mixed in one litre vessel. At equilibrium, 0.4 mol of PCl_(3) is present. The value of K_(c ) for the reaction PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) would be

PCl_(5), PCl_(3) and Cl_(2) are at equilibrium at 500 K and having concentration 1.59 M PCl_(3), 1.59 M Cl_(2) and 1.41 M PCl_(5) . Calculate K_(c) for the reaction, PCl_(5) hArr PCl_(3) + Cl_(2)