Home
Class 11
CHEMISTRY
PCl(5) dissociates as follows in a close...

`PCl_(5)` dissociates as follows in a closed reaction vessel `PCI_(5(g)) harr PCl_(3(g)) + Cl_((g))`. If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of `PCl_(5)` is x, the partial pressure of `PCI_(3)` will be

A

`[(x)/(x-1)]P`

B

`[(x)/(1-x)]P`

C

`[(x)/(x+1)]P`

D

`[(2x)/(1-x)]P`

Text Solution

Verified by Experts

The correct Answer is:
C

`P_(Cl_(3))=(x)/(1-x)P`
Promotional Banner

Similar Questions

Explore conceptually related problems

At 200^(@) C PCl_(5) dissociates as follow , PCl_(s(g)) hArr PCl_(3(g)) + Cl_(2(g)) . It was found that the equilibrium vapour. are 62 times as heavy as hydrogen. The degree of dissociation of PCl_(5) at 200^(@) C is nearly.

For PCl_(5 (g)) hArr PCl_(3 (g)) + Cl_(2 (g)) at equilibrium , K_(P) = P//3 . Then degree of dissociation of PCl_(5) at that temperature is

For the reaction PCl_(5(g)) harr PCl_(3(g)) + Cl_(2(g)) . Which of the graph "is"//"are" correct?

For the reaction PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)) the forward reaction at constant temperature is favoured by

Consider the following reaction, PCl_(5) (g) hArr PCl_(3) (g) + Cl_(2)(g) the forward reaction at constant temperature is favoured by

For the reaction N_2O_4 harr 2NO_(2(g)) , the degree of dissociation at equilibrium is 0.2 at 1 atm. Then K_p will be

AB_(2(g)) dissociates as AB_(2(g)) harr AB_(2(g))+B_((g)) . The initial pressure of AB_2 is 600 mm Hg and equilibrium pressure of the mixture of gases is 800 mm Hg. Then

Calculate the extent of dissociation of PCl_5 if the equilibrium pressure is numerically 3 times of its K_p .