Home
Class 11
CHEMISTRY
Equal moles of F(2(g)) and Cl(2(g)) are...

Equal moles of `F_(2(g)) and Cl_(2(g))` are introduced into a sealed container and heated to temperature T to attain the following equilibria. `Cl_(2(g)) + F_(2(g)) hArr 2C//F_((g)) , Kp_(I)= 3.2 , Cl_(2(g)) + 3 F_(2(g)) hArr 2 CIF_(2(g)) , Kp_(2) `. If partial pressures of CIF and `CIF_(3)` at equilibria are 0.2 and 0.04 atm respectively. What is the value of `Kp_(2)` ?

A

14.66 `atm^(-2)`

B

1.466 `atm^(-2)`

C

14.66 `atm^(-2)`

D

0.1466 `atm^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
A

`:. 3.2=((0.2)^(2))/((P-0.12)(P-0.16)) implies P=0.2535` `:. 0.2535-0.12" "0.2535-0.16" "0.04`
`=0.1335" "=0.0935`
`Kp_(2)=((0.04)^(2))/((0.1335)(0.0952)^(3))=14.66 atm^(-2)`
Promotional Banner

Similar Questions

Explore conceptually related problems

For the reaction C(s) + CO_(2) (g) hArr 2CO(g). the partial pressures of CO_(2) and CO are 2.0 and 4.0 atm respectively at equilibrium. What is the value of K_(p) for this reaction?

For the reaction CO_((g)) + Cl_(2(g)) hArr COCl_(2(g)). The K_(p)//K_(c) is equal to

The K_(p) values for the reaction H_(2)(g)+I_(2)(g)hArr2HI(g) at 460^(@)C is 49. If the initial pressure of H_(2) and I_(2) are 0.5 atm respectively, determine the partial pressure of each gases at equilibrium.

Variation of K_(P) with temperature T is given by for the following equlibrium 2NH_(3(g)) harr N_(2(g))+3H_(2(g)) Thus, decomposition of NH_(3(g)) can be prevented by

Variation of K_(p) with temperature T is given by for the following cqulibrium 2NH_(3(g)) hArr N_(2(g)) + 3H_(2(g)) Thus, decomposition of NH_(3(g)) can be prevented by

For the reaction CO_((g))+CI_(2(g)) harr COCI_(2(g)) . The K_p // K_c is equal to

Process (A) : F_(2(g)) + 2e^(-) rarr 2F^(-)_((g)) , Process (B) : Cl_(2(g)) +2e^(-) rarr 2Cl_(g)^(-) which of these process is easy ? Why ?