Home
Class 11
CHEMISTRY
The pressure of iodine gas at a particul...

The pressure of iodine gas at a particular temperature is found to be `0.111atm`, where as the expected pressure is `0.074 atm`, the increased pressure is due to `I_(2)hArr 2I`. Calculate `K_(p)` for this equilibrium.

A

`0.074`

B

`0.148`

C

`0.05`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
2

`{:(l_(2)(g),hArr,,2l(g),),(0.074,,,,),(0.074-P,,,2P,):}`
`0.074-P+2P=0.111`
`P=0.037`
`K_(p)=(4P^(2))/(0.074-P)=(4(0.037)^(2))/(0.037)=0.148`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIUM

    RESONANCE|Exercise Solved examples|9 Videos
  • CHEMICAL EQUILIBRIUM

    RESONANCE|Exercise (MSPs)|8 Videos
  • CHEMICAL BONDING

    RESONANCE|Exercise Inorganic chemistry (Chemistry Bonding)|49 Videos
  • D & F-BLOCK ELEMENTS & THEIR IMPORTANT COMPOUNDS

    RESONANCE|Exercise Match the column|1 Videos

Similar Questions

Explore conceptually related problems

The pressure of iodine gas at 1273 K is found to be 0.112 atm whereas the expected pressure is 0.074 atm. The increased pressure is due to dissociation I_(2) hArr 2I . Calculate K_(p) .

At 1000^@C the pressure of iodine gas is found to be 0.112 atm whereas the expected pressure is 0.074 atm. The increased pressure is due to dissociation, I_2 hArr 2I . Calculate K_p . Also find out pressure at which I_2 will be 90% dissociation at 1000^@C .

At a certain temperature and a total pressure of 10^(5) Pa, iodine vaour contain 40% by volume of iodine atmos [I_(2) (g) hArr 2 I (g) ]. " Calculate " K_(p) for the equilibrium.

At 1000K, the pressure of iodine gas is found to be 0.112 atm due to partial dissociation of I_(2)(g) into I(g). Had there been no dissociation, the pressure would have been 0.074 atm. Calculate the value of K_(p) for the reaction: I_(2)(g) hArr 2I(g) .

At a certain temperature and a total pressure of 10^(5) Pa , iodine vapour contains 40% by volume of I "atoms" , Calculate K_(p) for the equilibrium. I_(2(g))hArr2I_((g))

Osmotic pressure of insulin solution at 298 K is found to be 0.0072atm . Hence, height of water Column due to this pressure is

At 1000K , the pressure of iodine gas is found to be 0.1 atm due to partial dissociation of I_(2)(g) into I(g) . Had there been no dissociation, the pressure would have been 0.07 atm . Calculate the value of K_(p) for the reaction: I_(2)(g)hArr2I(g) .