At `800^(@)C,` the following equilibrium is established as `F_(2)(g)hArr2F(g)` The cojmpositionof equilibrium may be determinded by measuring the rate of effusion of theh kmixture through a pin hole. It is found that at `800^(@)C` and `1` atm mixture effuses `1.6` times as fast as `SO_(2)` effuse under the similar conditions. (At. mass of F =`19`) what is the value of `K_(p)` (in atm) ?
A
`0.315`
B
`0.685`
C
`0.46`
D
`1.49`
Text Solution
Verified by Experts
The correct Answer is:
D
Topper's Solved these Questions
CHEMICAL EQUILIBRIUM
NARENDRA AWASTHI|Exercise Level 1 (Q.93 To Q.122)|1 Videos
CHEMICAL EQUILIBRIUM
NARENDRA AWASTHI|Exercise Level 2|1 Videos
ATOMIC STUCTURE
NARENDRA AWASTHI|Exercise Exercise|273 Videos
DILUTE SOLUTION
NARENDRA AWASTHI|Exercise Level 3 - Match The Column|1 Videos
Similar Questions
Explore conceptually related problems
The composition of the equilibrium mixture for the equilibrium Cl_(2) hArr 2 Cl at 1400 K may be determined by the rate of diffusion of mixture through a pin hole. It is found that at 1400 K, the mixture diffuses 1.16 times as fast as krypton diffuses under the same conditions. Find the degree of dissociation of Cl_(2) equilibrium
The composition of the equilibrium mixture for the equilibrium Cl_(2)hArr2Cl at 1470^(@)K , may be determined by the rate of diffusion of mixture through a pin hole. It is found that at 1470^(@)K , the mixture diffuses 1.16 times as fast as krypton (83.8) diffuses under the same conditions. Calculate the % degree of dissociation of Cl2 at equilibrium.
The composition of the equilibrium mixture ( Cl_(2) 2Cl ) , which is attained at 1200^(@)C , is determined by measuring the rate of effusion through a pin hole. It is observed that a 1.80 mm Hg pressure, the mixture effuses 1.16 times as fact as krypton effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms (atomic weight of Kr is 84 ).
At what temperature, the rate of effusion of N_(2) would be 1.625 times that of SO_(2) at 50^(@)C ?
