`9.2` grams of `N_(2)O_(4(g))` is taken in a closed one litre vessel and heated till the following equilibrium is reached `N_(2)O_(4(g))hArr2NO_(2(g))`. At equilibrium, `50% N_(2)O_(4(g))` is dissociated. What is the equilibrium constant (in mol `litre^(-1)`) (Molecular weight of `N_(2)O_(4) = 92`) ?
A
`0.1`
B
`0.4`
C
`0.2`
D
`2`
Text Solution
Verified by Experts
The correct Answer is:
c
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
9.2gm of N_(2)O_(4) (g) is taken is a closed one litre vessel and heated till the following equilibrium is reached N_(2)O_(4) (g) harr 2 NO_(2) (g) . At equilibrium 50% of N_(2)O_(4)(g) is dissociated. What is the equilibrium constant (in mole lit^(-1) ) ?.(M.wt.of N_(2)O_(4) is 92)
18.4 g of N_(2)O_(4) is taken in a 1 L closed vessel and heated till the equilibrium is reached. N_(2)O_(4(g))rArr2NO_(2(g)) At equilibrium it is found that 50% of N_(2)O_(4) is dissociated . What will be the value of equilibrium constant?
For the following gases equilibrium, N_(2)O_(4)(g)hArr2NO_(2)(g) K_(p) is found to be equal to K_(P) . This is attained when:
For the following equilibrium N_(2)O_(4)(g)hArr 2NO_(2)(g) K_(p) is found to be equal to K_(c) . This is attained when :
In the given reaction N_(2)(g)+O_(2)(g) hArr 2NO(g) , equilibrium means that
For the reaction N_(2)O_(4)(g)hArr2NO_(2)(g) , the degree of dissociation at equilibrium is 0.2 at 1 atm pressure. The equilibrium constant K_(p) will be
For the following equilibrium in gaseous phase, N_(2)O_(4)hArr 2NO_(2) NO_(2) is 50% of the total volume, when equilibrium is set up. Hence, percent dissociation of N_(2)O_(4) is :
For the following equilibrium reaction N_(2)O_(4)(g)hArr 2NO_(2)(g) , NO_(2) is 50% of the total volume at a given temperature. Hence, vapour density of the equilibrium mixture is :