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9.2 grams of N(2)O(4(g)) is taken in a c...

`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

0.2

C

0.4

D

2

Text Solution

Verified by Experts

The correct Answer is:
B
`N_(2)O_(4)" moles "=(9.2)/(92) ="0.1 moles"`
`N_(2)O_(4) Leftrightarrow 2NO_(2)`
`{:(,"Initial",0.1,0),(,"Reacted & formed",0.05,0.1),(,"At equilibrium",0.05, 0.1):}`
`K_(c)=([NO_(2)]^2)/([N_(2)O_(4))`
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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?

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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

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