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For the equilibrium N(2)+3H(2)hArr2NH(3)...

For the equilibrium `N_(2)+3H_(2)hArr2NH_(3),K_(c)` at 1000K is `2.73xx10^(-3)` if at equlibrium `[N_(2)]=2M,[H_(2)]=3M,` the concentraion of `NH_(3)` is

A

`0.00358 M `

B

`0.0358 M`

C

`0.358 M `

D

`3.58 M`

Text Solution

Verified by Experts

The correct Answer is:
C
`K_c =([NH_3]^(2))/([N_2][H_2]^3)`
`implies 2.37 xx10^(-3) = (x^2)/([2][3]^3 )implies x^2 = 0.12798`
x = 0.358 M.
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