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15 moles of H(2) and 5.2 moles of I(2) a...

15 moles of `H_(2)` and `5.2` moles of `I_(2)` are mixed are allowed to attain equilibrium at `500^(@)C`. At equilibrium the concentration of HI is found to be 10 moles. The equilibrium constant for the formation of HI is

A

50

B

15

C

100

D

25

Text Solution

Verified by Experts

The correct Answer is:
A
`H_2+I_2 hArr 2HI`
`{:(t =0 , 15 , 5.2,0),("at eqm",(15-5),(5.2-5),10):}`
`K_C = ([HI]^2)/([H_2][I_2])=(10 xx10)/(10xx0.2) = 50`
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