1.

A mixiture of 0.3 mole of H_(2) and 0.3 mole of I_(2) is allowed to ract in a 10 lotre evacuated falsk at 500^(@)C. The reaction is H_(2)+I_(2)hArr2Hl, the K is found to be 64. The amount of unrecacted I_(2) at equilibrium is

Answer»

`0.15` MOLE
`0.06` mole
`0.03` mole
`0.2` mole

Solution :`K_(C)=([HI]^(2))/([H_(2)][I_(2)]),64=(4x^(2))/((0.3-X)(0.3-x))`
`x^(2)=64xx9xx10^(-4)`
`x=8xx3xx10^(-2)=0.24`
x is the amount of HI at equlibrium amount of `I_(2)` at equlibrium will be `0.30-0.24=0.06`


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