A mixture of `0.3` mole of `H_(2)` and `0.3` mole of `I_(2)` is allowed to react in a 10 litre evacuated flask at `500^(@)`C. The reaction is `H_(2) + I_(2) hArr 2H`, the K is found to be 64. The amount of undreacted `I_(2)` at equilibrium isA. `0.15` moleB. `0.06` moleC. `0.03` moleD. `0.2` mole

A mixture of `0.3` mole of `H_(2)` and `0.3` mole of `I_(2)` is allowe

1.	A mixture of `0.3` mole of `H_(2)` and `0.3` mole of `I_(2)` is allowed to react in a 10 litre evacuated flask at `500^(@)`C. The reaction is `H_(2) + I_(2) hArr 2H`, the K is found to be 64. The amount of undreacted `I_(2)` at equilibrium isA. `0.15` moleB. `0.06` moleC. `0.03` moleD. `0.2` mole
Answer» Correct Answer - B `K_(c) = ([HI]^(2))/([H_(2)][I_(2)]), 64 = (x^(2))/(0.03xx0.03)` `x^(2) = 64xx9xx10^(-4)` `x = 8xx3xx10^(-2) = 0.24` x is the amount of HI at equilibrium amount of `I_(2)` at equilibrium will be `0.30 - 0.24 = 0.06`

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A mixture of `0.3` mole of `H_(2)` and `0.3` mole of `I_(2)` is allowed to react in a 10 litre evacuated flask at `500^(@)`C. The reaction is `H_(2) + I_(2) hArr 2H`, the K is found to be 64. The amount of undreacted `I_(2)` at equilibrium isA. `0.15` moleB. `0.06` moleC. `0.03` moleD. `0.2` mole

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