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At 1000K, the pressure of iodine gas is ...

At `1000K`, the pressure of iodine gas is found to be `0.1 atm` due to partial dissociation of `I_(2)(g)` into `I(g)`. Had there been no dissociation, the pressure would have been `0.07 atm`. Calculate the value of `K_(p)` for the reaction:
`I_(2)(g)hArr2I(g)` .

Text Solution

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To solve the problem, we need to calculate the equilibrium constant \( K_p \) for the reaction: \[ I_2(g) \rightleftharpoons 2I(g) \] ### Step-by-Step Solution ...
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