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

At 1000K, the pressure of iodine gas is found to be 0.112 atm due to partial dissociation of `I_(2)(g)` into I(g). Had there been no dissociation, the pressure would have been 0.074 atm. Calculate the value of `K_(p)` for the reaction:
`I_(2)(g) hArr 2I(g)`.

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To find the equilibrium constant \( K_p \) for the reaction \[ I_2(g) \rightleftharpoons 2I(g) \] we will follow these steps: ...
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