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The equilibrium constant of the reaction...

The equilibrium constant of the reaction
`H_(2)(g)+I_(2)(g)hArr2HI(g)`
at `426^(@)C` is `55.3`, what will be the value of equilibrium constant
a. if the reaction is reversed and
b. if the given reaction is represented as
`3H_(2)+3I_(2)hArr6HI`?

Text Solution

Verified by Experts

The reverse reaction of the given reaction is
`2HIhArrH_(2)+I_(2)`
:. Equilibrium constant `=1/55.3`
b. The reaction
`3H_(2)+3I_(2)hArr6HI`
has been obtained by multiplying the reaction,
`H_(2)I_(2)hArr2HI`
by `3`, hence, `K=(55.3)^(3)`
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