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At 298K, the standard free energy of for...

At 298K, the standard free energy of formation of `H_(2)O(l)=-237.13kJ*mol^(-1)`. Calcualte the value of equilibrium constant at that temperature for the following reaction: `2H_(2)O(l)to2H_(2)(g)+O_(2)(g)`.

Text Solution

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Standard free energy change for the given reaction.
`DeltaG^(0)=2DeltaG_(f)^(0)[H_(2)(g)]+DeltaG_(f)^(0)[O_(2)(g)]-2DeltaG_(f)^(0)[H_(2)O(l)]`
The standard free energy of formation of an element is taken as zero, so, `DeltaG_(f)^(0)[O_(2)(g)]=0 and DeltaG_(f)^(0)[H_(2)(g)]=0`.
Therefore, `DeltaG^(0)=[0+0-2xx(-237.13)]kJ=+474.26kJ`
We know, `DeltaG^(0)=-RTlnK`.
`therefore +474.26xx10^(3)=-8.314xx298lnK or, ln K=-191.42`
`therefore K=7.36xx10^(-84)`
so, equilibrium constant for the reaction`=7.36xx10^(-84)`.
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