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The equilibrium constant for the reactio...

The equilibrium constant for the reaction
`CO_(2)(g) +H_(2)(g) hArr CO(g) +H_(2)O(g) at 298 K` is `73`. Calculate the value of the standard free enegry change `(R =8.314 J K^(-1)mol^(-1))`

Text Solution

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we know that, `DeltaG^(@)=-2.303RT" log "K_(C)`
given, `K_(C)=73,R=8.314JK^(-1)mol^(-1),T=298K`
`therefore`From above equation,
`DeltaG^(@)=-2.303xx8.314xx298log73`
`=-10.632kJ" "mol^(-1)`
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