1.

The standard free energy change of a reaction is Delta G^(@)=-115kJ at 298 K. Calculate the equlibrium constant K_(p)"in"logK_(p)(R=8.314jK^(-1)mol^(-1))

Answer»

<P>`20.16`
`2.303`
`2.016`
`13.83`

SOLUTION :The STANDARD free energy change of a reaction `DELTAG^(@)=-2.303RTlogK_(p)`
`-115xx10^(3)=-2.303RT log K_(p)`
`logK_(p)=(115xx10^(3))/(2.303xx8.314xx298)logK_(p)=20.16.`


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