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At 298 k, the equilibrium constant of th...

At 298 k, the equilibrium constant of the process `1.5O_2 (g) hArr O_3 (g) "is" 3xx10^(-29)`. Standard free energy change (in K. J `mol^(-1)` ) of the process is approximately `(R= 8.314 J mol^(-1) k^(-1), log 3 = 0.47)`

A

724

B

612

C

247

D

163

Text Solution

Verified by Experts

The correct Answer is:
D
Free energy change is linked to equilibrium constant `(K_(p))` by the formula.
`DeltaG^(@)=-2.303" RT "log K_(p)`
Value of `R=8.314" J mol"^(-1)K^(-1)`
`T=298" K"`
`K_(p)=3xx10^(-29)`
Substituting these value in equation and solving the value, `Delta G^(@)` obtained is 163 kJ `"mol"^(-1)`.
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