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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
Standard free energy change `= - 2.303 RT [log_(10) k_(eq)]`
`= -2.303 (8.314) (298) [log_(10) ( 3xx 10^(-29))]`
`=-5705.8 xx 0.47 xx (-29)`
`=162.747.21`
`=162.74 xx 10^3 ~~ 163 KJ`
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