Question

Equilibrium constant of a reaction is 0.008. Calculate the free energy charge at 298K.​

Answer 1

Equilibrium constant = 0.008 at 298 k

∆G° = -RTIn K

= -8.314 × 298 × In 0.008

= 11962 joule

here R = ideal gas constant

value of R = -8.314 J/mol/K

your answer is + 11.96 kJ

THIS IS YOUR ANSWER....

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Answer 2

The standard free energy change of the reaction is 11.6 kJ

Explanation:

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = 298 K

$K_{eq}$ = equilibrium constant = 0.008

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/Kmol)\times 298K\times \ln (0.008)\\\\\Delta G^o=11962.5J=11.9kJ$

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