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The reaction `CH_(3)COOH (l) + C_(2) H_(5) OH (l) hArr CH_(3) COOC_(2)H_(5) (l) + H_(2) O (l)` was carried out at `27^(@)C` by taking one mole of each of the reactants . The reaction reached equilibrium when 2/3 rd of the reactants were consumed. Calculate the free energy change for the reaction `(R= 8*314 " JK "^(-1) " mol "^(-1)).`

Answer» Correct Answer - `-3457*9 " j K"^(-1) " mol"^(-1)`
`K_(c) = (2//3 xx 2//3)/(1//3 xx 1//3)=4. "Then " Delta G=-2*303 " RT log " K_(c).`


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