When 1 mole of pure ethyl alcohol `(C_(2)H_(5)OH)` is mixed with `1` mole of acetic acid at `25^(@)C.` the equilibrium mixture contains `2//3` mole each of ester and water
`C_(2)h_(5)OH(l)+CH_(3)COOH(l)hArrCH_(3)COOC_(2)H_(5)(l)+H_(2)O(l)`
The `DeltaG^(@)` for the reaction at `298 K` is :

When I mole of pure ethyl alcohol (C_(2)H_(5)OH) is mixed with 1 mole of acetic acid at 25^(@)C with one lit volume, the equilibrium mixture contains 2//3 mole each of ester and water. CH_(2)OH_((l)) +CH_(3)COOH_((l)) harr CH_(3)COOC_(2)H_(5(l))+H_(2)O_((l)) The DeltaG^(@) for the reaction at 298 K is:

One mole of ethyl alcohol (C_(2)H_(5)OH) was treated with one mole of acetic acid at 25^(0)C . (2)/(3) of the acid changes in to ester at equilibrium.The equilibrium constant for the reaction is

CH_(3)COOH + C_(2)H_(5)OH overset(X) to CH_(3)COOC_(2)H_(5) +H_(2)O ,X is

When alcohol (C_(2)H_(5)OH(l)) "and acetic acid" (CH_(3)COOH(l)) are mixed together in equimolar ratio at 27^(@)C,33% of each is converted into ester. Then the K_(c) for the equilibrium C_(2)H_(5)OH(l)+CH_(3)COOH(l)hArrCH_(3)COOC_(2)H_(5)(l)+H_(2)O(l) is:

When alcohol (C_(2)H_(5)OH(l)) "and acetic acid" (CH_(3)COOH(l)) are mixed together in equimolar ratio at 27^(@)C,33% of each is converted into ester. Then the K_(c) for the equilibrium C_(2)H_(5)OH(l)+CH_(3)COOH(l)hArrCH_(3)COOC_(2)H_(5)(l)+H_(2)O(l) is:

The equilibrium constant for the reaction: CH_(3)COOH(l) +C_(2)H_(5)OH(l) hArr CH_(3)COOC_(2)H_(5)(l)+H_(2)O(l) has been found to be equal to 4 at 25^(@)C . Calculate the free energy change for the reaction.

The equilibrium constant for the reaction: CH_(3)COOH(l) +C_(2)H_(5)OH(l) hArr CH_(3)COOC_(2)H_(5)(l)+H_(2)O(l) has been found to be equal to 4 at 25^(@)C . Calculate the free energy change for the reaction.

Mention two ways by which the equilibrium of the given reaction can be shifted to the right. CH_(3)COOH(l)+C_(2)H_(5)OH(l) overset(H^(+))hArr CH_(3)COOC_(2)H_(5)(l)+H_(2)O(l)