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 :
A
`3435` J
B
`4` J
C
`-3435` J
D
zero
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The correct Answer is:
C
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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:
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.
When ethyl alcohol (C_2H_5OH(l)) and acetic acid (CH_3COOH(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_2H_5OH(l))+CH_3CHOOH(l) hArrCH_3COOC_2H_5(l)+H_2O(l)
The ester , ethyl acetate is formed by the reaction of ethanol and acetic acid and the equilibrium is represented as : CH_(3) COOH(l) +C_(2)H_(5)OH(l)hArr CH_(3)COOC_(2)H_(5)(l) +H_(2)O(l) (i) Write the concentration ratio (concentration quotient) Q for this reaction. Note that water is not in excess and is not a solvent in this reaction. (ii) At 293 K, if one starts with 1.000 mol of acetic acid 0.180 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture . Calculate the equilibrium constant. (iii) Starting with 0.50 mol of ethanol and 1.000 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after some time. Has equilibrium been reached?
The equilibrium constant for the reaction: CH_(3)COOH(l) + C_(2)H_(5)OH(l) iff CH_(3)COOC_(2)H_(5)(l)+ H_(2)O(l) has been found to be equal to 4 at 25^(@)C . Calcuclate the free energy chnage for the reaction.
