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The ester ethyl acetate is formed by the...

The ester ethyl acetate is formed by the reaction of ethanol and acetic acid and the equilibrium is represented as
`CH_3COOH (l) + C_2H_5OH(l) hArr CH_3COOC_2H_5(l) +H_2O (l)`
Write the concentration ratio , Q for this reaction. Note that water is not in excess and is not a solvent in this reactions .

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` Q_c = ([CH_3COOC_2H_5][H_2O(l)])/([CH_3COOH(l)][C_2H_5OH(l) ])`
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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 ester ethyl acetate is formed by the reaction of ethanol and acetic acid and the equilibrium is represented as CH_3COOH (l) + C_2H_5OH(l) hArr CH_3COOC_2H_5(l) +H_2O (l) At 293 K, if one starts with 1,000 mole of acetic acid and 1.80 moles of ethanol , there are 0.171 moles of ethyl acetate in the final equilibrium mixture , Calculate the equilibrium constant.

Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as : CH_3COOH(l) +C_2H_5OH(l) hArr CH_3COOC_2H_5(l) +H_2O(l) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime . Has equilibrium been reached?

Ethyl acetate is formed by the reaction between ethanol and acetic and the equiblibrium is represented as : CH_3COOH(l) +C_2H_5OH(l) hArr CH_3COOC_2H_5(l) +H_2O(l) At 293 K , if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol there is 0.171 mol of ethyl acetate in the final equilibrium mixture . Calculate the equilibrium constant.

sort out the homogeneous and netrogeneous equilibrium from the following CH_3COOH (l) + C_2H_5(l) hArr CH_3COOC_2H_5 (l)+ H_2O(l)

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.

Write the equilibrium constant expressions for the following reactions. CH_3 COOH(aq) +H_2O(l) hArr CH_3 COO^(-) (aq) + H_3O^(+) (aq)

For the reaction CH_3COOH(l)+C_2H_5(l)hArrCH_3COOC_2H_5(l)+H_2O(l) the value of equilibrium constant (K) is 4 at 298 K. The standard free energy change (DeltaG^@) is equal to

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 :

The equilibrium constant for the reaction CH_(3)COOH+C_(2)H_(5)OH hArr CH_(3)COOC_(2)H_(5)+H_(2)O is 4.0 at 25^(@)C . Calculate the weight of ethyl acetate that will be obtained when 120g of acetic acid are reacted with 92 g of alcohol.

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