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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 18 of 73
Marks: +1, -0
Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as :
CH3COOH(l)+C2H5OH(l)\mathrm{CH_3COOH}_{(l)} + \mathrm{C_2H_5OH}_{(l)} ⇌ CH3COOC2H5(l)+H2O(l)\mathrm{CH_3COOC_2H_5}_{(l)} + \mathrm{H_2O}_{(l)}
(i) Write the concentration ratio (reaction quotient), QcQ_c, for this reaction (note : water is not in excess and is not a solvent in this reaction).
(ii) 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.
(iii) 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?
Solution:  
(i) QcQ_c = [CH3COOC2H5][H2O][CH3COOH][C2H5OH]\frac{[\mathrm{CH_3COOC_2H_5}][\mathrm{H_2O}]}{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}
(ii) CH3COOH+C5H5OH\mathrm{CH_3COOH} + \mathrm{C_5H_5OH} ⇌ CH2COOC2H5+H2O\mathrm{CH_2COOC_2H_5} + \mathrm{H_2O}
Initially
nCH3COOHn_{\mathrm{CH_3COOH}} = 1.00 ; nC2H5OHn_{\mathrm{C_2H_5OH}} = 0.180 and nCH3COOC2H5n_{\mathrm{CH_3COOC_2H_5}} = nh2On_{h_2O} = 0
At equilibrium :
nCH3COOHn_{\mathrm{CH_3COOH}} = (1.00 − 0.171) = 0.829, nC2H5OHn_{\mathrm{C_2H_5OH}} = (0.180 − 0.171) = 0.009
nCH3COOC2H5n_{\mathrm{CH_3COOC_2H_5}} = nH2On_{\mathrm{H_2O}} = 0.171
K = [CH3COOC2H5][H2O][CH3COOH][C2H5OH]\frac{[\mathrm{CH_3COOC_2H_5}][\mathrm{H_2O}]}{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]} = (0.171)(0.171)(0.829)(0.009)\frac{(0.171)(0.171)}{(0.829)(0.009)} = 3.919
(iii) CH3COOH+C2H5OH\mathrm{CH_3COOH} + \mathrm{C_2H_5OH} ⇌ CH3COOC2H5+H2O\mathrm{CH_3COOC_2H_5} + \mathrm{H_2O}
Initial
nCH3COOHn_{\mathrm{CH_3COOH}} = 1.00 mol, nC2H5OHn_{\mathrm{C_2H_5OH}} = 0.500 mol , nCH3COOC2H5n_{\mathrm{CH_3COOC_2H_5}} = nH2On_{\mathrm{H_2O}} = 0
At equilibrium,
nCH3COOHn_{\mathrm{CH_3COOH}} = (1.00 - 0.214) = 0.786 mol,
nC2H5OHn_{\mathrm{C_2H_5OH}} = (0.500 - 0.214) = 0.286 mol,
nCH3COOC2H5n_{\mathrm{CH_3COOC_2H_5}} = 0.214 and nH2On_{\mathrm{H_2O}} = 0.214
QcQ_c = [CH3COOC2H5][H2O][CH3COOH][C2H5OH]\frac{[\mathrm{CH_3COOC_2H_5}][\mathrm{H_2O}]}{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]} = (0.214)(0.214)(0.786)(0.286)\frac{(0.214)(0.214)}{(0.786)(0.286)} = 0.2037
As QcQ_c < K, the equilibrium has not reached. The reaction tends to proceed right, towards products.
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