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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)) (i)Write the concentration ratio (reaction quotient), Q_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 ? |
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Answer» Solution :(i)`Q_c=([CH_3COOC_2H_5][H_2O])/([CH_3COOH][C_2H_5OH])` `{:("EQUILIBRIUM reaction :",CH_3COOH_((l))+,C_2H_5OH_((l)) hArr, CH_3COOC_2H_(5(l))+, H_2O_((l))),("INITIAL mol:",1.0,0.18,0,0),("Change in reaction :",-x,-x,+x,+x),("Mol at equilibrium :",(1.0-x),(0.18-x),x=0.171,x),(,=(1.0-0.171),=(0.18-0.171),0.171,=0.171),(,=0.829,=0.009,0.171,0.171),("Suppose Volume =V So, M:", 0.829/V, 0.009/V,0.171/V,0.171/V):}` `therefore K_c=([CH_3COOC_2H_5][H_2O])/([CH_3COOH][C_2H_5OH])=(0.171/V)(0.171/V)(V/(0.829))(V/0.009)=(0.171xx0.171)/(0.829xx0.09)`=3.9192 `{:("Equilibrium reaction :",CH_3COOH_((l)) + , C_2H_5OH_((l)) hArr , CH_3COOC_2H_(5(l))+,H_2O_((l))),("Initial mol:", 1.0,0.5,0,0),("Change in reaction :",-x,-x,+x,+x),("Mol at equilibrium:",(1.0-x),(0.5-x),x,x),(,=(1.0-0.214),=(0.5-0.214),=0.214,0.214),(,=0.786,=0.286,,),("At equilibrium M:",0.786/V,0.286/V,0.214/V,0.214/V):}` `therefore Q_c=([CH_3COOC_2H_5][H_2O])/([CH_3COOH][C_2H_5OH])=(0.214/V)(0.214/V)(V/0.786)(V/0.286)=(0.214xx0.214)/(0.786xx0.286)`=0.2037 `(Q_c = 0.2037) lt (K_c=3.9192)` `therefore Q_c NE K_c`, So, equilibrium is not . `Q_c lt K_c` , The VALUE of `Q_c` is not equal to `K_c` , The reaction is continue. `Q_c` is less ,So, forward reaction occurs and more product is obtain. |
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