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One mole of ethyl alcohol was treated with one mole of acetic acid at 25^(@)C. Two-third of the alcohol change into easter at equlibrium. The equlibrium constant for the reaction will be

Answer»

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Solution :`C_(2)H_(2)OH+COOHtoCH_(3)COOC_(2)H_(5)+H_(2)O`
`{:("Initial CONC.","1 mol","1 mol","0 MOLE","0 mole"),("At equlibrium", 1-2/3=1/3 "mol",1/3"mol",2/3"mol",2/3"mol"):}`
`K_(C)=((2)/(3)xx(2)/(3))/(1/2xx1/3)=4`


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