If 1 mole of CH_(3)COOH and 1 mole of C_(2)H_(5)OH are taken in 1 litre flask, 50% of CH_(3)COOH is converted into ester as, CH_(3)COOH_((l))+C_(2)H_(5)OH_((l))hArrCH_(3)COOC_(2)H_(5)(l)+H_(2)O_((l)) There is 33% conversion of CH_(3)COOH into ester, if CH_(3)COOH "and" C_(2)H_(5)OH have been taken initially in molar ratio x:1, find x.

If 1 mole of CH_(3)COOH and 1 mole of C_(2)H_(5)OH are taken in 1 litr

1.	If 1 mole of CH_(3)COOH and 1 mole of C_(2)H_(5)OH are taken in 1 litre flask, 50% of CH_(3)COOH is converted into ester as, CH_(3)COOH_((l))+C_(2)H_(5)OH_((l))hArrCH_(3)COOC_(2)H_(5)(l)+H_(2)O_((l)) There is 33% conversion of CH_(3)COOH into ester, if CH_(3)COOH "and" C_(2)H_(5)OH have been taken initially in molar ratio x:1, find x.
Answer» Solution :`CH_(3)COOH_((l))+C_(2)H_(5)OH_((l))hArrCH_(3)COOC_(2)H_(5(l)+H_(2)O_((l))` `1-0.5 1-0.5 0.5 0.5` So, `K_(C)=(0.5xx0.5)/(0.5xx0.5)=1` Now let a moles of `CH_(3)COOH "and" b "moles of" C_(2)H_(5)OH` are TAKEN: `a-(a)/(3) b-(a)/(3) (a)/(3) (a)/(3)` So, `K_(c)=((a//3)XX(a//3))/(2a//3xx(b-(a)/(3))) or 2(b-(a)/(3))=(a)/(3)` or `2b=a` or `(a)/(b)=(2)/(1)`