# Given,60 grams of $C{{H}_{3}}COOH$and 46 grams of ${{C}_{2}}{{H}_{5}}OH$ reacts in 5L flask to form 44 grams $C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}$at equilibrium on taking 120 grams of $C{{H}_{3}}COOH$ and 46 grams of ${{C}_{2}}{{H}_{5}}OH$,$C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}$ formed at equilibrium is:(A).44 g(B).20.33 g(C).22 g(D).58.66 g

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Hint: The activation energy refers to the minimum amount of energy which is required for the reaction to occur. If the amount of activation energy is less which does not meet the required need of activation energy for a reaction the process did not get successful which means that the reaction does not occur.

Molar mass of $C{{H}_{3}}COOH$= 60g/mol
Molar mass of ${{C}_{2}}{{H}_{5}}OH$= 46g/mol
Molar mass of $C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}$= 88g/mol
Initial concentration of $C{{H}_{3}}COOH$=$\dfrac{60}{(60)(5)}$= 0.2M
Initial concentration of ${{C}_{2}}{{H}_{5}}OH$=$\dfrac{40}{(40)(5)}$= 0.2M
Initial concentration of $C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}$=$\dfrac{44}{(88)(5)}$= 0.1M
Reaction involved:
$C{{H}_{3}}COOH+{{C}_{2}}{{H}_{5}}OH\rightleftharpoons C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O$

 0.2M 0.2M 0 0 Initial concentration $\left( 0.2-0.1 \right)$ $\left( 0.2-0.1 \right)$ 0.1M 0.1M Equilibrium concentration

Value of equilibrium constant =
${{K}_{c}}=\dfrac{[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}][{{H}_{2}}0]}{[C{{H}_{3}}COOH][{{C}_{2}}{{H}_{5}}OH]}=\dfrac{(0.1)(0.1)}{(0.1)(0.1)}$
Equilibrium constant = 1
Similarly for second case,
$C{{H}_{3}}COOH+{{C}_{2}}{{H}_{5}}OH\rightleftharpoons C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O$

 0.4M 0.2M 0 0 Initial concentration $\left( 0.40-x \right)$ $\left( 0.20-x \right)$ X M X M Equilibrium concentration

${{K}_{c}}=\dfrac{[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}][{{H}_{2}}0]}{[C{{H}_{3}}COOH][{{C}_{2}}{{H}_{5}}OH]}=\dfrac{{{x}^{2}}}{(0.4-x)(0.2-x)}$
The value of x = $\dfrac{8}{60}$M
Moles of ethyl acetate produced = $(\dfrac{8}{60})(5)=\dfrac{2}{3}$
So the mass of ethyl acetate produced = $(\dfrac{2}{3})(88)$= 58.66g

Hence the correct answer is option (D).