Question

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

Answer 1

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.

Complete answer:
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).

Additional information:
Activation energy is introduced by a scientist named Svante Arrhenius from Sweden. In the presence of the catalyst the activation energy gets lowered because the catalyst increases the rate of the reaction. Slower the chemical reaction higher will be the activation energy of the reaction. The release of heat also lowers the activation energy which is required by the reaction.

Note:
In terms of the transition state theory, the activation energy is the difference between the energy content of the atoms or the molecules in an activated or the transition state configuration. If the value of activation energy becomes zero there will be no effective collision and no product will be formed.