Consider the reaction, $C{H_3}COOH{\rm{ }} + {\rm{ }}{C_2}{H_5}OH\;\;\leftrightharpoons\;\;C{H_3}COO{C_2}{H_5} + {\rm{ }}{H_2}O.$ One mole each of $C{H_3}COOH$ and ${C_2}{H_5}OH$ are heated in the presence of little concentrated ${H_2}S{O_4}.$ On equilibrium:

1 mole of ethyl acetate is formed

2 mole of ethyl acetate are formed

$1/2$ mole of ethyl acetate is formed

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JEE CHEMISTRY MCQ: Equilibrium

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Application +4 | -1

Consider the reaction,

C{H_3}COOH\mathrm{ } + \mathrm{ }{C_2}{H_5}OH\;\;\leftrightharpoons\;\;C{H_3}COO{C_2}{H_5} + \mathrm{ }{H_2}O.

One mole each of

C{H_3}COOH

and

{C_2}{H_5}OH

are heated in the presence of little concentrated

{H_2}S{O_4}.

On equilibrium:

View answer and explanation

Correct option: D

2/3

mole of ethyl acetate is formed

Option D (

2/3

mole of ethyl acetate is formed) is correct because the equilibrium constant for this reaction is 4, and solving for the extent of reaction gives 2/3.

More Questions from Equilibrium

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The indicator used in titrating oxalic acid with caustic soda solution is:

\mathrm{K}_1

and

\mathrm{K}_2

are equilibrium constant for reactions (i) and (ii)

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

(i)

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

(ii)

Then,

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

are the respective ionisation constants for the following reactions

\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}

The correct relationship between

K{a_1},\,\,K{a_2}\,,K{a_3}

is The

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

solution is (For

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

) Reaction

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

In equilibrium condition pressure of

\text{O}_2

depends on: A : At equilibrium concentration of the reactant and product does not change with time for a chemical reaction.
R: The rate of reaction is zero at equilibrium.