$K_{\mathrm{a}}$ of $\mathrm{HCN}$ and $\mathrm{CH}_3 \mathrm{COOH}$ are respectively $4.5 \times 10^{-10}$ and $1.5 \times 10^{-5}$ at $298 \mathrm{~K}$. Thus, equilibrium constant for $\mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH}\rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{HCN}$ would be

JEE CHEMISTRY MCQ: Equilibrium

1 Hard

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Application +4 | -1

K_{\mathrm{a}}

\mathrm{HCN}

and

\mathrm{CH}_3 \mathrm{COOH}

are respectively

4.5 \times 10^{-10}

and

1.5 \times 10^{-5}

298 \mathrm{~K}

.
Thus, equilibrium constant for

\mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH}\rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{HCN}

would be

View answer and explanation

Correct option: D

3.3 \times 10^4

Option D (

3.3 \times 10^4

) is correct because the equilibrium constant is calculated by multiplying the acid dissociation constant of acetic acid by the reciprocal of the acid dissociation constant of HCN.

More Questions from Equilibrium

Practice All Chapter MCQs →

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.