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The dissociation constants for acetic and HCN at 25^{o}C\ are \ 1.5\times 10^{-5} \ and \ 4.5\times 10^{-10} respectively. The equilibruim constant for the equilibruim.CN^{-}+CH_{3}COOH\rightleftharpoons HCN+CH_3COO^{-} would be:

  • Option 1)

    3.0\times 10^{-5}

  • Option 2)

    3.0\times 10^{-4}

  • Option 3)

    3.0\times 10^{4}

  • Option 4)

    3.0\times 10^{5}

 

Answers (1)

best_answer

As we discussed in concept

Law of Chemical equilibrium -

At a given temperature, the product of concentration of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentration of the reactants raised to their individual stoichiometric coefficients has a constant value.

- wherein

aA+bB\rightleftharpoons cC+dD


K_{c}=\frac{[C]^{c\:[D]^{d}}}{[A]^{a}\:[B]^{b}}

[A],\:[B],\:[C]\:[D]

are equilibrium concentration

 

 CH_{3}COOH\rightleftharpoons CH_{3}COO^{-}+H^{+}

K_{1}=\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}=1.5\times 10^{-5}

HCN\rightleftharpoons H^{+}+CN^-

K_{2}=\frac{[H^{+}][CN^{-}]}{[HCN]}=4.5\times 10^{10}

CN^{-}+CH_{3}COOH\rightleftharpoons CH_{3}COO^{-}+HCN

K=\frac{[HCN][CH_{3}COO^{-}]}{[CN^{-}][CH_{3}COOH]}=>K=\frac{K_{1}}{K_{2}}=3\times 10^{4}

 


Option 1)

3.0\times 10^{-5}

This option is incorrect.

Option 2)

3.0\times 10^{-4}

This option is incorrect.

Option 3)

3.0\times 10^{4}

This option is correct.

Option 4)

3.0\times 10^{5}

This option is incorrect.

Posted by

prateek

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