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For the reaction equilibrium,

N_{2}O_{4(g)}\rightleftharpoons 2NO_{2(g)} the concentrations of  N_{2}O_{4}\; and \; NO_{2}  at equilibrium are 4.8\times 10^{-2}\; and\; 1.2\times 10^{-2}\; mol\; L^{-1}\; respectively.

The\; value\; of\; K_{c}\; for\; the\; reaction \; is

  • Option 1)

    3.3\times 10^{2}\; mol\; L^{-1}

  • Option 2)

    3\times 10^{-1}\; mol\; L^{-1}

  • Option 3)

    3\times 10^{-3}\; mol\; L^{-1}

  • Option 4)

    3\times 10^{3}\; mol\; L^{-1}

 

Answers (1)

As we learnt in 

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

 

 N_{2}O_{4} \:_{(g)} \rightleftharpoons 2NO_{2} _{(g)}\:

4.8x10-2                       1.2\times 10^{-2}

K_{c}= \frac{(NO_{2})^{2}}{(N_{2O_{4})}}= \frac{(1.2\times 10^{-2})}{(4.8\times 10^{-2})}= 3\times 10^{-3} MoleL^{-1}


Option 1)

3.3\times 10^{2}\; mol\; L^{-1}

This option is incorrect 

Option 2)

3\times 10^{-1}\; mol\; L^{-1}

This option is incorrect 

Option 3)

3\times 10^{-3}\; mol\; L^{-1}

This option is correct 

Option 4)

3\times 10^{3}\; mol\; L^{-1}

This option is incorrect 

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Vakul

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