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If 0.1 mole of I2, is introduced into 1 litre flask at 1000 k, at equilibrium (Kc = 10-6), which one is correct? 

  • Option 1)

    [ I2 (g) ] > [I (g)]

  • Option 2)

    [I2 (g)] < [1 (g)]

  • Option 3)

    [I2 (g)] = [ 1(g) ]

  • Option 4)

    [I(g)] = \frac{1}{2}[I (g)]

 

Answers (1)

best_answer

As we learnt in

Law of Chemical equilibrium -

A+B\rightleftharpoons C+D

where  A & B are the reactants, C & D are the product in balanced chemical equations.

- wherein

K_{c}=\frac{[C]\:[D]}{[A]\:[B]}

Kc is the equilibrium constant.

 

 The reaction takes place as follows

I_{2}\left ( g \right )\leftrightharpoons 2I (g), K_{c}=10^{-6}

Clearly, the equilibrium constant favours I_{2}

\therefore \left [ I_{2} \right ]> \left [ I \right ]

 

 


Option 1)

[ I2 (g) ] > [I (g)]

correct

Option 2)

[I2 (g)] < [1 (g)]

incorrect

Option 3)

[I2 (g)] = [ 1(g) ]

incorrect

Option 4)

[I(g)] = \frac{1}{2}[I (g)]

incorrect

Posted by

divya.saini

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