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The equilibrium constant for the reaction

N_{2(g)}+O_{2(g)}\rightleftharpoons 2NO(g) at temperature T is 4\times 10^{-4} The value of K_{c}  for the reaction : NO_{(g)}\rightleftharpoons \frac{1}{2}N_{2(g)}+\frac{1}{2}O_{2(g)} at the same temperature is

  • Option 1)

    2.5\times 10^{2}

  • Option 2)

    50

  • Option 3)

    4\times 10^{-4}

  • Option 4)

    0.02

 

Answers (1)

best_answer

As we learnt in

 

Equilibrium constant for the reverse reaction -

Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.

- wherein

K'_{c}=\frac{1}{K_{c}}

{K'_{c}}   is  equilibrium constant for reverse direction.

 

 

N_{2(g)}+O_{2(g)}\rightleftharpoons 2NO(g)

K_{c}= \frac{\left [ NO \right ]^{2}}{\left [ N_{2} \right ]\left [ O_{2} \right ]}= 4\times 10^{-4}

NO(g)\rightleftharpoons\frac{1}{2} N_{2(g)}+\frac{1}{2}O_{2(g)}

K{}'_{c}= \frac{ \left [ N_{2} \right ]^{1/2}\left [ O_{2} \right ]^{1/2}}{NO}= \frac{1}{\sqrt{K_{c}}}=\frac{1}{\sqrt{4\times 10^{-4}}}

= \frac{1}{2\times 10^{-2}}= \frac{100}{2}= 50


Option 1)

2.5\times 10^{2}

This option is incorrect

Option 2)

50

This option is correct

Option 3)

4\times 10^{-4}

This option is incorrect

Option 4)

0.02

This option is incorrect

Posted by

divya.saini

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