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  • The equilibrium constant (KC) for the reaction <img alt="\mathrm{N_{2}\left ( g \right )+O_{2}(g)\rightarrow 2NO(g)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BN_%

The equilibrium constant (KC) for the reaction \mathrm{N_{2}\left ( g \right )+O_{2}(g)\rightarrow 2NO(g)} at temperature T is  4\times 10^{-4}. The value of KC for the given reaction at the same temperature is : 

\mathrm{NO(g)\rightarrow \frac{1}{2} \: N_{2}(g)+\frac{1}{2}\: O_{2}(g)}

Option: 1

0.02


Option: 2

2.5 x 10 2


Option: 3

4 x 10 -4


Option: 4

50


Answers (1)

best_answer

As we discussed in the concept

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.

\mathrm{N_{2(g)}+O_{2(g)}\rightarrow 2NO_{(g)}\ \ \ \rightarrow(i)}

\mathrm{K_{c}=4\times 10^{-4}}

By multiplying the equation (i) by \frac{1}{2}

\mathrm{\frac{1}{2}N_{2(g)}+\frac{1}{2}O_{2(g)}\rightarrow NO_{(g)}\ \ \ \rightarrow(ii)}

\mathrm{K_{c}^{'}=\sqrt{K_{c}}=\sqrt{4\times 10^{-4}}=2\times 10^{-2}}

By reversing the equation (ii), we get

\mathrm{NO_{(g)}\rightarrow \frac{1}{2}N_{2(g)}+\frac{1}{2}O_{2(g)}}

\mathrm{K_{c}^{''}=\frac{1}{K_{c}^{'}}=\frac{1}{2\times 10^{-2}}=50}

Hence, the correct answer is Option (4)

Posted by

Pankaj

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