9.2 grams of N_(2)O_(4) (g) is taken in a closed one litre vessel and heated till the following equilibrium is reachedN_(2) O_(4) (g) hArr 2 NO_(2) (g) At equilibrium , 50% N_(2)O_(4) (g) is dissociated . What is the equilibrium constant ( in mol lit^(-1)) (" Molecular weughtof" N_(2)O_(4) = 92)

9.2 grams of N_(2)O_(4) (g) is taken in a closed one litre vessel and

1.	9.2 grams of N_(2)O_(4) (g) is taken in a closed one litre vessel and heated till the following equilibrium is reachedN_(2) O_(4) (g) hArr 2 NO_(2) (g) At equilibrium , 50% N_(2)O_(4) (g) is dissociated . What is the equilibrium constant ( in mol lit^(-1)) (" Molecular weughtof" N_(2)O_(4) = 92)
Answer» <html><body><p>`0.1`<br/>`0.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>`<br/>`0.4`<br/>2</p>Solution :Intial` [N_(2)O_(4)] = 9.2//92 " mol"<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>^(-1) = 0.1 " mol "L^(-1)` <br/> At. Eqm. ( after 50% dissociation ), <br/> `[N_(2)O_(4)] = 0.05 M, [NO_(2)] = 0.1 M` <br/> ` <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a> = ([NO_(2)])/([N_(2)O_(4)])=(0.1)^(2)/0.05 = 0.2`</body></html>