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13*8 g " of "N_(2)O_(4) was placed in 1 L reaction vessel in 1 L reaction vessel at 400 K and allowed to attain equilibrium : N_(2)O_(4) (g) hArr 2 NO(g).The total pressure at equilibrium was found to be 9* 15 bar. Calculate K_(c) , K_(p) and partial pressures at equilibrium.

Answer» <html><body><p></p>Solution :`13*8 "g"N_(2)O_(4) = (13*8)/<a href="https://interviewquestions.tuteehub.com/tag/92-342088" style="font-weight:bold;" target="_blank" title="Click to know more about 92">92</a>" mol"=0*15" mol""` `13*8 N_(2)O_(4) = (13*8)/92" mol"=0*15" mol""(Molar mass of "N_(2)O_(4) = 92 g " mol"^(-1))` <br/> PV= nRT <br/> ` :. P <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 1 L = 0* 15 mol xx 0* 083 " bar"L mol^(-1) K^(-1) xx 400 K or P = 4*98 " bar"` <br/> ` {:(,N_(2)O_(4)(g),hArr,2NO_(2)),("<a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> pressures",4*98 "bar",,0),(" At equilibrium",(4*98-p),,2p):}` <br/>` :. 4*98 - p+ 2 p = 9*15 " bar"or p= 4*17 " bar"` <br/> `:. (p_(N_(2)O_(4)))eq = 4* 98 - 4*17 = 0*<a href="https://interviewquestions.tuteehub.com/tag/81-338942" style="font-weight:bold;" target="_blank" title="Click to know more about 81">81</a> "bar", (p_(NO_(2)))eq = 2 xx 4*17 = 8*34 "bar"` <br/> `K_(p) = p_(NO_(2))^(2)//_(P_(N_(2)O_(4)))=(8*34)^(2)//)*81=<a href="https://interviewquestions.tuteehub.com/tag/85-339428" style="font-weight:bold;" target="_blank" title="Click to know more about 85">85</a>*87,` <br/> ` K_(p) = K_(c) (RT)^(Delta n) :. 85*87 = K_(c) (0*083 xx400)^(1) or K_(c)= 2*586 = 2*6`</body></html>


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