3C_(2)H_(2) hArr C_(6) H_(6) the above reaction is performed in a I lit vessel. Equilibrium is established when 0.5 mole of benzene is present at certain tempe-rature. If equilibrium constant is 4 lit^(2) "mole"^(-2) The total number ot moles of the substances present at cquilibrium,

3C_(2)H_(2) hArr C_(6) H_(6) the above reaction is performed in a I li

1.	3C_(2)H_(2) hArr C_(6) H_(6) the above reaction is performed in a I lit vessel. Equilibrium is established when 0.5 mole of benzene is present at certain tempe-rature. If equilibrium constant is 4 lit^(2) "mole"^(-2) The total number ot moles of the substances present at cquilibrium,
Answer» <html><body><p>0.5<br/><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a><br/>1.5<br/>2</p>Solution :`3C_(2)H_(2(g)) <a href="https://interviewquestions.tuteehub.com/tag/harr-2692945" style="font-weight:bold;" target="_blank" title="Click to know more about HARR">HARR</a> C_(6)H_(6(g))` <br/> `K_(<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>)=([C_(6)H_(6)])/([C_(2)H_(2)]^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)), 4=(0.5)/([C_(2)H_(2)]^(3)) implies [C_(2)H_(2)]^(3)=(1)/(8)` <br/> `[C_(2)H_(5)]=0.5=(n)/(v) implies n=0.5` <br/> Total no. of moles of `C_(2)H_(2)` & `C_(6)H_(6)` at equilibrium = 0.5+0.5=1</body></html>