64 gm of CH_(4) "and" 68gm "of" H_(2)S was placed in an close container and heated up to 727^(@)C following equilibrium is established in gaseous phase reaction is: CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g) The total pressure at equilibrium is 1.6 atm and partial pressure of H_(2) "is" 0.8 atm. Then

64 gm of CH_(4) "and" 68gm "of" H_(2)S was placed in an close containe

1.	64 gm of CH_(4) "and" 68gm "of" H_(2)S was placed in an close container and heated up to 727^(@)C following equilibrium is established in gaseous phase reaction is: CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g) The total pressure at equilibrium is 1.6 atm and partial pressure of H_(2) "is" 0.8 atm. Then
Answer» <html><body><p>Total moles at equilibrium `4.8`<br/>`K_(P)=K_(C)(<a href="https://interviewquestions.tuteehub.com/tag/rt-615359" style="font-weight:bold;" target="_blank" title="Click to know more about RT">RT</a>)^(2)`<br/>Mole fraction `H_(2)` at equilibrium`=0.5`<br/>On increasing moles of `H_(2)S` equilibrium constant increases.</p>Solution :`CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)` <br/> `{:(t=0,64gm,68gm),("Moles",4"mole",2"mole"):}` <br/> `4-x 2-2x x 4x` <br/> Total moles at equilibrium`=4-x+2-2x+x+4x=(6-2x)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/acc-846605" style="font-weight:bold;" target="_blank" title="Click to know more about ACC">ACC</a>. To `PV=nRT,T=727+273=1000K` <br/> `1.84xxV=(6-2x)xxRxx1000 `....(1) <br/> For `H_(2) "gas" V,P_(H_(2))=n_(H_(2))RTto0.8xxV=4xxRxx1000` ....(2) <br/> From <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a> (1) and(2) `(1.6)/(0.8)=((6-2x))/(4x)` <br/> `x=0.6`</body></html>