`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

Options

A. Total moles at equilibrium `4.8`
B. `K_(P)=K_(C)(RT)^(2)`
C. Mole fraction `H_(2)` at equilibrium`=0.5`
D. On increasing moles of `H_(2)S` equilibrium constant increases.

Answer

A, B, C, D

Explanation

`CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)`
`{:(t=0,64gm,68gm),("Moles",4"mole",2"mole"):}`
`4-x 2-2x x 4x`
Total moles at equilibrium`=4-x+2-2x+x+4x=(6-2x)`
Acc. To `PV=nRT,T=727+273=1000K`
`1.84xxV=(6-2x)xxRxx1000 `....(1)
For `H_(2) "gas" V,P_(H_(2))=n_(H_(2))RTto0.8xxV=4xxRxx1000` ....(2)
From eq (1) and(2) `(1.6)/(0.8)=((6-2x))/(4x)`
`x=0.6`