`NH_(3)` is heated at `15` at, from `25^(@)C` to `347^(@)C` assuming volume constant. The new pressure becomes `50` atm at equilibrium of the reaction `2NH_(3) hArr N_(2)+3H_(2)`. Calculate `%` moles of `NH_(3)` actually decomposed.

`NH_(3)` is heated at `15` at, from `25^(@)C` to `347^(@)C` assuming v

1.	`NH_(3)` is heated at `15` at, from `25^(@)C` to `347^(@)C` assuming volume constant. The new pressure becomes `50` atm at equilibrium of the reaction `2NH_(3) hArr N_(2)+3H_(2)`. Calculate `%` moles of `NH_(3)` actually decomposed.
Answer» Correct Answer - A::C::D Pressure of `NH_(3)` at `27^(@)C=15 "atm"` Pressure of `NH_(3)` at `347^(@)C=P "atm"` Using relation `P_(1)/T_(1)=P_(2)/T_(2)` `P/620=15/300` `P=31 "atm"` Let a mol of ammonia be present. Total pressure at equilibrium`=50` atm. `{:(,2NH_(3)(g),hArr,N_(2)(g),+,3H_(2)(g)),("At equilibrium",(a-2x),,x,,3x),("Total moles" =,a-2x+x+3x=a+2x,,,,):}` `("Initial number of moles")/("moles at equilibrium")=("Initial pressure")/("Equilibrium pressure")` `a/((a+2x))=31/50` `x=16/62 a` Amount of ammonia decomposed `=2x=2xx19/62a=19/31a` `%` of ammonia decomposed `=(19xxa)/(31xxa)xx100=61.3`

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`NH_(3)` is heated at `15` at, from `25^(@)C` to `347^(@)C` assuming volume constant. The new pressure becomes `50` atm at equilibrium of the reaction `2NH_(3) hArr N_(2)+3H_(2)`. Calculate `%` moles of `NH_(3)` actually decomposed.

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