1.

1 mole of N_(2) and 2 moles of H_(2) are allowed to react in a 1 dm^(3) vessel. At equilibrium, 0.8 mole of NH_(3) is formed. The concentrationof H_(2) in the vessel is

Answer»

0.6 mole
0.8 mole
0.2 mole
0.4 mole

Solution :`N_(2)+3H_(2)iff2NH_(3)`
FORMATION of 0.8 mole of `NH_(3)` MEANS that `(3)/(2)xx0.8` mole of `H_(2)` have reacted.
HENCE, concentration of `H_(2)=2-(3)/(2)xx0.8`
`=2-1.2=0.8" mole"`


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