1.

For which change DeltaH ne DeltaE :

Answer»

`H_(2)+I_(2)rarr2HI`
`HCl+NaOHrarrNaCl`
`C_((s))+O_(2(g))rarrCO_(2(g))`
`N_(2)+3H_(2)rarr2NH_(3)`

Solution :`DELTAH=DeltaE+DeltanRT`
`therefore` For `DeltaH NE DeltaE. DELTAN ne 0`
`Deltan=0` for reaction (a) as both reactants & PRODUCTS have same no. of molecules
`Deltan=0` for (c ) as C is in solid state. Further `DeltaH=DeltaE` for reaction (B) as it is carried out in liquid state
Moreover `Deltan=0` for this reaction also,
`HCl+NaOHrarrNaCl+H_(2)O`.


Discussion

No Comment Found

Related InterviewSolutions