Calculate `DeltaH^(@)` for the reaction
`Na_(2)O(g)+SO_(3)(g)toNa_(2)SO_(4)(s)`
Given the following
`(i) Na(s)+H_(2)O(l)toNaOH(s)+(1)/(2)H_(2)(g)`, `DeltaH^(@)=-146 kJ`
`(ii) Na_(2)SO_(4)(s)+H_(2)O(l)to2NaOH(s)+SO_(3)(g)`, `DeltaH=+418 kJ`
`(iii) 2Na_(2)O(s)+2H_(2)(g)to4Na(s)+2H_(2)O(l)`, `DeltaH=+259kJ`

Explanation

(a) ` oveset(N)_(2)+overset(0)O_(2)(g)rarr2 overset(+2)Boverset(-2)O(g)`
in this reaction the compound nitric oxide is formed combination of elemental substances like nitrogen and oxygen therefore it is a combination reaction since in this reaction nitrogen increase form 0 in `N_(2)` to +2 NO and that of oxygen decrease from 0 in `O_(2)` gas to -2 NO therefore it is combination redox reaction
(b) `overset(+2)Pboverset(+5)NO_(3)_(2)overset(triangle)rarrPbO(s)+2NO_(2)+1/2O_(2)(g)`
In this reaction lead nitrate decomposes to from three product viz oxide nitrogen dioxide and oxygen therefore it is decompostion reaction since in this reaction the oxidation number of nitrogen and that of oxygen increase from -2 in lead nitrate to 0 in `O_(2)` therefore it is a decompostion redox reaction
(c ) `overset(+)Naoverset(-1)H(s)+overset(+1)H_(2)overset(-2)O(l)rarroverset(+1)Naoverset(-2)Ooverset(+1)H_(2)(g)`
in this reaction hydrogen of water has been displaced by hydride ion to from dihydrogen gas therefore it is a displacement reaction since in this reaction the oxidation number of hydrogen increase from -1 in hydride ion to 0 in the displacement redox reaction
(d)
this is a disproportionation reaction since here the oxidatoin of nitrogen decreases from +4 in `NO_(2)` to +3 in `NO_(2)^(-)` ion as well as increase from +4 `NO_(2)` to +5 `NO_(3)^(-)` ion