For the neutralisation of `1` mol of `H_(2)SO_(4)` with `2` mols of `NaOH` in dilute solutions the heat evolved is

Options

A. `=57.0 kJ`
B. ` lt 57.0 kJ`
C. ` gt 57.0 kJ`
D. Unpredictable

Answer

C

Explanation

`1 "mol"H_(2)SO_(4)=2"mol"H^(+)`
`2"mol"NaOH=2"mol"OH^(-)`
`2H^(+)(aq)+2OH^(-)(aq)to2H_(2)O(l)`
`DeltaH=57kJmol^(-1)xx2"mol"`
`=114kJ gt 57kJ`