Question
The enthalpy of neutralization of HCl and NaOH is −57 kJ mol−1. The heat evolved at constant pressure (in kJ) when 0.5 mole of H2SO4 react with 0.75 mole of NaOH is equal to
Answer: Option A
:
A
H−+OH−→H2O+57kJmol−1
nH+=2×nH2SO4=2×0.5=1.0
nOH=nNaOH=0.75
Heat evolved = 0.75×57=34×57kJ
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:
A
H−+OH−→H2O+57kJmol−1
nH+=2×nH2SO4=2×0.5=1.0
nOH=nNaOH=0.75
Heat evolved = 0.75×57=34×57kJ
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