Question

If $H^{+}+{OH}^{-}\leftrightharpoons H_{2}O+13.7kcal$, then heat of complete neutralisation of one gram mole of $H_2{SO}_4$ with strong base will be:

• A. $13.7kcal$
• B. $27.4kcal$
• C. $6.85kcal$
• D. $3.42kcal$

Answer 1

The correct option is B $27.4kcal$

Since, complete neutralisation of 1 mole of $H_{2}S{O}_4$ involves reaction of 2 moles of $H^{+}$ ion with 2 moles of $O{H}^{-}$ ions to form 2 moles of $H_{2}O$.

And since the reaction of 1 mole of $H^{+}$ ion with 1 mole of $O{H}^{-}$ ions to form 1 mole of $H_{2}O$ liberates $13.7Kcal$ of heat.
Hence, the reaction of 2 moles of $H^{+}$ ion with 2 moles of $O{H}^{-}$ ions to form 2 moles of $H_{2}O$ will liberate twice the heat.

i.e, $\Delta H\text{for complete neutralisation of 1 mole}{H}_{2}S{O}_4=2\times 13.7=27.4Kcal$

Hence, the answer is option B.