Question

If $H^{\oplus }+O{H}^{\ominus }⟶H_{2}O+13.6kcal$ then the heat of neutralization for complete neutralization of one mole of $H_{2}S{O}_4$ by a base (NaOH) will be:

• A. $-13.7kcal$
• B. $-27.2kcal$
• C. $-31.8kcal$
• D. $-3.425kcal$

Answer 1

The correct option is B $-27.2kcal$

Solution:-

$H_{2}S{O}_4+2NaOH⟶{Na}_{2}S{O}_4+2H_{2}O$

According to reaction,

$1$ mole $H_{2}S{O}_4$ completely neutralised by $2$ mole of $NaOH$.

As we know that,

Gram equivalent $=$ no. of moles $\times$ Valency factor

Valency factor of $H_{2}S{O}_4=2$

Therefore,

Gram equivalent of $H_{2}S{O}_4=1\times 2=2$

As we know that,

Heat of neutralisation of $1$ gm eq. of strong acid $=-13.6kJ$

Heat of neutralisation of $2$ gm eq. of strong acid $=-13.6\times 2=-27.2kcal$

Hence the heat evolved will be $-27.2kcal$.