Question

What is enthalpy change when $100mL$ of $1M{H}_{2}S{O}_4$ is completely neutralized by $100mL$ of $1MCa(OH{)}_2$?

• A. $-13.7kcal$
• B. $-27.4kcal$
• C. $+1.37kcal$
• D. $-2.74kcal$

Answer 1

The correct option is D $-2.74kcal$

$1$ mole of $H_{2}S{O}_4$ give $2$ mole of $H^{+}$ ions.

$100mL$ of $1M$ $H_{2}S{O}_4=1\times 2mole$ of $H^{+}$ ions.

$100mL$ of $1M{H}_{2}S{O}_4=\frac{100\times 1\times 2}{1000}$ mole of $H^{+}$ ions $=0.2mole$ of $H^{+}$

Similarly, $100mL$ of $1MCa(OH{)}_2$ gives $=\frac{100\times 1\times 2}{1000}$ mole of $O{H}^{-}$ ions $=0.2$ mole of $O{H}^{-}$

Since $1$ mole of $H^{+}$ and $1$ mole of $O{H}^{-}$ react to give heat of neutralization $=13.7Kcal$

$\Rightarrow 0.2mole$ of each $H^{+}$ and $O{H}^{-}$ react to give heat of neutralization $=0.2\times -13.7Kcal$

$=-2.74Kcal$