Question

In the reactions,
$HCl+NaOH\to NaCl+H_{2}O+xcal.$
$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O+ycal$

• A. $x=y$
• B. $x=2y$
• C. $x=\frac{y}{2}$
• D. $x=\sqrt{y}$

Answer 1

The correct option is C $x=\frac{y}{2}$

Given reactions are neutralisation reactions.

Neutralisation reaction, is more precisely, a reaction between $H^{+}$ ions and $O{H}^{\ominus }$ ions to form $H_{2}O$ molecules.

Reaction -1

$HCl+NaOH⟶NaCl+H_{2}O+xCal$

$\Rightarrow H^{+}+C{l}^{\ominus }+N{a}^{+}+O{H}^{\ominus }\to N{a}^{+}+C{l}^{\ominus }+H_{2}O+xCal$

$\Rightarrow H^{+}+O{H}^{\ominus }⟶H_{2}O+xcal$

for $1$ mole of $H_{2}O$ molecule, Energy Releases $=xCal$.......(I)

Reaction-2

$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O+ycal$

$\Rightarrow 2H^{+}+S{O}_4^{2-}+2N{a}^{+}+2O{H}^{\ominus }⟶2N{a}^{+}+S{O}_4^{2-}+2H_{2}O+yCal$

$\Rightarrow 2H^{+}+2O{H}^{\ominus }⟶2H_{2}O+yCal$

For $2$ mole of $H_{2}O$ molecule, Energy Releases $=yCal$

$\therefore$ For $1mole⟶$ Energy Released $=\frac{y}{2}Cal$.........(II)

from (1) and (II) $\Rightarrow x=\frac{y}{2}$

Correct Ans :- (c)