The enthalpy of neutralization of $HCl$ and is $NaOH$ is $- 57 KJ {mol}^{- 1}$ . The heat evolved at constant pressure (in $KJ$ ) when $0.5$ mole of $H_{2} {SO}_{4}$ reacts with $0.75$ mole of $NaOH$ is equal to:

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$57 \times \frac{3}{4}$

$57 \times 0.5$

$57 \times 0.25$

$57$

answer is A.

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Detailed Solution

$\begin{array}{l} 57 \times 3 / 4 \\ H^{-} + {OH}^{-} \to H_{2} O + 57 {kJmol}^{- 1} \\ n_{H^{+}} = 2 \times n_{H_{2} {SO}_{4}} = 2 \times 0 .5 = 1 .0 \\ n_{{OH}^{-}} = n_{NaOH} = 0 .75 \end{array}$

Heat evolved $= 0.75 \times 57 = \frac{3}{4} \times 57 kJ$

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