Question

$HCl+NaOH⟶NaCl+H_{2}O;\Delta H=-57.1kJmo{l}^{-1}$
$\Delta H$ is the heat of neutralisation of $HCl$ and $NaOH$ solution.If true enter 1 else o

• A. 1

Answer 1

The correct option is A 1
When a solution of hydrochloric acid is added to a solution of sodium hydroxide dissolved in water, heat is released. This is called enthalpy of neutralisation.
$HCl+NaOH⟶NaCl+H_{2}O;\Delta H=-57.1kJmo{l}^{-1}$
$\Delta H$ is the heat of neutralisation of $HCl$ and $NaOH$ solution.