Question

50 cm3 of 2.50 mol dm–3 hydrochloric acid was placed in a polystyrene beaker of negligible heat capacity.
Its temperature was recorded and then 50 cm3 of 2.50 mol dm–3 NaOH at the same temperature was quickly added, with stirring. The temperature rose by 17 °C.
The resulting solution may be considered to have a specific heat capacity of 4.2 J g–1K–1. What is an approximate value for the molar enthalpy change of neutralisation of hydrochloric acid and sodium hydroxide from this experiment?

Answer 1

$\begin{aligned} HCl + NaOH& \to NaCl + H_2O\\ \\ \textsf{moles of HCl}&= C×V =\frac{50}{1000}dm^3× 2.5mol\\ &=0.125mol\\ \\ \triangle T &= 17°C\\ c &= 4.2\ J/g.K\\ \\ \textsf{Total Volume}&= 50cm^3+ 50cm^3\\ &= 100cm^3 \end{aligned}$ Since the solution is mostly water, we can assume that;

$\textsf{density} = 1g/cm^3$

Using the formula;

$q_1 + q_2=0$

where $q_1 = \textsf{heat of neutralization}\\ q_2 = \textsf{heat released as }\triangle T$

$\begin{aligned} &\therefore n\triangle H + mc\triangle T = 0\\ &0.125×\triangle H+ (1g/cm^3×100cm^3)×4.2×17= 0\\ &0.125×\triangle H = -(1g/cm^3×100cm^3)×4.2×17\\ &0.125\triangle H =-7140J\\ &\triangle H = -57120Jmol^{-1} \end{aligned}$