# How can I calculate enthalpy of neutralization?

Jul 23, 2014

It's a calorimetry calculation. Here's how you do it.

EXAMPLE

When 25.0 mL of 0.700 mol/L NaOH was mixed in a calorimeter with 25.0 mL of 0.700 mol/L HCl, both initially at 20.0 °C, the temperature increased to 22.1 °C. The heat capacity of the calorimeter is 279 J/°C. What is the molar enthalpy of neutralization per mole of HCl?

Solution

The equation for the reaction is

NaOH + HCl → NaCl + H₂O

Moles of HCl = 0.0250 L HCl × $\left(0.700 \text{mol HCl")/(1"L HCl}\right)$ = 0.0175 mol HCl

Volume of solution = (25.0 + 25.0) mL = 50.0 mL

Mass of solution = 50.0 mL soln ×$\left(1.00 \text{g")/(1"mL soln}\right)$ = 50.0 g soln

ΔT = T_2 –T_1 = (22.1 – 20.0) °C = 2.1 °C

The heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

${q}_{1} + {q}_{2} + {q}_{3}$ = 0

nΔH + mcΔT + CΔT = 0

0.0175 mol × ΔH + 50.0 g × 4.184 J·g⁻¹°C⁻¹ × 2.1 °C + 279 J°C⁻¹ × 2.1 °C = 0

0.0175 mol × ΔH + 439.32 J + 585.9 J = 0

0.0175 mol × ΔH = -1025.22 J

ΔH = (-1025.22"J")/( 0.0175"mol") = -58 600 J/mol = -58.6 kJ/mol