How can I calculate enthalpy of neutralization?

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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 × #(0.700"mol HCl")/(1"L HCl")# = 0.0175 mol HCl

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

Mass of solution = 50.0 mL soln ×#(1.00"g")/(1"mL soln")# = 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

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