# How can I calculate enthalpy of neutralization?

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 ×

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

Mass of solution = 50.0 mL soln ×

The heats involved are

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

0.0175 mol ×

0.0175 mol ×

0.0175 mol ×

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The enthalpy of neutralization can be calculated using the formula:

ΔH = q / n

where: ΔH = enthalpy of neutralization (in kJ/mol) q = heat released or absorbed during the neutralization reaction (in kJ) n = number of moles of the limiting reactant involved in the neutralization reaction

To calculate q, you can use the formula:

q = m × c × ΔT

where: m = mass of the solution (in g) c = specific heat capacity of the solution (usually 4.18 J/g°C for water) ΔT = change in temperature (in °C)

Once you have calculated q, you can determine the number of moles of the limiting reactant involved in the neutralization reaction and use it to find the enthalpy of neutralization using the first formula.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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