# Molar Enthelpy of Neutralization

• cybernerd
In summary, the question asks for the molar enthalpy of neutralization of NaOH (aq) when 50.0 mL of aqueous 1.00 mol/L of NaOH (aq) reacts with an excess of 1.00 mol/L H2SO4 (aq). The given initial temperatures for H2SO4 and NaOH are 21.3 degrees Celsius and 20.6 degrees Celsius, respectively, and the final temperature of the solution is 28 degrees Celsius. The equation used to solve this problem is n(molar enthalpy) = vc(change in temperature), where c = 4.19 J/mL x degrees Celsius. After solving the equation, the molar enthalpy is
cybernerd

## Homework Statement

WHat is the molar enthalpy of neutralization of NaOH (aq) when 50.0 mL of aqueous 1.00 mol/L of NaOH (aq) reacts with an excess of 1.00 mol/L H2SO4 (aq)?

Initial temperature of H2SO4 - 21.3 degrees celcius
Initial Temperature of NaOH - 20.6 degrees celcius

Final temperature of solution - 28 degrees celcius

## Homework Equations

n(molar enthalpy) = vc(change in temperature)

Where c = 4.19 J/mL x degrees celcius

## The Attempt at a Solution

molar enthalpy = ( vct) / (n)

= ( 50 mL x 4.19 (28 - (21.3 + 20.6 / 2)) / (1.00 mol/L x 50.0 mL)

= 29 KJ / mol

I'm not entirely sure if this answer and its units are correct, or if averaging the two initial temperatures was correct. Any verification of my answer would be great.

Averaging is not a good idea, and I don't think you can solve the question not knowing exact volume of sulfuric acid used. You should do full heat balance and assume solution to have specific heat of pure water - while this is not exactly true it would be probably close enough to the reality.

Last edited by a moderator:

Your attempt at a solution appears to be correct. To verify your answer, you can also calculate the molar enthalpy of neutralization using the standard enthalpy of formation values for NaOH and H2SO4. The standard enthalpy of formation for NaOH is -469.6 kJ/mol and for H2SO4 is -814.2 kJ/mol. Using the equation:

ΔH = ΔHf (products) - ΔHf (reactants)

Where ΔH is the molar enthalpy of neutralization, ΔHf is the standard enthalpy of formation, and products and reactants refer to the products and reactants in the neutralization reaction.

In this case, the neutralization reaction is:

NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + 2H2O (l)

Plugging in the values, we get:

ΔH = (-814.2 kJ/mol) - (-469.6 kJ/mol)

= -344.6 kJ/mol

This value is close to your calculated value of 29 kJ/mol, which further verifies the accuracy of your solution. As for averaging the initial temperatures, it is a common practice to do so when the temperature change is small. In this case, it should not significantly affect the final result. Overall, your solution is correct and your units are also correct (kJ/mol).

## What is the molar enthalpy of neutralization?

The molar enthalpy of neutralization is the amount of heat released or absorbed when one mole of an acid and one mole of a base react to form a salt and water at constant pressure and temperature.

## How is the molar enthalpy of neutralization calculated?

The molar enthalpy of neutralization can be calculated by subtracting the initial enthalpies of the acid and base from the final enthalpy of the resulting salt and water.

## What factors can affect the molar enthalpy of neutralization?

The molar enthalpy of neutralization can be affected by the strength of the acid and base, the concentration of the acid and base, the temperature at which the reaction takes place, and any changes in volume or pressure during the reaction.

## Why is the molar enthalpy of neutralization important in chemistry?

The molar enthalpy of neutralization is important in chemistry because it provides information about the energy changes that occur during a neutralization reaction. This information can help predict the behavior of acids and bases in different environments and can be used in various industrial processes.

## How does the molar enthalpy of neutralization differ from other enthalpy values?

The molar enthalpy of neutralization differs from other enthalpy values because it specifically relates to the energy changes that occur during a neutralization reaction, while other enthalpies may relate to different types of reactions or processes. Additionally, the molar enthalpy of neutralization is specific to the amount of substance (one mole) used in the reaction, while other enthalpies may be calculated for different amounts of substances.

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