What Is the Molar Enthalpy of Neutralization of NaOH?

[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.

 

[Borek]

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.

 

[Blueshift5]

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).