# What is the Molar Heat of Neutralization for Sodium Hydroxide?

• XxphysicsxX
In summary, the conversation discusses calculating the molar heat of reaction for NaOH(aq) and the percent difference in value obtained compared to the neutralization value of -57 kJ/mol. The specific heat capacity and density of both substances are assumed to be equal to that of water. The process for conducting the experiment is also outlined. However, the attempt at finding the molar heat of reaction is incorrect as the mass and molar mass used do not correspond to the reacting substance. The specific substance being reacted is not specified.
XxphysicsxX

## Homework Statement

Calculate the molar heat of reaction for the NaOH(aq)
calculate the percent difference for the value u obtained to the value of neautralization of sodium hdroxide, -57 kJ/mol
"assume the specific heat capacity of both these substances equal to that of water and that each of these solutions has a density equal to water."

Initial temp. of H2SO4(aq)=24.2
initial temp. of NaOH(aq) =23.9
average initial temp. of solutions = 24.1(degree Celsius)
final temperature of solutions = 32.9 (degree Celsius)

Step 1: Nest two polystyrene cups.

Step 2: Measure 30.0 mL of 1.0 mol/L H2SO4(aq) solution using a 50-mL graduated cylinder and pour it into the calorimeter.

Step 3: Using another 50-mL graduated cylinder, measure 50.0 mL of NaOH(aq) solution.

Step 4: Pour the NaOH solution into the calorimeter. Use the third cup as a lid. Insert the thermometer into a hole in the lid, and slowly stir the mixture with the thermometer.

Step 5: Pour the reaction mixture down the sink. The reaction mixture contains dilute excess sulfuric acid and acqueous sodium sulfate, and it does not require specific treatment prior to disposal. Rinse the calorimeter and other glassware used. Put all equipment away.

Q= ΔrH
mcΔt=ΔrH

nΔrHm=ΔrH

## The Attempt at a Solution

Q=mcΔt
Q=(80g)(4.19J/g.°C)(8.80°C)

Q=2.95 KJ

Q= ΔrH
ΔrH= 2.95 KJ

ΔrHm= ΔrH / n
n=m/M
n=(30g+50g) / {22.99g/mol+16.00g/m+1.01g/mol+2.02g/mol+32.07g/mol+64.00g/mol}

n= 0.579mol <--- the unit is not the KJ/mol like the accepted value for the molar heat of sodium hydroxide...

ΔrHm= ΔrH / n
= 2.95 KJ/ 0.579mol
= 5.09 KJ/mol<------- not anywhere close to -57KJ /mol

XxphysicsxX said:
n=m/M
n=(30g+50g) / {22.99g/mol+16.00g/m+1.01g/mol+2.02g/mol+32.07g/mol+64.00g/mol}

No idea what you did here, but it is twice wrong. 80g is not mass of the reacting substance but mass of the solution, sum in teh denominator is not molar mass of the reacting substance, so number you calculated has nothing to do with the number of moles of the reacting substance.

Which leads us to the question: - what is the reacting substance here? Can you write reaction equation?

## What is molar heat of neutralization?

Molar heat of neutralization is the amount of heat energy released or absorbed when one mole of an acid and one mole of a base react to form water and a salt.

## How is molar heat of neutralization calculated?

The molar heat of neutralization is calculated by dividing the amount of heat released or absorbed by the number of moles of the acid or base used in the reaction.

## What factors influence the molar heat of neutralization?

The strength and concentration of the acid and base, as well as the temperature of the reaction, can influence the molar heat of neutralization. Stronger acids and bases tend to have higher molar heats of neutralization compared to weaker ones.

## What is the relationship between molar heat of neutralization and enthalpy change?

The molar heat of neutralization is equal to the enthalpy change of the reaction. Enthalpy change is a measure of the heat energy released or absorbed during a chemical reaction.

## How is molar heat of neutralization used in real-world applications?

Molar heat of neutralization is used in industries such as pharmaceuticals and food production to calculate and control the amount of heat released during chemical reactions. It is also used in thermodynamic calculations and in the design of heating and cooling systems.

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