What is the Molar Heat of Neutralization for Sodium Hydroxide?

[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.

Homework Equations

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

 

[Borek]

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?