Enthalpy of neutralization for H2SO4 and NaOH

I was trying to determine the standard enthaply change of neutralization for H2SO4 and NaOH. In my reaction 58 cm3 of H2SO4 , c=(1,80 mol/dm3) reacted with 1dm3 NaOH c=0,162mol/dm3. That means that 0,104 mol of H2SO4 reacted with 0,162 mol of NaOH.  In my experiment I calculated standard enthaply change of neutralization and my result was -67241,79 J/mol. If i understand correctly my result should have been around -55.8 kJ/mol.

I would like to know what could have affected my result? 

We used tehnicqe of calorimetry.

Recheck calculations :

Limiting reactant is NaOH  and if you check balanced reaction...

2NaOH + H2SO4---->2H2O + Na2SO4 

so the number of moles of H2O formed  0.162 mol 

generally experimentallly calculated values are less than 55.8 KJ because some heat is always given off  even in insulated system.Heat lost to the stirrer / thermometer  and  coffee cup calorimeter is also not 100 % insulated.So experimental value of heat is generally less than theoretical value.Please recheck the calulations.

a ten.70 mL pattern of sulfuric acid from an vehicle battery requires 35.68 mL of two.eleven M sodium hydroxide solution for whole neutralization. UK Essay Writing Service writers says what's the molarity of the sulfuric acid? Sulfuric acid incorporates two acidic hydrogens.

Related Questions