Question

Please help!? I do not understand how to do this at all.

Volume of HCl solution 100.5 mL HCl
Volume of NaOH solution 100.0 mL NaOH
Initial temperature in calorimeter 25.2 °C
Final temperature in calorimeter 28.2 °C



3. Determine the number of moles of NaOH.
4. Determine enthalpy per mole of NaOH. Show all of your work.

enthalpy change of this reaction= 1.258 kj/mol

Answer 1

Are we to assume that there was no reagent in excess?

Answer 2

Assuming there is no reagent in excess,

Find the energy change of the system using
\[\Delta Q = mc \theta\]
m = mass of liquid in the beaker (assuming they're diluted, 1 mL of liquid = 1g)
c = specific heat capacity of water
theta = temperature change (initial temp - final temp)
(Note the energy change, the enthalpy change of the reaction and temperature change are all negative values, so enthalpy change of this reaction = -1.258 kJ/mol)
The enthalpy change of this reaction would be the enthalpy change of neutralisation

\[\Delta H [neutralisation] = \frac{ \Delta Q }{ n(water) }\]

n(water) = moles of water

Note: Convert the enthalpy change of neutralisation to J/mol first

You can then find the moles of water, and using stoichiometry, you can find the moles of NaOH.

The enthalpy per mole of NaOH would be \[\Delta H [NaOH] = \frac{ \Delta Q }{ n (NaOH) }\]
Whicih is the same as the enthalpy change of the reaction