Why does solid NaOH produce more heat than NaOH solution in neutralization?

[borderline182]

here's the procedure and calculations to a lab i just did... the questions i need help answering are at the bottom.

Part 1
1. rinse the graduated cylinder w/ a small quantity of 1.00mol/L NaOH(aq) then add 50mL of 1.00mol/L NaOH(aq) to the coffee cup calorimeter. Record the initial temp.
2. 2. Rinse the graduated cylinder with tape water and then with a small quantity of 1.00mol/L HCl(aq). Quickly and carefully add 50mL of 1.00mol/L HCl(aq) to the NaOH(aq) in the calorimeter.
3. Record the highest temp observed, stirring gently and continuously.

Part 2
1. rinse the graduated cylinder with a small quantity of 1.00mol/L HCl(aq) quickly and carefully add 50mL of 1.00mol/L HCl(aq) to the calorimeter and record the initial temp
2. quickly and carefully measure out 2g of NaOH pellets. Add these to the 1.00mol/L HCl(aq) as soon as possible
3. stir the pellets until they have completely dissovled and record the highest temp.

Calculations

Reaction 1
Initial temp=21.5
Final temp=28
Change in temp= 6.5

Qsol’n = mct
Q = (100)(4.184)(6.5)
Q = 2719.6 J
Qrxn = -2719.6 J

Reaction 2
Initial temp= 21.5
Final temp = 43
Change in temp = 21.5

Qsol’n=mct
Q = (50)(4.184)(21.5)
Q = 4496.8 J
Qrxn = -4497.8 J

# of mol’s of HCl used = 0.05
# of mol’s of NaOH used = 0.05


QUESTIONS I NEED HELP ANSWERING 

WHY was the change in temperature much more significant when using solid NaOH? And why is the enthalpy change (Qrxn) much lower?

What changes do you need to make to the thermochemical equation if you perform the investigation using solid sodium?

 

[wais]

The change in temperature was more significant when using solid NaOH because it requires more energy to break down the bonds between the solid particles and dissolve them in the solution. This results in a larger change in temperature compared to using already dissolved NaOH. The enthalpy change (Qrxn) is lower because the reaction is not happening between two aqueous solutions, but between a solid and a solution. This means that there is less heat released or absorbed during the reaction.

To perform the investigation using solid sodium, the thermochemical equation would need to be adjusted to include the enthalpy of fusion for solid sodium. This would account for the energy needed to melt the solid sodium before it can react with the solution. The equation would now be:

Na(s) + HCl(aq) → NaCl(aq) + H2O(l)

ΔHrxn = ΔHfus + ΔHsol’n + ΔHsol’n

Where ΔHfus is the enthalpy of fusion for solid sodium.