See the attached document and find the heat of reaction for each set of reactions.
Reaction 1: Heat of Dissolution of NaOH
Reaction 2: Heat of Reaction Between Aqueous NaOH and Aqueous HCl
Reaction 3: Heat of Reaction Between Solid NaOH and Aqueous HCl
1. Write the net ionic equation for each reaction, and note the value of ▲H for each reaction.
e.g. H+(aq) + OH-(aq) → H2O(l) ▲H2 = -45kJ/mol HaOH
2. Explain why ▲H is negative in all the reactions.
3. Describe what each equation represents.
4. Add the net ionic equation for Reaction 1 to the net ionic equation for Reaction 2. How does the result compare with the net ionic equation for Reaction 3?
5. Add ▲H1 and ▲H2. How does the sum compare with ▲H3? Should they be the same? Why?
6. Calculate the percent difference between the sum of (▲H1 + ▲H2) and ▲H3, assuming that ▲H3 is the correct value.
7. Other then being unable to accurately measure the quantities of reagents, what are the possible major sources of error in the procedure for this experiment?
8. If I had used 11.0 g NaOH(s) instead of using about 5.5 g of NaOH(s), i.e. twice the mass, in Reaction 1,
a) What effect would this have had on the amount of heat evolved in Reaction 1,
b) What effect would this have had on the value of ▲H1, the heat evolved per mole of NaOH reacted?
9. In Reaction 1, Na+ ions were separated from OH- ions, a process of dissociation that absorbs energy, yet, in the end, heat was liberated by the overall process. Explain.
10. Make a statement regarding the additivity of heats of reaction as it specifically applies to the results of this experiment.
The answers are in the attached document. here is a brief summary:
Temperature of 0.25 M HCl before mixing °C : 25
Temperature after mixing (°C): 25.8
Temperature difference (K): 0.8
Mass of solution = 100g
Total heat change = 100 x 0.00418 x 0.8 = 0.3344
Mass of NaOH = ...
This solution provides calculation for the heat of reaction for different sets of reactions for dissolution of NaOH. It then answers questions mentioned in the problem.