Thermochemistry
Discussion Questions

1. Report the mean ΔHrxn for each part with standard deviation and % error.

The mean ΔHrxn for the heat of neutralization was calculated to be -49 kj/mol with a standard deviation of ±2 and a % error of -12%. The mean ΔHrxn for the heat of solution was calculated to be -23 with a standard deviation of ±2 and a % error of
-22%.

2. Looking at the % errors, comment on the accuracy of each experiment.

For the heat of neutralization, the percent error was calculated to be -12%. For the heat of solution, the percent error was calculated to be -18%. Because both percent errors are greater than the accuracy threshold of ± 10%, neither measurement is accurate.

3. Looking at the standard deviations, comment on the precision of each experiment.

The standard deviation for the heat of neutralization, which was ±2, was precise because the mean ΔHrxn was calculated to be -47kj/mol. Because a significant figure did not have to be dropped, the measurement was precise.
The standard deviation for the heat of solution, which was also ±2, was precise because the mean ΔHrxn was calculated to be -49 kj/mol. Because a significant figure did not have to be dropped, the measurement was also precise.

4. Why is the sign of heat gained or lost by a reaction the opposite sign of the heat gained or lost by the solution?

The sign of heat gained or lost by a reaction is the opposite sign of the heat gained or lost by the solution because as the reaction commences, the reactants will either release heat or absorb heat. When this happens, the products will either contain more heat or less heat than the reactants because of the energy transfer. Since the product either absorbs or evolves the heat energy, the qproducts will have the opposite sign of the qreactants.

5. Classify each of the two reactions you studied as being either endothermic or exothermic.

In the heat of neutralization experiment, the reaction was endothermic because the mean ΔHrxn = -49 kj/mol which would mean that the ΔHproducts = 49 kj/mol because the heat gained or lost by a reaction is the opposite sign of the heat gained or lost by the solution. Since the ΔHproducts = 49 kj/mol > 0, the reaction was endothermic.
In the heat of solution experiment, the reaction was theoretically exothermic because the theoretical ΔHrxn = 28 kj/mol which would mean that the ΔHproducts = -28 kj/mol which is 0, making the reaction endothermic. This error could be due to the imperfection of the calorimeter, which may have absorbed some of the heat from the reaction.

6. If the NH4NO3 were not completely dry, how would this affect the value of:
a. qrxn If the NH4NO3 were not completely dry, the qrxn would be less because the salt would have already reacted with the water in the hydrate. This would mean that the NH4NO3 would not react with the excess water, which would significantly lower the qrxn.

b. molar ΔHrxn If the NH4NO3 were not completely dry, the molar ΔHrxn would be less because the qrxn would have been lowered due to the inactivity of the salt with the water because the salt would have already reacted with the water in the hydrate.

7. Suppose the concentrations of both reactants were only 0.50 M. Explain how this would affect the value of: (think about each as intensive or extensive properties)
a. qrxn If the concentrations of both reactants were only 0.50 M, then the qrxn would also be lower because qsoln=-qrxn , and qsoln=mc Δ T. If the molarity was 0.50 M, then the amount of moles present would be less, and therefore the mass would be less. So if the molarity were lower, the qrxn would also be lower.

b. ΔHrxn If the concentrations of both reactants were only 0.50 M, then the molar ΔHrxn would be lower. This is because the qrxn would also be lower, and since molar ΔHrxn = qrxn/moles of solution, the molar ΔHrxn would be lower.

8. Is the accuracy of the digital thermometer a vital component for this experiment?

No, the accuracy of the digital thermometer is not a vital component for the experiment because the accuracy of the temperature is insignificant in this experiment.

9. Discuss the assumptions made in this experiment.
a. The use of the specific heat of water when actually dealing with a solution.

The assumption that the solution had the same specific heat as that of water allows for error to occur because specific heat is a value that is unique to all solutions. Because we assumed that the solution had the same specific heat as water, the calculations would be off.

b. Using a homemade calorimeter and assuming a heat capacity of zero.

The assumption that a homemade calorimeter with a heat capacity of zero, which would make this calorimeter perfect, allows for error to occur. Because the calorimeter is actually not perfect, some of the heat may have been lost to the calorimeter causing a smaller qrxn, smaller molar ΔHrxn, and a greater percent error than if we had used a perfect calorimeter.