Exam Review and Calorimetry Lab Notes

Quiz Question Discussion

Questions 8 and 9 involve substances not on the table, making them unsolvable without additional information.
If the substances and chemical reaction are provided, you can solve for \Delta H using the equation: \Delta G = \Delta H - T\Delta S.
This involves plugging in known values for Gibbs free energy, temperature, and entropy to solve for enthalpy, or vice versa.
These are considered simpler "plug and chug" problems.

A student calculated the energy using two methods and obtained slightly different results. One calculation yielded a result around -514.8, while another using H and S gave -514.4.
The value -514.4 is considered the correct one.
Another student obtained 2878.2 using all given values and -3967410.7 initially, then corrected to 3367 for another calculation. Not bad.

If the temperature is not explicitly given, assume the data was gathered at 298.15 K. This is the standard temperature to use unless specified otherwise.

For problems where you need to find the temperature at which a reaction becomes spontaneous, set \Delta G to zero.
\Delta G = 0 represents the tipping point of spontaneity.
Any temperature above the calculated temperature will result in a spontaneous reaction.
For example, if 200 K is the calculated temperature, then 200.00000001 K will be spontaneous
Calculate using the definition of \Delta G using \Delta H and \Delta S. This calculation is described as literally super duper simple.

\Delta H represents the heat of reaction, which can be calculated using heats of formation (products minus reactants).
\Delta S represents the change in entropy, also calculated as products minus reactants, using values from a table.
The Gibbs free energy of reaction is distinct from the Gibbs free energy of formation unless you are forming a single product from its elements.
If glucose is the only product, then the Gibbs free energy of formation and the heat of reaction would be the same.

If data is missing from the provided table (e.g., silver oxide in number seven), it may be necessary to search for the information online.

Example calculation: H_2S value is 33.6, so 3 ims 33 ilda 100; potassium chloride 2.96, so 4 ims 2.96 \tilda 1000.

Spontaneous reactions always have a negative \Delta G.
This occurs when \Delta H is negative (exothermic) and \Delta S is positive (increased entropy).
Exothermic reactions that increase entropy are always spontaneous. This is a potential true/false question.

The lab will focus on calorimetry and understanding how it works, emphasizing conceptual understanding over simply copying answers.
Calorimeters will be used to measure the heat of reaction between hydrochloric acid and either sodium bicarbonate or sodium carbonate.
Hess's law will be applied to determine the \Delta H of the reaction:
2 NaHCO3 \rightarrow CO2 + H2O + Na2CO_3
This reaction's \Delta H cannot be directly measured with the available equipment, so Hess's law is used.

React sodium bicarbonate and sodium carbonate with hydrochloric acid separately.
Flip one of the equations to allow for the cancellation of hydrochloric acid, adjusting the sign of \Delta H.
Double the first equation, meaning the obtained energy value will have to be doubled.

Weigh the amount of sodium carbonate used to convert mass to moles.
The hydrochloric acid solution is treated as water for calculations.
Hydrochloric acid is mostly water, so the mass of the hydrochloric acid solution is used as the mass of water in the calculations.
Weigh the solution before and after to determine the mass of the water.
Use 4.18 J/g^{\circ}C as the specific heat of water for the hydrochloric acid solution.
Calculate q for the water using: q = mc\Delta T, where:
- m is the mass of the water (hydrochloric acid solution).
- c is the specific heat of water ( ilda 4.18).
- \Delta T is the change in temperature.
If the water gains energy (positive q), the reaction is endothermic, and the heat of reaction will be the negative of q.

The calculated heat of reaction is for a small fraction of a mole of sodium carbonate or sodium bicarbonate (approximately 2.25 to 2.5 grams).
Convert grams to moles to find the energy per mole.
Units guide the calculation: joules per mole (j/mol) or kilojoules per mole (kj/mol).
Divide the energy in kilojoules by the moles of sodium carbonate or sodium bicarbonate to get the energy per mole.

If the reaction requires doubling (as indicated in the balanced equation), double the \Delta H value.

The instructor emphasizes ethical concerns:
- Cautioning against cheating and copying, as it will eventually be revealed.
- Mentioning statistical analysis to identify potential instances of academic dishonesty.
- The instructor stresses the importance of wearing goggles for safety during the lab.

The mass of hydrochloric acid should be used for calculations rather than the mass of the solution.
Values for the variables are: mass = \tilda 49 grams. Should be around \tilda 48 to \tilda 50 grams splitting the difference.
In calculation with provided values, the heat of the reaction is negative for the second reaction.
The \Delta H value does eventually have to be multiplied by 2.