Termokimia (2) | Menentukan Perubahan Entalpi Reaksi berdasarkan Data Kalorimeter | Kimia kelas 11

Main Ideas

• A calorimeter measures heat changes during a reaction. The system is the chemical reaction; the surroundings include the solution (often water) and the calorimeter itself.

• In an exothermic reaction, the system releases heat and the surroundings absorb it. The heat-balance principle is: [ q_{\text{reaction}}=-q_{\text{surroundings}} ]

• The heat absorbed by a solution is calculated as: [ q_{\text{solution}}=mc\Delta T ] where (m) is the solution’s mass, (c) is its specific heat capacity, and (\Delta T=T_{\text{final}}-T_{\text{initial}}).

• The heat absorbed by the calorimeter is: [ q_{\text{calorimeter}}=C_{\text{cal}}\Delta T ] where (C_{\text{cal}}) is the calorimeter’s heat capacity.

• Reaction enthalpy per mole is found by dividing the reaction heat by the moles of reaction. For an exothermic reaction: [ \Delta H=-\frac{q_{\text{solution}}+q_{\text{calorimeter}}}{n} ] If the calorimeter’s heat capacity is ignored, use: [ \Delta H=-\frac{mc\Delta T}{n} ]

• When the solution’s volume is given instead of its mass, the lesson assumes a density of (1\text{ g/mL}). Thus, the volume in milliliters has the same numerical value as the mass in grams.

• For neutralization reactions, (n) is based on the amount of water formed, accounting for the limiting reactant and the reaction’s stoichiometric coefficients. For combustion, use the moles of fuel burned.

Method

1. Decide whether to include or ignore the calorimeter’s heat capacity.
2. Choose the matching heat-balance formula.
3. If needed, determine the solution’s mass from its volume and density.
4. Calculate the temperature change, (\Delta T).
5. Find the relevant amount in moles:
   • For neutralization, balance the reaction and identify the limiting reactant and the moles of water formed.
   • For combustion, calculate the moles of fuel from its mass and molar mass.
6. Substitute the heat and mole values into the formula, and report the result in energy per mole.

Worked Examples and Subtitle Inconsistencies

HCl and NaOH Neutralization

The video gives 50 mL of 0.4 M HCl and 50 mL of 0.1 M NaOH, with the temperature rising from 25°C to 28°C. It calculates the solution mass as 100 g and uses (c=4.2\text{ J g}^{-1}\text{K}^{-1}).

The spoken calculation appears to use 20 mmol of water and gives about (-64\text{ kJ/mol}). However, that does not match the stated concentrations: 50 mL of 0.1 M NaOH is 5 mmol, which limits the reaction. Using the stated data gives about (-252\text{ kJ/mol}). The concentration or mole calculation in the subtitles or video may therefore be incorrect.

HNO₃ and KOH Neutralization

The stated quantities are 33 mL of 1.4 M HNO₃ and 42 mL of 1 M KOH, so KOH limits the reaction at 42 mmol.

The video uses a temperature rise of 7°C and reports (-52.5\text{ kJ/mol}), but the question as transcribed says the temperature rises from 25°C to 30°C, a 5°C increase. With a 5°C rise, the result would be about (-37.5\text{ kJ/mol}). The video’s result corresponds to a 7°C rise, suggesting the stated final temperature may have been transcribed incorrectly.

Methane Combustion

The video uses 250 mL of water, a 10°C temperature rise, a calorimeter heat capacity of 75 J/°C, and 200 mg of methane. It calculates a total absorbed heat of 11,250 J and 0.0125 mol of methane, giving a combustion value of about (-900\text{ kJ/mol}) using its method.

Speakers and Sources

• Speaker: One presenter, who introduces himself as “Brother” on the channel; no personal name is provided.
• Source/channel: Kimatika, as identified in the video metadata.