P E R F O R M A N C E 2 n d S e m e s t e r - T A S K 4 t h General 02 Chemistry 0 1 Q u a r t e r Aldueza, Eldridge - Bobier, Angela - Caraig, Ashly Maldo, Atasha - Sumabat, Sheena ENTROPY, ENTHALPY, & FREE ENERGY Table 01 ∆Η° (kJ/mol) S° (J/K mol) ∆G° (kJ/mol) HCO3- -691.1 94.94 -587.1 H+ 0 0 0 H2O (l) -285.8 69.9 -237.2 CO2 (g) -393.5 213.6 -394.4 Materials: vinegar baking soda thin-walled cup tablespoon measure teaspoon measure P E R F O R M A N C E 2 n d S e m e s t e r - T A S K 4 t h 0 1 Q u a r t e r Aldueza, Eldridge - Bobier, Angela - Caraig, Ashly Maldo, Atasha - Sumabat, Sheena General 02 Chemistry ENTROPY, ENTHALPY, & FREE ENERGY Exploration STEP 1 - Put about 2 tablespoons of vinegar in a cup. Add a teaspoon or two of baking soda to the cup. (a) What do you observe through sight, sound, and touch? Answer: Sight: The mixture or formation causes the liquid to form a foam and result in a rise in the cup. Sound: We observed hissing sounds. This housing sound is caused by the formation and mixture of carbon dioxide gas. Touch: It feels slightly cool (b) What kind of change is occurring? Answer: This experiment or mixture shows a chemical charge. Observing bubbles and the formation of a mixture is our example of a chemical change. (c) What are the formulas of the 2 major components of vinegar and of the one component of baking soda? Answer: The major components of vinegar are acetic acid (CH3COOH) and water (H2O), while baking soda is primarily composed of sodium bicarbonate (NaHCO3). (d) Write the overall equation and the net ionic equation for the process. Answer: Overall equation: NaHCO3(aq) + HC2H3O2(aq) ➝ H2O(l) + CO2(g) + NaC2H3O2(aq) Net ionic equation: HCO3- + H+ ➝ H2O(l) + CO2(g) P E R F O R M A N C E 2 n d S e m e s t e r - T A S K 4 t h General 02 Chemistry 0 1 Q u a r t e r Aldueza, Eldridge - Bobier, Angela - Caraig, Ashly Maldo, Atasha - Sumabat, Sheena ENTROPY, ENTHALPY, & FREE ENERGY STEP 2 (a) Define entropy and the significance of the sign of its value. Answer: Entropy is the measurement of the amount of disorder/disorganization in the system. Entropy helps predict the possible outcomes of the chemical reaction process. Also, Entropy helps to explain the processes occurring behavior of living things. Because living things absorb energy from their different surroundings. That's why entropy preserves the low level of entropy, living things radiant excess heat. (b) Based on your observations, explain the entropy change for the system observed in Step 1. Answer: in step 1, The entropy of the system was increased because gas was formed and produced. Gas is more disarranged than liquid matter. (c) Use the entropy data from Table 1 to calculate the entropy change for the net ionic equation from Step 1. Answer: ∆S° = ∑S°(products) - ∑S°(reactants) ∆S° = (69.9 + 213.6) - (94.94 + 0) ∆S° = 188.56 J/K (d) Is your calculated value consistent with your conclusions from observation? Explain. Answer: Yes, the value is consistent with the observations made during the reaction. The calculated positive value aligns with the observation that the system becomes more disordered, supporting the conclusion drawn from the observation. P E R F O R M A N C E 2 n d S e m e s t e r - T A S K 4 t h General 02 Chemistry 0 1 Q u a r t e r Aldueza, Eldridge - Bobier, Angela - Caraig, Ashly Maldo, Atasha - Sumabat, Sheena ENTROPY, ENTHALPY, & FREE ENERGY STEP 3 (a) Define enthalpy and the significance of the sign of its value. Answer: Enthalpy is the heat energy exchanged during a reaction, and a negative value indicates an exothermic reaction. For an endothermic process (where the system absorbs heat from the surroundings), the enthalpy is positive. Conversely, for an exothermic process (where the system releases heat to the surroundings), the enthalpy is negative. (b) Based on your observations, explain the enthalpy change for the system observed in Step 1. Answer: The system feels cooler, indicating an exothermic reaction where heat is released to the surroundings. (c) Use the enthalpy data from Table 1 to calculate the enthalpy change for the net ionic equation from Step 1. Answer: ∆H° = ∑H°(products) - ∑H°(reactants) ΔH =(ΔH° H2O +ΔH° CO2)−(ΔH° NaHCO3∘+ΔH° HC2H3O2∘) ΔH =((−285.8 kJ/mol)+(−393.5 kJ/mol))−((0 kJ/mol)+(−691.1 kJ/mol)) ΔH =(−679.3 kJ/mol)−(−691.1 kJ/mol) ΔH=11.8 kJ/mol The enthalpy change for the reaction is ΔH =11.8 kJ/mol (d) Is your calculated value consistent with your conclusions from observation? Explain. Answer: This value is consistent with our conclusion that the reaction is exothermic. A negative enthalpy change signifies the release of heat, confirming our observation of the system cooling down. P E R F O R M A N C E 2 n d S e m e s t e r - T A S K 4 t h General 02 Chemistry 0 1 Q u a r t e r Aldueza, Eldridge - Bobier, Angela - Caraig, Ashly Maldo, Atasha - Sumabat, Sheena ENTROPY, ENTHALPY, & FREE ENERGY STEP 4 (a) Define free energy and the significance of the sign of its value. There are two ways to calculate the free energy of the system observed in Step 1 from the available data. Answer: The amount of energy that is accessible for work is known as free energy. A process is considered to be spontaneous if the value is negative; otherwise, it is said to be non-spontaneous. (b) Calculate free energy both ways using the 2 equations and Table 1 data and calculated values from Steps 2 and 3 (assume that the reaction was performed at 25°C). Answer: - Free energy values = -44.5 kJ; - Entropy and enthalpy values = -44.4 kJ. (c) Compare the values. Answer: Both computed values are nearly equivalent. (d) What do these values tell you about the type of change that occurred? Explain. Answer: The reliability and uniformity of the calculating techniques are indicated by the minimal distinction between the two computed values. Therefore, the process is said to be spontaneous according to the negative free energy values. (e) A change in what condition would make which one of these two equations invalid? Explain. Answer: When the temperature changes, the Gibbs free energy equation ∆G = ∆H–T∆S becomes the only applicable equation, as it only benefits processes occurring at 25°C. standard circumstances. P E R F O R M A N C E 2 n d S e m e s t e r - T A S K 4 t h General 02 Chemistry 0 1 Q u a r t e r Aldueza, Eldridge - Bobier, Angela - Caraig, Ashly Maldo, Atasha - Sumabat, Sheena ENTROPY, ENTHALPY, & FREE ENERGY STEP 5 Now think about the process of an ice cube melting at room temperature. (a) What kind of change (physical/chemical) is this? Answer: The process of an ice cube melting at room temperature is a physical change, specifically a phase change from solid to liquid. (b) Without doing any calculations, predict the signs of entropy, enthalpy, and free energy and explain your reasoning. Answer: Based on the indications, we can foretell specific results when an ice cube melts. Entropy (∆S): Positive, as the solid (ice) becomes a more disordered liquid. Enthalpy (∆H): Positive, as heat is absorbed from the surroundings to overcome the intermolecular forces holding the ice together. Free energy (∆G): Negative, as the process is spontaneous under standard conditions. P E R F O R M A N C E 2 n d S e m e s t e r - T A S K 4 t h General 02 Chemistry 0 1 Q u a r t e r Aldueza, Eldridge - Bobier, Angela - Caraig, Ashly Maldo, Atasha - Sumabat, Sheena ENTROPY, ENTHALPY, & FREE ENERGY CHALLENGE Based on the consideration of two real processes, what can be concluded about how the sign and/or magnitude of entropy and enthalpy affect free energy? Answer the challenge by considering the equation: ∆G = ∆H–T∆S, to complete Table 2: Table 02 ∆H ∆S ∆G positive negative + positive negative positive - negative large and negative small and negative + positive small and positive large and positive - negative P E R F O R M A N C E 2 n d S e m e s t e r - T A S K 4 t h 0 1 Q u a r t e r Aldueza, Eldridge - Bobier, Angela - Caraig, Ashly Maldo, Atasha - Sumabat, Sheena General 02 Chemistry ENTROPY, ENTHALPY, & FREE ENERGY Add a little bit ohttps://chemistrytalk.org/howto-write-net-ionic-equations/f body text https://www.youtube.com/watch?v=biLiciae8J0
