AP Chemistry 9.2 Absolute Entropy and Entropy Change Study Notes - New Syllabus Effective fall 2024
AP Chemistry 9.2 Absolute Entropy and Entropy Change Study Notes- New syllabus
AP Chemistry 9.2 Absolute Entropy and Entropy Change Study Notes – AP Chemistry – per latest AP Chemistry Syllabus.
LEARNING OBJECTIVE
Calculate the standard entropy change for a chemical or physical process based on the absolute entropies (standard molar entropies) of the species involvedinthe process.
Key Concepts:
- Calculating Standard Entropy Change (\( \mathrm{\Delta S^\circ} \))
Calculating Standard Entropy Change (\( \mathrm{\Delta S^\circ} \))
The standard entropy change of a reaction (\( \mathrm{\Delta S^\circ_{reaction}} \)) is the difference between the total entropy of the products and the total entropy of the reactants, each multiplied by their respective stoichiometric coefficients from the balanced chemical equation.
Formula:
\( \mathrm{\Delta S^\circ = \sum S^\circ(products) – \sum S^\circ(reactants)} \)
where:
\( \mathrm{S^\circ} \) = standard molar entropy (in J mol\(^{-1}\) K\(^{-1}\))
\( \mathrm{\sum S^\circ(products)} \) = sum of the standard entropies of all products
\( \mathrm{\sum S^\circ(reactants)} \) = sum of the standard entropies of all reactants
Important Notes:
- Use the balanced chemical equation to determine the number of moles of each species.
- The units of entropy are usually J mol\(^{-1}\) K\(^{-1}\).
- If \( \mathrm{\Delta S^\circ > 0} \): the process leads to greater disorder (entropy increases).
- If \( \mathrm{\Delta S^\circ < 0} \): the process becomes more ordered (entropy decreases).
Step-by-Step Method:
- Write the balanced equation.
- List the standard molar entropies (\( \mathrm{S^\circ} \)) for all reactants and products.
- Multiply each \( \mathrm{S^\circ} \) value by its coefficient from the equation.
- Subtract the total entropy of the reactants from that of the products.
Example :
Calculate the standard entropy change for the combustion of hydrogen gas:
\( \mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(l)} \)
Given data:
\( \mathrm{S^\circ(H_2(g)) = 131.0\ J\ mol^{-1}\ K^{-1}} \)
\( \mathrm{S^\circ(O_2(g)) = 205.0\ J\ mol^{-1}\ K^{-1}} \)
\( \mathrm{S^\circ(H_2O(l)) = 70.0\ J\ mol^{-1}\ K^{-1}} \)
▶️ Answer/Explanation
Step 1: Apply the formula:
\( \mathrm{\Delta S^\circ = \sum S^\circ(products) – \sum S^\circ(reactants)} \)
Step 2: Substitute the values:
\( \mathrm{\Delta S^\circ = [2(70.0)] – [2(131.0) + 205.0]} \)
Step 3: Calculate:
\( \mathrm{\Delta S^\circ = 140.0 – 467.0 = -327.0\ J\ mol^{-1}\ K^{-1}} \)
Step 4: Interpretation:
The entropy change is negative, indicating that the products are more ordered than the reactants. This is consistent with gases forming a liquid, which has less particle freedom.
Final Answer: \( \mathrm{\Delta S^\circ = -327\ J\ mol^{-1}\ K^{-1}} \)
Example :
Calculate \( \mathrm{\Delta S^\circ} \) for the decomposition of calcium carbonate:
\( \mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)} \)
Given:
\( \mathrm{S^\circ(CaCO_3(s)) = 93.0\ J\ mol^{-1}\ K^{-1}} \)
\( \mathrm{S^\circ(CaO(s)) = 39.7\ J\ mol^{-1}\ K^{-1}} \)
\( \mathrm{S^\circ(CO_2(g)) = 213.7\ J\ mol^{-1}\ K^{-1}} \)
▶️ Answer/Explanation
Step 1: Use the formula:
\( \mathrm{\Delta S^\circ = [S^\circ(CaO) + S^\circ(CO_2)] – [S^\circ(CaCO_3)]} \)
Step 2: Substitute values:
\( \mathrm{\Delta S^\circ = (39.7 + 213.7) – 93.0} \)
Step 3: Calculate:
\( \mathrm{\Delta S^\circ = 253.4 – 93.0 = +160.4\ J\ mol^{-1}\ K^{-1}} \)
Step 4: Interpretation:
The positive entropy change indicates increased disorder due to the formation of a gas from a solid reactant.
Final Answer: \( \mathrm{\Delta S^\circ = +160.4\ J\ mol^{-1}\ K^{-1}} \)