Find ΔS° for the formation of CH 2 Cl 2 (g) from its gaseous elements in their standard states. Rationalize the sign of ΔS°.

Hello everyone today. We are being asked to use the entropy of formation of the standard states of its elements to identify the change in entropy for the formation of phosphorus oxy chloride. So the first thing we wanna do is we want to reference our standard states of its elements for the following compounds. So we have our phosphorus oxy chloride here Which is in the gas form of course and that's going to be equal to 325.5 jewels per mole kelvin. We then have phosphorus as a solid in white form Which equals 41. jewels per mole kelvin. We of course have oxygen, molecular oxygen and the gas form which is going to be equal to 205.2 jewels per mole kelvin. And lastly we have molecular chloride or cl two in the gas form which is going to have an entropy of 223.1 jewels per mole kelvin. And once again these reference values may be found in your textbook. So the next thing you wanna do put that as step one. The next thing I wanna do is we want to write our out our equation. So we're going to have our phosphorus in the solid form react with half a mole of molecular oxygen in the gas form As well as 3/2 of our chlorine, Our Cl two in the gas form to finally form our phosphorus oxy chloride in the gas form as well. And so to find our change in our entropy for the reaction, we're going to take the standard states of its elements, the entropy for our products and subtract them from our reactant. And so our only product is phosphorus oxy chloride that's going to be 300 and 0.5 jewels Promote Kelvin by itself. And then we're going to subtract that from our entropy values for our reactant. So first we have solid phosphorus which has an entropy value of 0.1 jewels per mole kelvin. We're going to add that to our half More of molecular oxygen entropy. We're just going to be equal to 205. jules promote kelvin. And lastly we're going to add that to our three has entropy for chlorine, gas is going to be 223.1 jewels per mole kelvin. When we plug that into our calculator, we get a final value of negative 152.9 jules per mole. As our final answer. I hope this helped until next time

Ch.19 - Free Energy & Thermodynamics

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Tro 5th Edition

Ch.19 - Free Energy & Thermodynamics

Problem 57

Chapter 19, Problem 57

Find ΔS° for the formation of CH₂Cl₂(g) from its gaseous elements in their standard states. Rationalize the sign of ΔS°.

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Identify the reaction for the formation of CH2Cl2(g) from its gaseous elements: C(graphite) + 2H2(g) + 2Cl2(g) -> CH2Cl2(g).

Use the standard entropy values (S°) for each substance involved in the reaction. These values are typically found in a table of thermodynamic data.

Calculate the change in entropy (ΔS°) for the reaction using the formula: ΔS° = ΣS°(products) - ΣS°(reactants).

Substitute the standard entropy values into the formula to find ΔS°.

Rationalize the sign of ΔS°: Consider the change in the number of moles of gas and the complexity of the molecules involved. Formation of CH2Cl2(g) from simpler gaseous elements may lead to a decrease in entropy due to a decrease in the number of gas molecules.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Standard Entropy (ΔS°)

Standard entropy (ΔS°) is a measure of the disorder or randomness of a system at standard conditions (1 bar, 25°C). It quantifies the amount of energy unavailable for doing work due to the dispersal of energy in a system. A positive ΔS° indicates an increase in disorder, while a negative ΔS° suggests a decrease in disorder.

Formation Reaction

A formation reaction is a chemical reaction in which one mole of a compound is formed from its constituent elements in their standard states. For example, the formation of dichloromethane (CH2Cl2) involves combining carbon, hydrogen, and chlorine gases. The standard enthalpy and entropy changes for this reaction are crucial for calculating thermodynamic properties.

Gibbs Free Energy and Spontaneity

Gibbs free energy (G) combines enthalpy and entropy to determine the spontaneity of a reaction at constant temperature and pressure. The change in Gibbs free energy (ΔG) is given by the equation ΔG = ΔH - TΔS. A negative ΔG indicates a spontaneous process, while the sign of ΔS helps rationalize whether the reaction increases or decreases disorder, influencing the overall spontaneity.

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Textbook Question

Use data from Appendix IIB to calculate ΔS°_rxnfor each of the reactions. In each case, try to rationalize the sign of ΔS°_rxn. b. Cr₂O₃(s) + 3 CO(g) → 2 Cr(s) + 3 CO₂(g)

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Use data from Appendix IIB to calculate ΔS°_rxnfor each of the reactions. In each case, try to rationalize the sign of ΔS°_rxn. c. SO₂(g) + 1/2 O₂(g) → SO₃(g)

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Use data from Appendix IIB to calculate ΔS°_rxnfor each of the reactions. In each case, try to rationalize the sign of ΔS°_rxn. d. N₂O₄(g) + 4 H₂(g) → N₂(g) + 4 H₂O(g)

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Textbook Question

Methanol (CH₃OH) burns in oxygen to form carbon dioxide and water. Write a balanced equation for the combustion of liquid methanol and calculate ΔH°_rxn, ΔS°_rxn, and ΔG°_rxnat 25 °C. Is the combustion of methanol spontaneous?

In photosynthesis, plants form glucose (C₆H₁₂O₆) and oxygen from carbon dioxide and water. Write a balanced equation for photosynthesis and calculate ΔH°_rxn, ΔS°_rxn, and ΔG°_rxnat 25 °C. Is photosynthesis spontaneous?

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Textbook Question

For each reaction, calculate ΔH°_rxn, ΔS°_rxn, and ΔG°_rxnat 25 °C and state whether or not the reaction is spontaneous. If the reaction is not spontaneous, would a change in temperature make it spontaneous? If so, should the temperature be raised or lowered from 25 °C? a. N₂O₄(g) → 2 NO₂(g)

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Find ΔS° for the formation of CH2Cl2(g) from its gaseous elements in their standard states. Rationalize the sign of ΔS°.

Verified Solution

Key Concepts

Standard Entropy (ΔS°)

Formation Reaction

Gibbs Free Energy and Spontaneity

Find ΔS° for the formation of CH₂Cl₂(g) from its gaseous elements in their standard states. Rationalize the sign of ΔS°.