Consider the following equilibrium between oxides of nitrogen 3 NO(g) ⇌ NO 2 (g) + N 2 O(g) (a) Use data in Appendix C to calculate ΔH° for this reaction.

hey everyone today, we're being asked to find the standard entropy of the reaction for the reaction of phosphorus becoming di phosphorus and tetra phosphorous at equilibrium. So to do this, we need to utilize the standard entropy of formation in order to find the standard entropy of the reaction as a whole. And we can use a general formula that goes as such. The standard entropy of the reaction reaction is equal to the difference between the standard entropy of the formation of the products and the standard entropy of formation of the reactant reactant. So with that in mind we actually need to find the specific values for the standard entropy of formation for each of the compounds present in the reaction. So let's go ahead and take a look at phosphorus first. So phosphorus, the standard entropy of formation, which is a experimentally calculated value and can be found in tables either online or in a textbook, but for phosphorus specifically phosphorus gas, The entropy of formation is 316. Kill it jules per mole for di phosphorus down to h the standard entropy of formation is 144 killer jewels per mole. And for tetra phosphorous, a standard entropy of formation is 58 0.9 kg joules per mole. So putting this into our products products minus reaction formula and let's keep our products in blue. We also need to consider how many moles of everything we have. So for the products we only have one mole produced of each. So we have one mole times 144 Kill it, jules per mole plus uh, one more Times 58.9 kg per mole. So this will be the standard entropy of formation for the products. And from here we will subtract the standard entropy of formation of the reactant. And we have six moles of loan phosphorus, that's what we have right here. So we actually factor that in as well. So we have six moles of p Times the 316.5 kila jules per mole and adding it all up. We get a final value. The final standard entropy of the reaction of negative 1696.1 kg jewels. I hope this helps. And I look forward to seeing you all in the next one.

Ch.15 - Chemical Equilibrium

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Brown 14th Edition

Ch.15 - Chemical Equilibrium

Problem 65a

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Chapter 15, Problem 65a

Consider the following equilibrium between oxides of nitrogen 3 NO(g) ⇌ NO₂(g) + N₂O(g) (a) Use data in Appendix C to calculate ΔH° for this reaction.

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Identify the standard enthalpies of formation (ΔH°f) for each compound involved in the reaction from Appendix C. You will need the values for NO(g), NO2(g), and N2O(g).

Write the balanced chemical equation for the reaction: 3 NO(g) ⇌ NO2(g) + N2O(g).

Apply the formula for the standard enthalpy change of the reaction (ΔH°rxn): ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants). Remember to multiply the ΔH°f of each substance by its stoichiometric coefficient in the balanced equation.

Calculate the sum of the standard enthalpies of formation for the products: (1 mole of NO2(g) × ΔH°f of NO2(g)) + (1 mole of N2O(g) × ΔH°f of N2O(g)).

Calculate the sum of the standard enthalpies of formation for the reactants: 3 moles of NO(g) × ΔH°f of NO(g).

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Key Concepts

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

Hess's Law

Hess's Law states that the total enthalpy change for a chemical reaction is the sum of the enthalpy changes for the individual steps of the reaction, regardless of the pathway taken. This principle allows for the calculation of ΔH° for reactions that may not be easily measured directly by using known enthalpy changes of related reactions.

Standard Enthalpy of Formation (ΔH°f)

The standard enthalpy of formation (ΔH°f) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. This value is crucial for calculating the enthalpy change of a reaction using Hess's Law, as it provides a reference point for the energy content of reactants and products.

Equilibrium Constant (K)

The equilibrium constant (K) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. Understanding K is essential for analyzing the position of equilibrium and how changes in conditions can affect the reaction, which is relevant when considering the enthalpy changes in reactions involving gases.

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Equilibrium Constant K

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Textbook Question

Consider the reaction 4 NH₃(𝑔) + 5 O₂(𝑔) ⇌ 4 NO(𝑔) + 6 H₂O(𝑔), Δ𝐻 = −904.4 kJ Does each of the following increase, decrease, or leave unchanged the yield of NO at equilibrium? (f) increase temperature.

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Open Question

How do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction: (a) removal of a reactant, (b) removal of a product?

Open Question

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction vessel. Does the balanced chemical equation have more molecules on the reactant side or product side?

Textbook Question

Methanol (CH₃OH) can be made by the reaction of CO with H₂: CO(𝑔) + 2 H₂(𝑔) ⇌ CH₃OH(𝑔) (a) Use thermochemical data in Appendix C to calculate ΔH° for this reaction.

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Textbook Question

Methanol (CH₃OH) can be made by the reaction of CO with H₂: CO(𝑔) + 2 H₂(𝑔) ⇌ CH₃OH(𝑔) (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature?

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Textbook Question

Methanol (CH₃OH) can be made by the reaction of CO with H₂: CO(𝑔) + 2 H₂(𝑔) ⇌ CH₃OH(𝑔) (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

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Consider the following equilibrium between oxides of nitrogen 3 NO(g) ⇌ NO2(g) + N2O(g) (a) Use data in Appendix C to calculate ΔH° for this reaction.

Verified Solution

Key Concepts

Hess's Law

Standard Enthalpy of Formation (ΔH°f)

Equilibrium Constant (K)

Consider the following equilibrium between oxides of nitrogen 3 NO(g) ⇌ NO₂(g) + N₂O(g) (a) Use data in Appendix C to calculate ΔH° for this reaction.