Using values from Appendix C, calculate the value of H for each of the following reactions: (a) CaO(s) + 2 HF(g) → CaF 2 (s) + H 2 O(g) (b) Fe 2 O 3 (s) + 3 C(s) → 2 Fe(s) + 3CO(g) (c) 2 CO(g) + 2 NO(g) → N 2 (s) + 2 CO 2 (g) (d) 4 NH 3 (g) + 5 O 2 (g) → 4 NO(g) + 6 H 2 Og)

hey everyone we're asked to calculate the standard entropy of reaction for the reduction of iron to oxide with carbon monoxide First. I'm going to go ahead and write out our change in entropy for each of our compounds. So the change in entropy of iron to oxide comes up to negative 272.0 kg jewels per mole. And our change in entropy of carbon monoxide comes up to negative 110.5 kg joules per mole. Our change in M V. P for iron comes up to zero kg jules, Permal. And lastly our change in entropy of carbon dioxide comes up to negative 393.5 kg joules per mole. And all of these values can be found in your textbook now to determine the standard entropy of our reaction. And as we've learned this is going to be the change in entropy of our products. Mine is the change in entropy of our reactant so far. Standard entropy of our reaction. This will be the change in entropy of our iron plus the change in entropy of our carbon dioxide minus the change in M. V. P of our iron to oxide plus the change in entropy of our carbon monoxide. So plugging in these values we get sarah killen joules per mole from our iron and we're going to add negative 393.5 kg joules per mole from our carbon dioxide. And we're going to subtract the sum of the change in entropy of our iron to oxide which is negative 272.0 killer joules per mole plus negative 110. kg joules per mole, which was the change in entropy of our carbon monoxide. So when we calculate this out, our standard envelop of our reaction comes up to negative 11 point oh kila jewels Permal, and this is going to be our final answer. Now, I hope that made sense and let us know if you have any questions.

Ch.5 - Thermochemistry

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Brown 14th Edition

Ch.5 - Thermochemistry

Problem 74

Chapter 5, Problem 74

Using values from Appendix C, calculate the value of H for each of the following reactions: (a) CaO(s) + 2 HF(g) → CaF₂(s) + H₂O(g) (b) Fe₂O₃(s) + 3 C(s) → 2 Fe(s) + 3CO(g) (c) 2 CO(g) + 2 NO(g) → N₂(s) + 2 CO₂(g) (d) 4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂Og)

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Identify the standard enthalpy of formation (ΔH_f^°) for each compound involved in the reaction from Appendix C.

Write the balanced chemical equation: CaO(s) + 2 HF(g) → CaF_2(s) + H_2O(g).

Apply Hess's Law, which states that the change in enthalpy for a reaction is the sum of the enthalpy changes for each step of the reaction.

Calculate the enthalpy change (ΔH) for the reaction using the formula: ΔH = Σ(ΔH_f^° of products) - Σ(ΔH_f^° of reactants).

Substitute the values of ΔH_f^° for each compound into the formula and perform the arithmetic to find the enthalpy change for the reaction.

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Key Concepts

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

Enthalpy Change (ΔH)

Enthalpy change (ΔH) is a measure of the heat content of a system at constant pressure. It indicates whether a reaction is exothermic (releases heat, ΔH < 0) or endothermic (absorbs heat, ΔH > 0). Calculating ΔH for a reaction often involves using standard enthalpy values from tables, which represent the enthalpy of formation for various substances.

Hess's Law

Hess's Law states that the total enthalpy change for a reaction is the same, regardless of the number of steps taken to achieve the reaction. This principle allows chemists to calculate the enthalpy change for complex reactions by summing the enthalpy changes of individual steps, making it easier to find ΔH using tabulated values.

Standard Enthalpy of Formation

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. These values are crucial for calculating the enthalpy change of a reaction using the formula ΔH = ΣΔH°f(products) - ΣΔH°f(reactants), allowing for the determination of the overall energy change during a chemical reaction.

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Enthalpy of Formation

Related Practice

Textbook Question

Acetylene (C₂H₂(g)) is used for welding because oxyacetylene is the hottest burning common fuel gas. Using standard enthalpies of formation, calculate the quantity of heat produced when 10 g of acetylene is completely combusted in air under standard conditions.

Textbook Question

Complete combustion of 1 mol of acetone (C₃H₆O) liberates 1790 kJ: C₃H₆O(l) + 4 O₂(g) → 3 CO₂(g) + 3 H₂O(l) ΔH° = -1790 kJ Using this information together with the standard enthalpies of formation of O₂(g), CO₂(g), and H₂O(l) from Appendix C, calculate the standard enthalpy of formation of acetone.

Textbook Question

Calcium carbide (CaC₂) reacts with water to form acetylene (C₂H₂) and Ca(OH)₂. From the following enthalpy of reaction data and data in Appendix C, calculate H°_f for CaC₂(s): CaC₂(s) + 2 H₂O(l) → Ca(OH₂)(s) + C₂H₂(g) ΔH° = -127.2 kJ

Textbook Question

Gasoline is composed primarily of hydrocarbons, including many with eight carbon atoms, called octanes. One of the cleanest–burning octanes is a compound called 2,3,4- trimethylpentane, which has the following structural formula: The complete combustion of one mole of this compound to CO₂(g) and H₂O(g) leads to ΔH° = -5064.9 kJ. (b) By using the information in this problem and data in Table 5.3, calculate H°f for 2,3,4-trimethylpentane.

Using values from Appendix C, calculate the value of H for each of the following reactions: (a) CaO(s) + 2 HF(g) → CaF2(s) + H2O(g) (b) Fe2O3(s) + 3 C(s) → 2 Fe(s) + 3CO(g) (c) 2 CO(g) + 2 NO(g) → N2(s) + 2 CO2(g) (d) 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2Og)

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Key Concepts

Enthalpy Change (ΔH)

Hess's Law

Standard Enthalpy of Formation

Using values from Appendix C, calculate the value of H for each of the following reactions: (a) CaO(s) + 2 HF(g) → CaF₂(s) + H₂O(g) (b) Fe₂O₃(s) + 3 C(s) → 2 Fe(s) + 3CO(g) (c) 2 CO(g) + 2 NO(g) → N₂(s) + 2 CO₂(g) (d) 4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂Og)