The mechanism for the oxidation of HBr by O 2 to form 2 H 2 O and Br 2 is shown in Exercise 14.74. (a) Calculate the overall standard enthalpy change for the reaction process.

Hi everyone for this problem we're told gaseous hydrochloric acid is oxidized by oxygen to form water vapor and chlorine gas, identify the standard entropy change of the reaction. Okay, so we're looking for standard entropy change. And in order for us to calculate the standard entropy change, we're going to first need to write out our reaction and so our reaction is going to look like this. Okay, so now that we have our reaction in order for us to calculate the standard entropy change, we're going to need the standard heats of formation for everything that's in our reaction. And so these are values that we're going to look up and each thing is going to have its own value. So the standard heat of formation of hydrochloric acid Is negative 92.30 killer jewels per mole four 02 gas. It's going to be zero kg jewels per mole and it's zero because elements in their standard states equal zero. Okay, next we have H 20 gas and this value is negative 241.8 kg joules per mole. And for chlorine gas Our value is zero because it's in its standard state. Okay, so now that we have those values in order to calculate the standard entropy change for our reaction, it's going to equal the sum of our products minus The sum of our reactant. So we're going to take these values and multiply them by the number of moles we have for each and that will give us our final standard entropy change for the reaction. So let's go ahead and start off with our products. Okay, so for our products we have two products, we have water and chlorine gas and we have two moles of each. So four our first one. Let's start off with this one. So we're going to take our two moles of water and multiply it by its standard heat of formation which is negative 241.8 kg joules per mole. And we're going to add our second reactant which is chlorine gas and this is zero. Okay, so we can just go ahead and write in zero kg joules per mole. So that's the sum of our products. And we're going to subtract the sum of our reactant. So let's take a look at our reactant. Our first reactant is hydrochloric acid and we have four moles of that. So we'll take four moles of hydrochloric acid and multiply it by its standard heat of formation, negative 92.30 kila jewels per mole plus our second reactant is gas. So this is going to be plus zero kila jewels per mole. Okay, let's move this over so we can see in full. Okay, so this is zero killer jewels Permal. So this is the sum of our products minus the sum of our reactant. Now, when we calculate this we'll get a final value That the standard entropy change for our reaction is going to equal negative 114 0. kila jules Permal. And this is our final answer. That's the end of this problem. I hope this was helpful.

Ch.14 - Chemical Kinetics

All textbooks

Brown 14th Edition

Ch.14 - Chemical Kinetics

Problem 121a

Chapter 14, Problem 121a

The mechanism for the oxidation of HBr by O₂ to form 2 H₂O and Br₂ is shown in Exercise 14.74. (a) Calculate the overall standard enthalpy change for the reaction process.

Verified step by step guidance

Identify the balanced chemical equation for the reaction: \(4 \text{HBr} + \text{O}_2 \rightarrow 2 \text{H}_2\text{O} + 2 \text{Br}_2\).

Determine the standard enthalpy of formation (\(\Delta H_f^\circ\)) for each reactant and product from a reliable source, such as a chemistry textbook or database.

Use the formula for the standard enthalpy change of the reaction: \(\Delta H_{reaction}^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})\).

Calculate the sum of the standard enthalpies of formation for the products: \(2 \times \Delta H_f^\circ (\text{H}_2\text{O}) + 2 \times \Delta H_f^\circ (\text{Br}_2)\).

Calculate the sum of the standard enthalpies of formation for the reactants: \(4 \times \Delta H_f^\circ (\text{HBr}) + 1 \times \Delta H_f^\circ (\text{O}_2)\), and then subtract the reactants' sum from the products' sum to find the overall \(\Delta H_{reaction}^\circ\).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

Standard Enthalpy Change

Standard enthalpy change (ΔH°) refers to the heat change that occurs at constant pressure when reactants are converted to products under standard conditions (1 atm pressure and a specified temperature, usually 25°C). It is a crucial concept in thermodynamics, allowing chemists to predict the energy changes associated with chemical reactions.

Hess's Law

Hess's Law states that the total enthalpy change for a 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 enthalpy changes for complex reactions by breaking them down into simpler steps for which enthalpy values are known.

Bond Enthalpies

Bond enthalpy is the energy required to break one mole of a specific type of bond in a gaseous molecule. Understanding bond enthalpies is essential for calculating the overall enthalpy change of a reaction, as it involves summing the energies required to break bonds in the reactants and the energies released when forming bonds in the products.

Recommended video:

Guided course

02:34

Enthalpy of Formation

Related Practice

Textbook Question

The reaction between ethyl iodide and hydroxide ion in ethanol (C₂H₅OH) solution, C₂H₅I(alc) + OH^-(alc) → C₂H₅OH(l) + I^-(alc), has an activation energy of 86.8 kJ/mol and a frequency factor of 2.10 × 10¹¹ M^-1 s^-1. (d) Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at 50 C.

Textbook Question

The gas-phase reaction of NO with F₂ to form NOF and F has an activation energy of Ea = 6.3 kJ/mol. and a frequency factor of A = 6.0 × 10⁸ M^-1 s^-1. The reaction is believed to be bimolecular: NO(g) + F₂(g) → NOF(g) + F(g) (b) Draw the Lewis structures for the NO and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule.

Textbook Question

The gas-phase reaction of NO with F2 to form NOF and F has an activation energy of Ea = 6.3 kJ>mol. and a frequency factor of A = 6.0 * 108 M-1 s-1. The reaction is believed to be bimolecular: NO1g2 + F21g2 ¡ NOF1g2 + F1g2 (e) Suggest a reason for the low activation energy for the reaction.

Textbook Question

The mechanism for the oxidation of HBr by O₂ to form 2 H₂O and Br₂ is shown in Exercise 14.74. (c) Draw a plausible Lewis structure for the intermediate HOOBr. To what familiar compound of hydrogen and oxygen does it appear similar?

Textbook Question

The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride:

Reaction 1: CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g)

This reaction is very slow in the absence of light. However, Cl₂(g) can absorb light to form Cl atoms:

Reaction 2: Cl₂(g) + hv → 2 Cl(g)

Once the Cl atoms are generated, they can catalyze the reaction of CH₄ and Cl₂, according to the following proposed mechanism:

Reaction 3: CH₄(g) + Cl(g) → CH₃(g) + HCl(g)

Reaction 4: CH₃(g) + Cl₂(g) → CH₃Cl(g) + Cl(g)

The enthalpy changes and activation energies for these two reactions are tabulated as follows:

Reaction ΔH° (kJ/mol) E_a (kJ/mol)

3 +4 17

4 -109 4

(b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4.

The mechanism for the oxidation of HBr by O2 to form 2 H2O and Br2 is shown in Exercise 14.74. (a) Calculate the overall standard enthalpy change for the reaction process.

Verified video answer for a similar problem:

Key Concepts

Standard Enthalpy Change

Hess's Law

Bond Enthalpies

The mechanism for the oxidation of HBr by O₂ to form 2 H₂O and Br₂ is shown in Exercise 14.74. (a) Calculate the overall standard enthalpy change for the reaction process.