Calculate ΔHrxn for the reaction: CH 4 (g) + 4 Cl 2 (g) → CCl 4 (g) + 4 HCl(g) Use the following reactions and given ΔH's: C(s) + 2 H 2 (g) → CH 4 (g) ΔH = –74.6 kJ C(s) + 2 Cl 2 (g) → CCl 4 ( g) ΔH = –95.7 kJ H 2 (g) + Cl 2 (g) → 2 HCl( g) ΔH = –92.3 kJ

Hello, everyone. Today we have the following problem. Calculate the ed for the reaction, use the following reactions and given changes in entropy. So we can use what is known as Hess's law that states some of the entropies of individual steps is equal to the entropy of the overall reaction. So let's look at reaction one, we see here that when compared to the overall reaction, we have our methane on the product side and our first reaction. But in our overall reaction, it is on the reactants. So we essentially have to reverse our sign of our entropy change. So that goes from an entropy of a negative 74.6 to positive 74.6 kilojoules. We look at our second reaction, we see that our fluorine gas is on the reactant side and the overall reaction and so is our carbon tetra fluoride that's on the product side. So there is no change. We then look at our third and final reaction. And we see here that while we do have our Hydrofluoric acid on the product side, we see that is or it has been halved. So what we need to do is we need to multiply our entropy change by that variable or two such that our new entropy will be negative 1084 kilojoules. And then to solve for hour entropy of the reaction, the total reaction, we simply add up the individual entropies. So we have our positive 74.6 kilojoules that will get added to our negative 933 kilojoules. And then lastly, we add by our negative 1084 kilojoules for a final value of negative 1942.4 kilojoules or WA C. And with that, we have solved the problem overall, I hope is helped. And until next time.

Ch.6 - Thermochemistry

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Tro 4th Edition

Ch.6 - Thermochemistry

Problem 82

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Chapter 6, Problem 82

Calculate ΔHrxn for the reaction:
CH₄(g) + 4 Cl₂(g) → CCl₄(g) + 4 HCl(g)
Use the following reactions and given ΔH's:
C(s) + 2 H₂(g) → CH₄(g) ΔH = –74.6 kJ
C(s) + 2 Cl₂(g) → CCl₄( g) ΔH = –95.7 kJ
H₂(g) + Cl₂(g) → 2 HCl( g) ΔH = –92.3 kJ

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Identify the target reaction: CH4(g) + 4 Cl2(g) \rightarrow CCl4(g) + 4 HCl(g).

List the given reactions and their enthalpy changes: 1) C(s) + 2 H2(g) \rightarrow CH4(g), \Delta H = -74.6 \text{ kJ}; 2) C(s) + 2 Cl2(g) \rightarrow CCl4(g), \Delta H = -95.7 \text{ kJ}; 3) H2(g) + Cl2(g) \rightarrow 2 HCl(g), \Delta H = -92.3 \text{ kJ}.

Use Hess's Law to manipulate the given reactions to derive the target reaction. This involves reversing and/or multiplying the given reactions to match the stoichiometry of the target reaction.

Reverse the first reaction to get CH4(g) \rightarrow C(s) + 2 H2(g), and change the sign of \Delta H to +74.6 \text{ kJ}.

Combine the manipulated reactions to form the target reaction, ensuring that all intermediate species cancel out, and sum the \Delta H values to find \Delta H_{\text{rxn}} for the target 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). Understanding ΔH is crucial for predicting the energy changes during chemical reactions, which is essential for calculating the overall enthalpy change for a reaction using Hess's law.

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 chemists to calculate the enthalpy change for a reaction that may be difficult to measure directly by using known enthalpy changes from related reactions.

Standard Enthalpy of Formation

The standard enthalpy of formation (ΔHf°) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. It provides a reference point for calculating the enthalpy changes of reactions. In the given problem, the ΔH values for the reactions provided can be used to derive the ΔHrxn for the overall reaction by applying Hess's Law.

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Enthalpy of Formation

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Related Practice

Textbook Question

Calculate ΔH_rxn for the reaction:

Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g)

Use the following reactions and given ΔH's:

2 Fe(s) + 3/2 O₂(g) → Fe₂O₃(s) ΔH = –824.2 kJ