From the enthalpies of reaction H 2 (g) + F 2 (g) → 2 HF(g) ΔH = -537 kJ C(s) + 2 F 2 (g) → CF 4 (g) ΔH = -680 kJ 2 C(s) + 2 H 2 (g) → C 2 H 4 (g) ΔH = +52.3 kJ Calculate H for the reaction of ethylene with F 2 : C 2 H 4 (g) + 6 F 2 (g) → 2 CF 4 (g) + 4 HF(g)

Hey everyone. So here it says to calculate the standard entropy of reaction for the following reaction. Here we have two moles of carbon solid reacting with one mole of hydrogen gas to produce one mole of a seedling gas. All right, so basically, we may have to manipulate one or more of these reactions given below to obtain this first one given to us. All right, So, let me just write it down here. So we want to see solid Plus H two gas Giving US C two, H 2. Gas. Doing that will help us find delta H of reaction. Alright, so, we need to find two moles of carbon solid. If we look up above here is carbon solid here, the issue is we need two of them. So, that means they're gonna multiply this entire thing. Times two. When I do that, I'm gonna get to see solid Plus 202. Give Me two c. 02. Multiplying by two. Means I also multiply delta H by two. And when I do that it becomes negative 787 kg joules. Alright, so now that I've done that, this is gone because it's now this new equation. Next I need to find one mole of H two. Alright, so where do we ch. Two? We see H two. Right here, it's they're both reactant. They're both one mole each. So that means I don't touch this equation at all. This equation stays as it is. Alright, so then I'm just gonna bring it over. So, I'm gonna have a church two gas Plus 002 gas gives me Htoo liquid. So the number stays negative to 85.8 killer jewels. Lastly I need one mole of C. Two H. Two. Okay, so if I need one more mole of c. two h. 2 as a product, I have to reverse the top equation. So when I reverse it I'm gonna have to C. 02. Gas plus H. Two. A liquid gives me C. Two H. Two. Gas Plus 5/2 is really just 2.5 02 gas. And when you reverse it you reverse the sign. So it's one positive 1129 9.5 kg. Now let's get rid of intermediates. Here we have water as a liquid in reacting form. And here it's in product form they cancel out two moles of C. 02. Gas as reacting cancel out with two moles of C. 02 as product. Now here we have half 02 plus 202. That's 2.502 as reactant. Those 2.5s cancel out with this 2.5. Remember 5/2 is equal to 2.5. Now, if we look what's left, we have two moles of C. Solid, one mole of H 20. And one mole of C. Two H. Two. Those make up are the equation. We need to find all we do. Now is we add up these three new delta H values, adding them all up together will give us our final answer. So when we add them all up together, I get my final answer as 2-6. killing jewels, so this would be my final answer for the given equation.

Ch.5 - Thermochemistry

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Brown 14th Edition

Ch.5 - Thermochemistry

Problem 65

Chapter 5, Problem 65

From the enthalpies of reaction H₂(g) + F₂(g) → 2 HF(g) ΔH = -537 kJ C(s) + 2 F₂(g) → CF₄(g) ΔH = -680 kJ 2 C(s) + 2 H₂(g) → C₂H₄(g) ΔH = +52.3 kJ Calculate H for the reaction of ethylene with F₂: C₂H₄(g) + 6 F₂(g) → 2 CF₄(g) + 4 HF(g)

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Identify the target reaction: \( \text{C}_2\text{H}_4(g) + 6 \text{F}_2(g) \rightarrow 2 \text{CF}_4(g) + 4 \text{HF}(g) \).

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

Write the given reactions and their enthalpy changes: \( \text{H}_2(g) + \text{F}_2(g) \rightarrow 2 \text{HF}(g) \), \( \Delta H = -537 \text{ kJ} \); \( \text{C}(s) + 2 \text{F}_2(g) \rightarrow \text{CF}_4(g) \), \( \Delta H = -680 \text{ kJ} \); \( 2 \text{C}(s) + 2 \text{H}_2(g) \rightarrow \text{C}_2\text{H}_4(g) \), \( \Delta H = +52.3 \text{ kJ} \).

Manipulate the given reactions to match the target reaction: Reverse the third reaction to form \( \text{C}_2\text{H}_4(g) \) as a reactant, which changes its \( \Delta H \) to \( -52.3 \text{ kJ} \).

Add the enthalpy changes of the manipulated reactions to find the total \( \Delta H \) 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 of Reaction

Enthalpy of reaction (ΔH) is the heat change associated with a chemical reaction 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 reactions and for applying Hess's law to calculate enthalpy changes for complex reactions.

Hess's Law

Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for individual steps, 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 of related reactions. It is particularly useful in multi-step reactions, like the one presented in the question.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on the balanced chemical equation. It provides the ratios in which substances react and form products, which is essential for determining the enthalpy changes in reactions. In the context of the question, stoichiometry helps in relating the amounts of C2H4 and F2 to the products CF4 and HF, allowing for accurate enthalpy calculations.

From the enthalpies of reaction H2(g) + F2(g) → 2 HF(g) ΔH = -537 kJ C(s) + 2 F2(g) → CF4(g) ΔH = -680 kJ 2 C(s) + 2 H2(g) → C2H4(g) ΔH = +52.3 kJ Calculate H for the reaction of ethylene with F2: C2H4(g) + 6 F2(g) → 2 CF4(g) + 4 HF(g)