Sulfur tetrafluoride (SF 4 ) reacts slowly with O 2 to form sulfur tetrafluoride monoxide (OSF 4 ) according to the following unbalanced reaction: SF 4 (g) + O 2 (g) → OSF 4 (g) The O atom and the four F atoms in OSF 4 are bonded to a central S atom. (c) Use average bond enthalpies (Table 8.3) to estimate the enthalpy of the reaction. Is it endothermic or exothermic?

hi everyone for this problem we're told to consider the two reactions below, calculate the entropy of reaction Using bond n therapies and identify if both reactions are eggs are thermic or not. So we're going to calculate the entropy ease of reaction for both and determine if it is eggs are thermic or not. So in order for us to calculate the entropy ease of reaction, it's going to be our ndp of reaction is going to equal the sum of the bond energy for our reactant minus the sum of the bond energy of our products. Okay, and so we're going to need to know what our heats of formation is for our products and our reactant. So let's go ahead and start off with what were the first reaction that were given? We'll go ahead and rewrite this. Okay, and so we're going to need the heats of formation for each thing that's in our reaction. So let's go ahead and look using bond entropy. So our heat of formation is going to be our bond entropy heat of formation. Okay, so for our hydro or hbr our standard heat of formation is 366 kil jewels per mole For H 20. Gas. The value is 463 Killah jewels per mole for I mean this is going to be zero because it is an element in its standard state. So any element in its standard state is going to have zero as that value and for H 30 plus Our bond entropy is 463 kg per mole. So now that we know the bond and trapeze we're going to multiply each bond entropy by the number of bonds that's in that reactant or product. Okay. And so what that will look like is our entropy for the reaction is going to equal, our first reaction is HBR So we have one bond in HbR and we're going to multiply that by its bond entropy which is kg joules per mole. Moving on to our second reactant, it is H 20 and H 20. We have two moles. So this is going to be plus two moles of H2. And we're going to multiply that by its bond entropy which is 463 kilograms per mole. So that's the sum of our reactant and we're going to subtract it by the sum of our products. So b R minus. We only this is going to be zero. So we can just put that in. And for our second reactant H30 plus this has three bonds. So we're going to multiply that by its bond entropy which is 463. Okay, so this is the sum of our reactant minus the sum of our products and will be left with a entropy of reaction equal to negative 97 kg joules per mole. So that is for our first reaction. Now we're going to go ahead and do the same thing. But for our second reaction. Okay so let's go ahead and Rewrite out that 2nd reaction. And our second reaction is this. Okay so we're going to need the bond entropy. So let's go ahead and write that down. So for H. I. This is equal to 299 killed Jules per mole. For the second reactant. H 2 0. gas is 463 killer jewels Permal for I minus. Since this is an Element in its standard state, it's going to equal zero and for H 30 plus It is 463 kg joules per mole. So now that we have those bond entropy is we can calculate our entropy of the reaction. Sorry entropy of the reaction. So our first product is H. I. This only has one bond. So we're going to multiply that by its Bond entropy which is 299 kg jewels per mole plus H +20. Has two bonds. And that multiplied by its bond entropy is 463 kg jewels Erase that and read it clear. 463 kila jules Permal. And this is going to be minus our products. So R I minus is zero kg jewels per mole plus our H +30 plus has three bonds. And that multiplied by its bond entropy of 463. Okay so once we do this calculation we're going to get An entropy of reaction equal to negative 164 killer jewels per mole. Okay so this is going to be our second and there'll be a reaction. The first one was negative 97 kg joules per mole. The second is a negative 1 64 kg joules per mole. Since both have a negative entropy of reaction, that means both our eggs ah thermic. Okay so we're right both exo thermic and this is because the entropy of reaction is negative. Okay so that's it for this problem And these are our final our final values. So for the first one our value is negative 97 kg joules per mole. And our second value is negative 164 kg joules per mole. And both our eggs are thermic. That's it. I hope this was helpful.

Ch.9 - Molecular Geometry and Bonding Theories

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Brown 14th Edition

Ch.9 - Molecular Geometry and Bonding Theories

Problem 114c

Chapter 9, Problem 114c

Sulfur tetrafluoride (SF₄) reacts slowly with O₂ to form sulfur tetrafluoride monoxide (OSF₄) according to the following unbalanced reaction: SF₄(g) + O₂(g) → OSF₄(g) The O atom and the four F atoms in OSF₄ are bonded to a central S atom. (c) Use average bond enthalpies (Table 8.3) to estimate the enthalpy of the reaction. Is it endothermic or exothermic?

Verified step by step guidance

Identify the bonds broken and formed in the reaction. In SF_4, the bonds are S-F, and in O_2, the bond is O=O. In OSF_4, the bonds are S=O and S-F.

Use the average bond enthalpies from Table 8.3 to find the energy required to break the bonds in the reactants. Calculate the total energy for breaking 4 S-F bonds and 1 O=O bond.

Use the average bond enthalpies to find the energy released when forming the bonds in the products. Calculate the total energy for forming 1 S=O bond and 4 S-F bonds.

Calculate the enthalpy change of the reaction (ΔH) by subtracting the total energy of bonds formed from the total energy of bonds broken: ΔH = (Energy of bonds broken) - (Energy of bonds formed).

Determine if the reaction is endothermic or exothermic. If ΔH is positive, the reaction is endothermic (absorbs heat). If ΔH is negative, the reaction is exothermic (releases heat).

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Key Concepts

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

Bond Enthalpy

Bond enthalpy, or bond dissociation energy, is the energy required to break one mole of a specific type of bond in a gaseous substance. It is a crucial concept in thermochemistry as it helps estimate the energy changes during chemical reactions. By using average bond enthalpies, one can calculate the total energy required to break bonds in reactants and the energy released when new bonds are formed in products.

Enthalpy Change (ΔH)

The enthalpy change (ΔH) of a reaction is the difference in enthalpy between the products and the reactants. It indicates whether a reaction absorbs heat (endothermic, ΔH > 0) or releases heat (exothermic, ΔH < 0). Calculating ΔH involves summing the bond enthalpies of the bonds broken and formed, providing insight into the energy dynamics of the reaction.

Exothermic vs. Endothermic Reactions

Exothermic reactions release energy to the surroundings, often in the form of heat, resulting in a temperature increase in the environment. In contrast, endothermic reactions absorb energy, leading to a temperature decrease. Understanding the nature of the reaction in question helps predict its behavior and the energy changes involved, which is essential for determining the reaction's enthalpy.