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Ch.9 - Molecular Geometry and Bonding Theories

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?

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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.
Related Practice
Textbook Question

The energy-level diagram in Figure 9.36 shows that the sideways overlap of a pair of p orbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding π orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the p2p to the p*2p molecular orbital. (c) Is the C¬C bond in ethylene stronger or weaker in the excited state than in the ground state? Why?

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Textbook Question

Sulfur tetrafluoride 1SF42 reacts slowly with O2 to form sulfur tetrafluoride monoxide 1OSF42 according to the following unbalanced reaction: SF41g2 + O21g2¡OSF41g2 The O atom and the four F atoms in OSF4 are bonded to a central S atom. (a) Balance the equation.

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Textbook Question

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. (b) Write a Lewis structure of OSF4 in which the formal charges of all atoms are zero.

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Textbook Question

Sulfur tetrafluoride 1SF42 reacts slowly with O2 to form sulfur

tetrafluoride monoxide 1OSF42 according to the following

unbalanced reaction:

SF41g2 + O21g2¡OSF41g2

The O atom and the four F atoms in OSF4 are bonded to a

central S atom.

(e) For each of the molecules you drew in part (d), state how many

fluorines are equatorial and how many are axial.

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Textbook Question
The phosphorus trihalides 1PX32 show the following variation in the bond angle X¬P¬X: PF3, 96.3°; PCl3, 100.3°; PBr3, 101.0°; PI3, 102.0°. The trend is generally attributed to the change in the electronegativity of the halogen. (b) What is the general trend in the X¬P¬X angle as the halide electronegativity increases?
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Textbook Question
Many compounds of the transition-metal elements contain direct bonds between metal atoms. We will assume that the z-axis is defined as the metal–metal bond axis. (d) Sketch the energylevel diagram for the Sc2 molecule, assuming that only the 3d orbital from part (a) is important.
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