The ΔH° f of TiI 3 (s) is –328 kJ/mol and the ΔH ° for the reaction 2 Ti(s) + 3 I 2 (g) → 2 TiI 3 (s) is –839 kJ. Calculate the ΔH of sublimation of I 2 (s), which is a solid at 25 °C.

hey everyone in this example, we need to calculate the change in entropy for our given reaction here, using the corresponding formation equations of sulfur dioxide gas and sulfur dioxide gas. So what we should do is write out these corresponding formation equations. So first, beginning with our sulfur trioxide gas, we would recognize that we would have the combination of sulfur as a solid, combining with oxygen gas to form our sulfur trioxide gas. Now we want to go ahead and make sure we have the proper coefficient so that things are balanced. And in order to balance this equation, we would place the coefficient of three halves in front of our oxygen, and that would give us a total of three oxygen's on both sides of our equation. And according to this question, we have an entropy of formation value equal to negative 95.7 kg jewels for this equation. Now, for our second formation equation, we want to form our sulfur dioxide. So we would have solid sulfur reacting with oxygen gas to form S. 0. 2 or sulfur dioxide gas, and again to make sure that this is balanced. We would go ahead and compare our oxygen and sulfur is on both sides. And as our equation is written, it is balanced. So we can leave it as is and we're going to write down our entropy a formation given for this equation as negative 2 96.8 kg joules. Now, our next step is to realize how these formation equations here, 4, 1 and two, sorry, that should be too Would equal our final equation, which in the problem has given us two moles of sulfur trioxide gas, which is producing two moles of sulfur dioxide gas and one mole of oxygen gas. And this is ultimately going to give us our n therapy for this reaction By determining how these two formation equations were manipulated to end up with this final equation. So analyzing our first equation here, we would recognize that we should have our sulfur trioxide actually on the reactant side, according to our final equation. So we're going to want to flip our sign of our entropy of formation here in order to flip our equation or rather than flip, we should just say reverse. We also should recognize that in our final equation, we want two moles of our sulfur trioxide. And so that means we're going to take our entire equation And multiply everything by two and that includes our entropy of formation. So we're going to multiply this by two and we're going to reverse our equation to get everything on the proper side. And so this is going to give us and I'll move this final equation down here. This is going to give us our manipulated first formation equation as two moles of our solid sulfur added to three moles of our oxygen gas. And sorry, this should be flipped. So this is actually going to be on the product side here, And this is produced from two moles of our sulfur trioxide gas. And then when we take that and they'll be a formation and multiply it by two. As well as flipping our sign there, we're going to get an entropy of formation equal to 91.4 kg jewels. And this is positive because we flip the sign due to us having to reverse our equation to get our reactant and products on the proper side. So this would be our new equation for the first equation there. And moving on to the second equation, we want to compare how this should be manipulated to end up with our final equation. So comparing it to our final equation, we do have all of our reactant and products on the proper side. However, we should end up with two moles of our sulfur dioxide gas. So we're going to actually multiply everything in this equation by two, including our entropy of formation. And so this would give us a new manipulated equation where we have two moles of our solid sulfur Added to two moles of our oxygen gas, producing two moles of our silver dioxide gas. And sorry, that should be a two there And then giving us an entropy of formation equal to a value of two times our original value, which gives us an amount of negative 5 93.6 kg jewels. And so now our next step is to add these two equations up to see how they give us our final equation. So we're going to go ahead and cancel out Our three moles of our oxygen gas Um with two moles of our oxygen gas and that's going to leave us with one mole of oxygen gas behind from our first equation. And we can also cancel out our two moles of our solid sulfur with the two moles of solid sulfur in the second equation. So everything that's left behind is going to give us our two moles of our sulfur trioxide gas, producing one mole of our oxygen gas and two moles of our silver dioxide gas. And now to find our final entropy, we're going to find that by taking our entropy from our first manipulated equation 7 91.4 kg jewels and adding that to our second manipulated entropy from the second equation, which is negative 5 93.6 kg jewels. And this gives us a final entropy value for our third equation equal to 1 97.8 kg jewels. So this is going to complete this example as our final answer for our change in entropy for our reaction, where two moles of sulfur trioxide produced two moles of sulfur dioxide and one mole of oxygen gas. So this completes this example. If you have any questions, please leave them down below and I will see everyone in the next practice video

Ch.7 - Thermochemistry

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Tro 6th Edition

Ch.7 - Thermochemistry

Problem 122

Chapter 7, Problem 122

The ΔH°_f of TiI₃(s) is –328 kJ/mol and the ΔH ° for the reaction 2 Ti(s) + 3 I₂(g) → 2 TiI₃(s) is –839 kJ. Calculate the ΔH of sublimation of I₂(s), which is a solid at 25 °C.

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Identify the given values: ΔH°f of TiI3(s) is -328 kJ/mol, and ΔH° for the reaction 2 Ti(s) + 3 I2(g) → 2 TiI3(s) is -839 kJ.

Write the formation reaction for TiI3(s) from its elements in their standard states: Ti(s) + 1.5 I2(g) → TiI3(s).

Calculate the ΔH° for the formation of 2 moles of TiI3(s) using the given ΔH°f: 2 moles × (-328 kJ/mol) = -656 kJ.

Set up an equation for the ΔH° of the reaction using Hess's Law: ΔH° reaction = [2 × ΔH°f(TiI3(s))] - [2 × ΔH°f(Ti(s)) + 3 × ΔH° sublimation(I2(s))].

Solve for ΔH° sublimation(I2(s)) by rearranging the equation and substituting the known values: ΔH° sublimation(I2(s)) = [2 × ΔH°f(TiI3(s)) - ΔH° reaction] / 3.

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Key Concepts

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

Enthalpy of Formation (ΔH°f)

The enthalpy of formation (ΔH°f) 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 energy changes in chemical reactions. In this question, the ΔH°f of TiI3(s) is given, which is essential for determining the overall energy change in the reaction involving titanium and iodine.

Sublimation

Sublimation is the process where a solid transitions directly into a gas without passing through the liquid phase. The enthalpy change associated with sublimation, known as the enthalpy of sublimation, is crucial for understanding the energy required to convert solid iodine (I2) into gaseous iodine. This concept is necessary to calculate the ΔH of sublimation for I2(s) in the context of the given reaction.

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 us to calculate unknown enthalpy changes by using known values from related reactions. In this problem, Hess's Law can be applied to relate the enthalpy of sublimation of I2(s) to the given enthalpy changes of formation and reaction.