Under certain nonstandard conditions, oxidation by O 2 (g) of 1 mol of SO 2 (g) to SO 3 (g) absorbs 89.5 kJ. The enthalpy of formation of SO 3 (g) is –204.2 kJ under these conditions. Find the enthalpy of formation of SO 2 (g).

Welcome back, everyone. Under certain non-standard conditions, the decomposition of sulfur trioxide into one half of the mole of oxygen gas and one mole of sulfur dioxide gas absorbs 227.8 kilojoules. The entropy of formation of sulfur trioxide gas is negative 204.2 kilojoules under these conditions, find the entropy of formation of sulfur dioxide gas. And we're given four answer choices. A negative 64.9 B 32.4 C negative 40.4 D 23.6. All of them are given in kilojoules per mole. So first of all, let's write down the reaction we have. So three gas decomposes as the problem suggests into one half of a mole of oxygen gas and a one mole of sulfur dioxide gas. We can essentially write down that the entropy change of this reaction is equal to the sum of the entropies of formation of products minus the sum of the entropies of formation of reactants. It might sound confusing but essentially what we're going to do is just write down the equation. We have one half of a mole of oxygen. So we're taking one half of a mole multiplied by the entropy of formation of oxygen gas plus. Now our next product is one mole of so two. So we take one mole multiplied by the entropy of formation of so two gas and we are subtracting the ample piece of formation of the reactants. So the only reactant is so two, we take one mole of it. It's actually so three, right? So we're taking the entropy affirmation of so three gas. Now, what do we know? Well, essentially we know that the reaction absorbs heat. What does that mean? Well, essentially we know that absorption of heat represents an endothermic reaction. So we know that the entropy change of the reaction would be equal to 227.8 kilojoules. We also know that the entropy of formation of oxygen, right. In this case, it would be zero because it is in a standard state auction is gas here. So we're going to see that it is equal to zero. What we're seeing from here is that we end up with the entropies of formation of so two and so three, what are we looking for? Well, we're looking for the entropy of formation of so two. So if we rearrange the equation, we can say that the entropy of formation of two gas would be equal to the entropy of the reaction minus the entropy of formation of. So three gas, we take 227.8 kilojoules. And now we need to use the thermodynamic tables to identify the entropy affirmation of. So three, right, according to the tables, we can essentially state that based on the fact that the entropy formation of so three gas is equal to negative 204 12 killer jules. For more, we can use this number in our equation. There's one more thing to consider actually, if we look at our equation and if we are solving for the entropy of formation of so two, that actually means that we must add the entropy of formation of so three, right? Because it was negative, we are moving it to the left. So my apologies here, we must have a positive sign. So we're taking 227.8 kilojoules and we're adding one mole multiplied by negative 204.2 kilojoules per mole. And that would be our calculation. Of course, it would be more accurate to say that here we're just adding one mole since this is exactly what we stated previously. So let's add that one mole for complete accuracy. And now let's evaluate the result. Well, essentially we end up with 23.6 killer jewels, right? We don't have any units per M. But if we're interested in the molar entropy change formation, we can essentially add and state that the entropy of formation based on this reaction because we already had one mole, we can add that unit. So the entropy formation of so two gas would be equal to 23.6 kilojoules per mole of. So two, you know, looking at the answer choices, we can state that the correct answer is option D. Thank you for watching.

Ch.6 - Thermochemistry

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Ch.6 - Thermochemistry

Problem 108

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Chapter 6, Problem 108

Under certain nonstandard conditions, oxidation by O₂(g) of 1 mol of SO₂(g) to SO₃(g) absorbs 89.5 kJ. The enthalpy of formation of SO₃(g) is –204.2 kJ under these conditions. Find the enthalpy of formation of SO₂(g).

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Identify the given information: the enthalpy change for the reaction \( \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{SO}_3(g) \) is +89.5 kJ, and the enthalpy of formation of \( \text{SO}_3(g) \) is -204.2 kJ.

Recall that the enthalpy change for a reaction (\( \Delta H_{\text{rxn}} \)) can be calculated using the enthalpies of formation: \( \Delta H_{\text{rxn}} = \Delta H_f(\text{products}) - \Delta H_f(\text{reactants}) \).

Set up the equation for the enthalpy change of the reaction: \( \Delta H_{\text{rxn}} = \Delta H_f(\text{SO}_3(g)) - \left( \Delta H_f(\text{SO}_2(g)) + \frac{1}{2} \Delta H_f(\text{O}_2(g)) \right) \).

Since the enthalpy of formation of \( \text{O}_2(g) \) is 0 (as it is in its standard state), simplify the equation to: \( 89.5 = -204.2 - \Delta H_f(\text{SO}_2(g)) \).

Solve for \( \Delta H_f(\text{SO}_2(g)) \) by rearranging the equation: \( \Delta H_f(\text{SO}_2(g)) = -204.2 - 89.5 \).

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Key Concepts

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

Enthalpy of Formation

The enthalpy of formation is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. It is a crucial concept in thermodynamics, allowing chemists to understand the energy changes associated with chemical reactions. For example, the enthalpy of formation of SO3 indicates how much energy is released or absorbed when SO3 is formed from sulfur and oxygen.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions involve the transfer of electrons between substances, leading to changes in oxidation states. In the given question, the oxidation of SO2 to SO3 involves the loss of electrons by sulfur, which is oxidized. Understanding redox reactions is essential for analyzing energy changes and determining the enthalpy associated with these processes.

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 for the calculation of unknown enthalpy changes by using known values. In this case, Hess's Law can be applied to relate the enthalpy of formation of SO2 to the enthalpy changes associated with the oxidation of SO2 to SO3.