Consider the combustion of liquid methanol, CH 3 OH(l): CH 3 OH(l) + 3/2 O 2 (g) → CO 2 (g) + 2 H 2 O(l) ΔH = -726.5 kJ (d) If the reaction were written to produce H 2 O(g) instead of H 2 O(l), would you expect the magnitude of ΔH to increase, decrease, or stay the same? Explain.

Hello. Everyone in this video, we're given a specific reaction right here and we're being asked if this hydrogen peroxide. So the starting reagent was in a solid phase instead of its liquid phase that it's currently in this equation. What the entropy of reaction change. So with this change, if it does change, is it going to increase or decrease? So let's take a deeper look into this. So entropy as we know, is the total heat content of the system at the moment. So we'll just put initial delta H. It is negative where we have a negative delta H. That means that this reaction is exhaust thermic. So if this hydrogen peroxide was in a solid state or phase then they will have less entropy. And why I say this delicious where it's solid ST equals resting state. And when I say resting state, although it's not a really formal way of putting this is in a solid state. All the atoms and molecules, they're just resting there just chilling and it's on all together, right there just being stuck in this kind of solid. As for the gashes face and liquid face, the molecules are always moving around, bouncing around whatever it is is constantly moving. But in solid state it mostly vibrates at most but doesn't really actively go anywhere. And because of this and it's just resting until heat is going to be added to this compound or molecule then it won't really fall or break apart. So in other words, there's going to be an increase of change. So we're going to increase change of entropy because heat is going to be required to go ahead and break this compound apart because right now it's in this resting state, it's in a solid state. If you want to go ahead and change this now, then we need heat or some sort of energy to go ahead and have that be broken apart. So this final answer is going to be this right here in highlight is that we do need an increase of change of

Ch.5 - Thermochemistry

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Brown 14th Edition

Ch.5 - Thermochemistry

Problem 47d

Chapter 5, Problem 47d

Consider the combustion of liquid methanol, CH₃OH(l): CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2 H₂O(l) ΔH = -726.5 kJ (d) If the reaction were written to produce H₂O(g) instead of H₂O(l), would you expect the magnitude of ΔH to increase, decrease, or stay the same? Explain.

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Understand that the enthalpy change (ΔH) given is for the formation of liquid water (H2O(l)) from the combustion of methanol.

Recall that converting liquid water to gaseous water (H2O(g)) requires additional energy, known as the enthalpy of vaporization.

Recognize that if the reaction produces H2O(g) instead of H2O(l), the system must supply this additional energy to vaporize the water.

Since energy is required to convert H2O(l) to H2O(g), the magnitude of ΔH for the reaction will increase (become less negative) because the system absorbs more energy.

Conclude that the enthalpy change for the reaction producing H2O(g) will be less exothermic compared to the reaction producing H2O(l), due to the added energy requirement for vaporization.

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Key Concepts

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

Enthalpy Change (ΔH)

Enthalpy change (ΔH) is a measure of the heat absorbed or released during a chemical reaction at constant pressure. In exothermic reactions, like the combustion of methanol, ΔH is negative, indicating that heat is released. The magnitude of ΔH reflects the energy difference between reactants and products, which can vary depending on the physical states of the products formed.

Phase Changes and Energy

The physical state of a substance (solid, liquid, gas) significantly affects its energy content. For example, converting liquid water (H2O(l)) to water vapor (H2O(g)) requires energy input, known as the enthalpy of vaporization. Therefore, if a reaction produces water in the gaseous state instead of the liquid state, the overall energy change (ΔH) will be influenced by this additional energy requirement.

Combustion Reactions

Combustion reactions involve the reaction of a substance with oxygen to produce heat and light, typically resulting in the formation of carbon dioxide and water. The products' states (gas or liquid) can affect the reaction's enthalpy change. In the case of methanol combustion, producing water vapor instead of liquid water would likely increase the energy released, as more energy is required to form gaseous products.

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Related Practice

Textbook Question

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO₃: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) ΔH = -89.4 kJ (c) The decomposition of KClO₃ proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of KClO₃ from KCl and O₂, is likely to be feasible under ordinary conditions? Explain your answer.

Textbook Question

Consider the combustion of liquid methanol, CH₃OH(l): CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2 H₂O(l) ΔH = -726.5 kJ (a) What is the enthalpy change for the reverse reaction?

Textbook Question

Consider the combustion of liquid methanol, CH₃OH(l): CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2 H₂O(l) ΔH = -726.5 kJ (b) Balance the forward reaction with whole-number coefficients. What is ΔH for the reaction represented by this equation?

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Textbook Question

Consider the decomposition of liquid benzene, C₆H₆(l), to gaseous acetylene, C₂H₂(g): C₆H₆(l) → 3 C2H₂(g) ΔH = +630 kJ (a) What is the enthalpy change for the reverse reaction?

Textbook Question

Consider the decomposition of liquid benzene, C₆H₆(l), to gaseous acetylene, C₂H₂(g): C₆H₆(l) → 3 C2H₂(g) ΔH = +630 kJ (b) What is H for the formation of 1 mol of acetylene?

Textbook Question

Consider the decomposition of liquid benzene, C₆H₆(l), to gaseous acetylene, C₂H₂(g): C₆H₆(l) → 3 C2H₂(g) ΔH = +630 kJ (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction?

Consider the combustion of liquid methanol, CH3OH(l): CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l) ΔH = -726.5 kJ (d) If the reaction were written to produce H2O(g) instead of H2O(l), would you expect the magnitude of ΔH to increase, decrease, or stay the same? Explain.

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Key Concepts

Enthalpy Change (ΔH)

Phase Changes and Energy

Combustion Reactions

Consider the combustion of liquid methanol, CH₃OH(l): CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2 H₂O(l) ΔH = -726.5 kJ (d) If the reaction were written to produce H₂O(g) instead of H₂O(l), would you expect the magnitude of ΔH to increase, decrease, or stay the same? Explain.