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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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

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.

Recommended similar problem, with video answer:

Verified Solution

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.