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Chapter 6: Problem 47

Which is the more negative quantity at \(25^{\circ} \mathrm{C}: \Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(l)\) or \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g) ?\)

Short Answer

Expert verified
The enthalpy of formation, \( \Delta H_{\mathrm{f}}^{\circ}\) for \mathrm{H}_{2} \mathrm{O}(l) [liquid water] at 25°C is the more negative quantity when compared to the \Delta H_{\mathrm{f}}^{\circ} for \mathrm{H}_{2} \mathrm{O}(g) [gaseous water] at the same temperature.

Step by step solution

01

Understanding Enthalpy of Formation

By definition, the enthalpy of formation, \( \Delta H_{\mathrm{f}}^{\circ} \), is the change in enthalpy (heat content) when 1 mole of a compound is formed from its elements under standard conditions (25°C, 1 atm), all substances in their standard state. Enthalpy of formation measures the energy involved in forming a compound.
02

Liquid Water vs Gaseous Water

Water in its liquid state and its gaseous state at the same temperature will have different enthalpy of formation values because transitioning to different states involves energy changes. It's known that gaseous water has higher enthalpy than liquid water, because energy is added to liquid water to overcome the attractive forces between water molecules and convert it to a gaseous state.
03

Comparing Enthalpy of Formation

So, given that the enthalpy of formation, \( \Delta H_{\mathrm{f}}^{\circ} \), for liquid water is -285.8 kJ/mol at 25°C and for gaseous water it is -241.8 kJ/mol at the same temperature, it's clear that the \Delta H_{\mathrm{f}}^{\circ} for \mathrm{H}_{2} \mathrm{O}(l) would be the more negative quantity.

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Most popular questions from this chapter

Decomposition reactions are usually endothermic, whereas combination reactions are usually exothermic. Give a qualitative explanation for these trends.

Define calorimetry and describe two commonly used calorimeters. In a calorimetric measurement, why is it important that we know the heat capacity of the calorimeter? How is this value determined?

Calculate the work done (in joules) when 1.0 mole of water is frozen at \(0^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm} .\) The volumes of one mole of water and ice at \(0^{\circ} \mathrm{C}\) are \(0.0180 \mathrm{~L}\) and \(0.0196 \mathrm{~L},\) respectively.

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Consider the following data: $$ \begin{array}{lcc} \text { Metal } & \text { Al } & \text { Cu } \\ \hline \text { Mass (g) } & 10 & 30 \\ \text { Specific heat }\left(\mathrm{J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right) & 0.900 & 0.385 \\ \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & 40 & 60 \end{array} $$ When these two metals are placed in contact, which of the following will take place? (a) Heat will flow from Al to Cu because Al has a larger specific heat. (b) Heat will flow from Cu to Al because Cu has a larger mass. (c) Heat will flow from Cu to Al because Cu has a larger heat capacity. (d) Heat will flow from Cu to Al because Cu is at a higher temperature. (e) No heat will flow in either direction.
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