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How is the standard state of an element defined? Why do elements always have \(\Delta H_{\mathrm{f}}^{\circ}=0 ?\)

Short Answer

Expert verified
Elements in their standard states have \(\Delta H_{\mathrm{f}}^{\circ}=0\) as they're already in their most stable form, needing no formation reaction.

Step by step solution

01

Understanding the Standard State

The standard state of an element is defined as its most stable form at a pressure of 1 bar and at a specified temperature, usually 298.15 K (25°C). For example, the standard state of oxygen is O₂ gas, and for carbon, it is graphite.
02

Concept of Enthalpy of Formation

The standard enthalpy of formation, denoted \(\Delta H_{\mathrm{f}}^{\circ}\), is defined as the heat change when one mole of a compound is formed from its elements in their standard states. For elements in their standard states, no formation reaction is needed.
03

Explaining \( \Delta H_{\mathrm{f}}^{\circ}=0 \)

Since elements in their standard states are already in their most stable form, forming them from themselves involves no chemical change. Therefore, no enthalpy change occurs, which means \(\Delta H_{\mathrm{f}}^{\circ}=0\).
04

Conclusion

Thus, any element in its standard state has a standard enthalpy of formation of zero because it is already in its most stable form, requiring no formation reaction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy of Formation
The enthalpy of formation is a key concept in thermodynamics. It provides insight into how substances are created from their constituent elements. The standard enthalpy of formation, denoted \( \Delta H_{\mathrm{f}}^{\circ} \), is defined as the heat exchange that occurs when one mole of a compound is synthesized from its elements in their standard states.
  • This process assumes a stable pressure of 1 bar.
  • The default temperature is often set to 298.15 K, which is 25°C.
Understanding this allows us to calculate how much energy is absorbed or released during the creation of specific compounds. Notably, for elements in their standard states, no enthalpy change is needed. There's no chemical reaction required, thus \( \Delta H_{\mathrm{f}}^{\circ} = 0 \).This is because these elements are already in their simplest, most stable form when they're in their standard state.
Standard Enthalpy Change
Standard enthalpy change is another fundamental principle related to energy transformations. It describes the heat absorbed or released during a reaction under standardized conditions.
  • These conditions include a pressure of 1 bar.
  • The temperature is usually set at 298.15 K (25°C).
Determining the standard enthalpy change helps us understand the energetic requirements or releases in specific reactions, providing an energy benchmark. It applies to reactions where matter is transformed from reactants to products. Knowing the standard enthalpy changes of reactions is crucial for predicting reaction feasibility and managing energy in industrial and laboratory processes.
Most Stable Form of Elements
The concept of the most stable form of elements is central to understanding why the standard enthalpy of formation is zero for these elements. Elements naturally exist in their most stable, low-energy configuration under standard conditions.For example:
  • Oxygen's most stable form is Oâ‚‚ gas.
  • Carbon’s most stable form is graphite, not diamond.
This idea is crucial because it's in these forms that elements are most often found in nature and used in chemical reactions.Knowing the most stable form assists in identifying the condition where no energy input is required to "form" the element from itself, reaffirming the rationale behind \( \Delta H_{\mathrm{f}}^{\circ} = 0 \) for elements in their standard states. This stability principle is fundamental in thermochemical calculations and efficient energy utilization in chemical processes.

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Most popular questions from this chapter

Write balanced equations for the formation of the following compounds from their elements: (a) iron(III) oxide (b) sucrose (table sugar, \(\left.\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) (c) uranium hexafluoride (a solid at \(25^{\circ} \mathrm{C}\) )

Methyl tert-butyl ether (MTBE), \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\), a gasoline additive used to boost octane ratings, has \(\Delta H_{\mathrm{f}}^{\circ}=-313.6 \mathrm{~kJ} / \mathrm{mol}\). Write a balanced equation for its combustion reaction, and calculate its standard heat of combustion in kilojoules.

The addition of \(\mathrm{H}_{2}\) to \(\mathrm{C}=\mathrm{C}\) double bonds is an important reaction used in the preparation of margarine from vegetable oils. If \(50.0 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(50.0 \mathrm{~mL}\) of ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) are allowed to react at \(1.5\) atm, the product ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) has a volume of \(50.0 \mathrm{~mL}\). Calculate the amount of \(P V\) work done, and tell the direction of the energy flow. $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)$$

Is it possible for a reaction to be nonspontaneous yet exothermic? Explain.

Citric acid has three dissociable hydrogens. When \(5.00 \mathrm{~mL}\) of 0.64 M citric acid and \(45.00 \mathrm{~mL}\) of \(0.77 \mathrm{M} \mathrm{NaOH}\) are mixed at an initial temperature of \(26.0^{\circ} \mathrm{C}\), the temperature rises to \(27.9^{\circ} \mathrm{C}\) as the citric acid is neutralized. The combined mixture has a mass of \(51.6 \mathrm{~g}\) and a specific heat of \(4.0 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\). Assuming that no heat is transferred to the surroundings, calculate the enthalpy change for the reaction of \(1.00 \mathrm{~mol}\) of citric acid in \(\mathrm{kJ}\). Is the reaction exothermic or endothermic?

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