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How is the standard state of an element defined? What is Hess's law, and why does it "work"?

Short Answer

Expert verified
The standard state is an element's most stable form at 298 K and 1 atm. Hess's Law states enthalpy change depends only on initial and final states, as it uses the state function property of enthalpy.

Step by step solution

01

Define Standard State

The standard state of an element is defined as its physical state under standard conditions of temperature and pressure, which are typically 298 K (25°C) and 1 atmosphere pressure. In this state, the element is in its most stable form, with an activity defined as one. For instance, the standard state of carbon is graphite.
02

Explain Hess's Law

Hess's Law states that the total enthalpy change of a chemical reaction is the same, regardless of the pathway or number of intermediate steps taken in the reaction, provided the initial and final states are the same. It is based on the principle of conservation of energy, emphasizing that energy changes are state functions.
03

Why Hess's Law Works

Hess's Law works because enthalpy is a state function, meaning its value depends only on the initial and final states of the system, not on the path taken between them. This allows enthalpy changes for complex reactions to be calculated using simpler, known reactions by summing their enthalpy changes.

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Key Concepts

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

Standard State
In chemistry, the concept of a "standard state" is important for understanding the properties of elements and compounds under controlled conditions. The standard state of an element refers to its physical state—solid, liquid, or gas—at a standard reference temperature and pressure. These standard conditions are usually set at 298 K (25°C) and 1 atmosphere of pressure. For example, the standard state of water is liquid, while the standard state of oxygen is a gas. This consistency allows chemists to compare and calculate thermodynamic data, such as enthalpy, more accurately across different chemical reactions.
  • Temperature and Pressure: Standard state requires conditions of 298 K and 1 atm.
  • Most Stable Form: Elements are in their most stable form, like graphite for carbon.
  • Reference for Calculations: It provides a baseline for measuring changes in thermodynamic quantities.
Hess's Law
Hess's Law is a fundamental principle in thermochemistry that helps us understand how energy changes in chemical reactions. It asserts that the total enthalpy change of a chemical reaction is the same, regardless of the method or pathway through which it occurs, as long as the initial and final conditions are identical. This law is underpinned by the conservation of energy, emphasizing that enthalpy, a type of energy, is a state function. Because enthalpy is a state function, it depends solely on the initial and final states of a system, not on how that change was achieved. This means you can calculate the enthalpy change for complex reactions using simpler steps. For example, if you know the enthalpy changes for smaller, known reactions, you can add these values to find the enthalpy change of the overall reaction.
  • State Function: The enthalpy change remains constant irrespective of the reaction path.
  • Conservation of Energy: Reflects the principle that energy cannot be created or destroyed.
  • Practical Calculations: Enables solving complex reaction enthalpies through simpler known changes.
Enthalpy
Enthalpy is a central concept in the study of thermochemistry. It is a measure of the total heat content in a system and is represented by the symbol \( H \). Enthalpy helps to express the heat change that occurs at constant pressure during a chemical reaction. When substances undergo chemical reactions, they absorb or release energy in the form of heat, and this energy change is recorded as the change in enthalpy, \( \Delta H \).The concept of enthalpy explains why some reactions release heat (exothermic reactions) while others absorb heat (endothermic reactions). Exothermic reactions, like combustion, give off heat to their surroundings and have a negative \( \Delta H \). In contrast, endothermic reactions take in heat from their surroundings, resulting in a positive \( \Delta H \).
  • Heat Content Measure: Enthalpy quantifies the total heat in a system.
  • Constant Pressure Condition: Enthalpy changes are particularly important for processes occurring at constant pressure.
  • Exothermic vs. Endothermic: Dictates whether the reaction releases or absorbs heat based on the sign of \( \Delta H \).

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Most popular questions from this chapter

Instant hot packs contain a solid and a pouch of water. When the pack is squeezed, the pouch breaks and the solid dissolves, increasing the temperature because of the exothermic reaction. The following reaction is used to make a hot pack: $$\mathrm{LiCl}(s) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{Li}^{+}(a q)+\mathrm{Cl}^{-}(a q) \quad \Delta H=-36.9 \mathrm{~kJ}$$ What is the final temperature in a squeezed hot pack that contains \(25.0 \mathrm{~g}\) of \(\mathrm{LiCl}\) dissolved in \(125 \mathrm{~mL}\) of water? Assume a specific heat of \(4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) for the solution, an initial temperature of \(25.0{ }^{\circ} \mathrm{C}\), and no heat transfer between the hot pack and the environment.

Styrene \(\left(\mathrm{C}_{8} \mathrm{H}_{8}\right)\), the precursor of polystyrene polymers, has a standard heat of combustion of \(-4395 \mathrm{~kJ} / \mathrm{mol}\). Write a balanced equation for the combustion reaction, and calculate \(\Delta H_{f}^{\circ}\) for styrene in \(\mathrm{kJ} / \mathrm{mol}\). $$\begin{aligned}\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{CO}_{2}(g)\right] &=-393.5 \mathrm{~kJ} / \mathrm{mol} ; \\ \Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{O}(l)\right] &=-285.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$

Which of the following are state functions, and which are not? (a) The distance from your dorm room to your chemistry class. (b) The temperature in the room of your chemistry class. (c) The balance in your bank account.

A \(110.0 \mathrm{~g}\) piece of molybdenum metal is heated to \(100.0^{\circ} \mathrm{C}\) and placed in a calorimeter that contains \(150.0 \mathrm{~g}\) of water at \(24.6^{\circ} \mathrm{C}\). The system reaches equilibrium at a final temperature of \(28.0^{\circ} \mathrm{C}\). Calculate the specific heat of molybdenum metal in \(\mathrm{J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\). The specific heat of water is \(4.184 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\).

A reaction is carried out in a cylinder fitted with a movable piston. The starting volume is \(V=5.00 \mathrm{~L}\), and the apparatus is held at constant temperature and pressure. Assuming that \(\Delta H=-35.0 \mathrm{~kJ}\) and \(\Delta E=-34.8 \mathrm{~kJ}\), redraw the piston to show its position after reaction. Does \(V\) increase, decrease, or remain the same?

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