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(a) Why are tables of standard enthalpies of formation so useful? (b) What is the value of the standard enthalpy of formation of an element in its most stable form? (c) Write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s), \Delta H_{f}^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right]\).

Short Answer

Expert verified
(a) Tables of standard enthalpies of formation are useful as they allow for the easy calculation of standard enthalpy changes of chemical reactions using Hess's law, aiding in the analysis of reaction thermodynamics. (b) The value of the standard enthalpy of formation for an element in its most stable form is zero, as there's no formation process required for the element in its standard state. (c) The chemical equation for the standard enthalpy of formation of glucose is: \[6 \mathrm{C}_{(s)} + 6 \mathrm{H}_{2(g)} + 3 \mathrm{O}_{2(g)} \rightarrow 1 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)\]

Step by step solution

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(a) Importance of Tables of Standard Enthalpies of Formation

Tables of standard enthalpies of formation are useful because they provide a convenient way to calculate the standard enthalpy change of any chemical reaction using Hess's law. Given the standard enthalpies of formation of all reactants and products in a reaction, one can easily determine the overall enthalpy change of the reaction, making it easier to analyze the thermodynamics of various chemical reactions.
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(b) Value of Standard Enthalpy of Formation for Most Stable Element

The value of the standard enthalpy of formation for an element in its most stable form is zero. This is because the standard enthalpy of formation is defined as the enthalpy change when one mole of compound is formed from its constituent elements in their standard states. For an element in its most stable form, there is no formation process required; the element is already in its standard state. Thus, the enthalpy change, by definition, is zero.
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(c) Chemical Equation for the Standard Enthalpy of Formation of Glucose

To write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of glucose, we need to consider the formation of one mole of glucose from its constituent elements in their standard states. In this case, the constituent elements are carbon (\(\mathrm{C}_{(s)}\)), hydrogen (\(\mathrm{H}_{2(g)}\)), and oxygen (\(\mathrm{O}_{2(g)}\)). The balanced chemical equation is as follows: \[6 \mathrm{C}_{(s)} + 6 \mathrm{H}_{2(g)} + 3 \mathrm{O}_{2(g)} \rightarrow 1 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)\] This equation represents the formation of one mole of glucose from its constituent elements in their standard states, with the enthalpy change being the standard enthalpy of formation of glucose, \(\Delta H_{f}^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right]\).

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Key Concepts

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

Hess's Law
Hess's Law is a foundational principle in chemical thermodynamics. It states that the total enthalpy change for a chemical reaction is independent of the path taken from reactants to products. This means that no matter how many steps a reaction is divided into, the total enthalpy change remains the same. The beauty of Hess's Law lies in its ability to simplify complex thermodynamic calculations by using standard enthalpies of formation.

When you have a table of standard enthalpies of formation, you can easily apply Hess's Law to calculate the enthalpy change of a reaction. Simply subtract the sum of the standard enthalpies of formation of the reactants from that of the products:

  • The enthalpy change of the reaction = (∑Enthalpies of formation of products) - (∑Enthalpies of formation of reactants)
This approach is incredibly useful for analyzing the energy dynamics in chemical reactions and helps scientists predict reaction behavior without needing to conduct extensive laboratory experiments.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In the context of chemistry, it is concerned with how energy is transferred in chemical reactions and changes of state. The concept of enthalpy, which measures the heat content of a system, plays a crucial role in thermodynamics.

Enthalpy ( H ) is a state function, indicating that its value is determined solely by the current state of the system, not the process used to arrive at that state. When a chemical reaction occurs, the change in enthalpy (denoted as ΔH ) reflects the heat absorbed or released under constant pressure.

In thermodynamics, we often refer to standard enthalpy changes. These are the enthalpy changes measured when reactants and products are in their standard states—usually at 1 atmospheric pressure and a specified temperature, often 25 degrees Celsius. Understanding and calculating these changes are essential for predicting whether a reaction is exothermic or endothermic:
  • Exothermic reactions release heat, resulting in a negative ΔH .
  • Endothermic reactions absorb heat, leading to a positive ΔH .
Thermodynamic principles help us understand energy transfers, providing insights into reaction feasibility and efficiency.
Standard State
The concept of a "Standard State" is essential when discussing thermodynamic quantities such as enthalpy. The standard state of a substance is its phase (solid, liquid, or gas) under standard conditions. This means a specified temperature, often 25°C, and a pressure of 1 atmosphere (101.3 kPa).

Units are crucial in reporting standard states, ensuring consistency and comparability of data across different experiments and publications. Here are some key points:

  • For a pure substance in a specific phase, the standard state is the pure substance itself.
  • For gaseous substances, the standard state is the ideal gas behavior at 1 atm.
  • For solutes in solution, the standard state is a concentration of 1 mol/L.
The importance of defining standard states lies in its role in calculating standard enthalpies of formation. By setting a common reference point, you can compute how much energy is needed or released during the formation of 1 mole of a substance from its elements in their respective standard states. This helps create a reliable groundwork for comparing different chemical reactions and substances.

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Most popular questions from this chapter

In a thermodynamic study a scientist focuses on the properties of a solution in an apparatus as illustrated. A solution is continuously flowing into the apparatus at the top and out at the bottom, such that the amount of solution in the apparatus is constant with time. (a) Is the solution in the apparatus a closed system, open system, or isolated system? Explain your choice. (b) If it is not a closed system, what could be done to make it a closed system?

(a) What is meant by the term standard conditions, with reference to enthalpy changes? (b) What is meant by the term enthalpy of formation? (c) What is meant by the term standard enthalpy of formation?

(a) When a \(0.235-\mathrm{g}\) sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises \(1.642^{\circ} \mathrm{C}\). When a \(0.265-\mathrm{g}\) sample of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{~N}_{4}\), is burned, the temperature rises \(1.525^{\circ} \mathrm{C}\). Using the value \(26.38 \mathrm{~kJ} / \mathrm{g}\) for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume. (b) Assuming that there is an uncertainty of \(0.002^{\circ} \mathrm{C}\) in each temperature reading and that the masses of samples are measured to \(0.001 \mathrm{~g}\), what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine?

When a mole of dry ice, \(\mathrm{CO}_{2}(s)\), is converted to \(\mathrm{CO}_{2}(g)\) at atmospheric pressure and \(-78^{\circ} \mathrm{C}\), the heat absorbed by the system exceeds the increase in internal energy of the \(\mathrm{CO}_{2}\). Why is this so? What happens to the remaining energy?

Consider the combustion of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(l)\) : $$ \begin{array}{r} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g)-\longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H=-726.5 \mathrm{~kJ} \end{array} $$ (a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number coefficients. What is \(\Delta H\) for the reaction represented by this equation? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If the reaction were written to produce \(\mathrm{H}_{2} \mathrm{O}(g)\) instead of \(\mathrm{H}_{2} \mathrm{O}(l)\), would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

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