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Chapter 6: Problem 43

State Hess's law. Explain, with one example, the usefulness of Hess's law in thermochemistry.

Short Answer

Expert verified
Hess's law states that whether a chemical reaction occurs in one step or several, the total enthalpy change is the same. It's useful in calculating the enthalpy change (\(螖H\)) of reactions by adding the enthalpies of intermediate steps. For example, with known enthalpies of formation, Hess's Law can compute the change in enthalpy of the combustion of methane (-802.3 kJ/mol), even if a direct measurement is difficult.

Step by step solution

01

State Hess's Law

Hess's law, also known as Hess's Law of Constant Heat Summation, states that the total enthalpy change during the complete course of a chemical reaction is the same whether the reaction is made in one step or in several steps. This law is a direct result of the law of conservation of energy.
02

Explain the usefulness of Hess's law in Thermochemistry

Hess's law is particularly useful in calculating the enthalpy change (\(螖H\)) of reactions where it is difficult to measure the enthalpy change directly. It allows us to determine the \(螖H\) of a reaction by adding the enthalpies of multiple intermediate reactions that lead to the overall process.
03

Example: Combustion of Methane

Let's take the combustion of methane (\(CH_4\)) as an example. The reaction is: \(CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)\). If we know the enthalpies of formation of \(CO_2\), \(H_2O\), and \(CH_4\), we can use Hess's law to calculate the enthalpy change of this reaction. \ The enthalpies of formation are given by: \(螖H_f [CH_4(g)] = -74.6 kJ/mol\) \ \(螖H_f [CO_2(g)] = -393.5 kJ/mol\) \(螖H_f [H_2O(g)] = -241.8 kJ/mol\) \ Using these data and Hess's law, the heat of combustion of methane is: \ \(螖H = 螖H_f [CO_2(g)] + 2 螖H_f [H_2O(g)] - 螖H_f [CH_4(g)]\) \(= -393.5 kJ/mol + 2(-241.8 kJ/mol) - -( -74.6 kJ/mol) = -802.3 kJ/mol\) Thus, the combustion of methane is exothermic with an enthalpy change of -802.3 kJ/mol.

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Key Concepts

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

Enthalpy Change
Enthalpy change refers to the amount of heat energy absorbed or released during a chemical reaction at constant pressure. It's symbolized as \(\Delta H\) and measured in kilojoules per mole (kJ/mol). The sign of \(\Delta H\) indicates whether a reaction is exothermic (releases heat, negative \(\Delta H\)) or endothermic (absorbs heat, positive \(\Delta H\)).

In the context of Hess's Law, enthalpy change allows us to understand that even though a reaction can take place through various pathways, the total enthalpy change will remain constant. This is extremely important as it offers flexibility in calculating the \(\Delta H\) for complex reactions by breaking them down into simpler, known reactions.
Law of Conservation of Energy
The law of conservation of energy is a fundamental principle stating that energy can neither be created nor destroyed, only transformed from one form to another. In thermochemistry, this translates to the fact that the total energy of an isolated system remains constant over time.

Hess鈥檚 Law is an application of this principle, asserting that the total change in enthalpy, a form of energy, for a chemical reaction is the same, no matter how the reaction is carried out. This reinforces the idea that energy in a chemical system is conserved and allows chemists to accurately calculate the enthalpy of reactions, even if they occur in multiple steps.
Enthalpy of Formation
The enthalpy of formation, denoted as \(\Delta H_f\), is defined as the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard state at 1 atmosphere of pressure. Standard values of \(\Delta H_f\) are vital in employing Hess's Law because they serve as the reference points for calculating the enthalpy changes of reactions.

For instance, the values of \(\Delta H_f\) for common substances like carbon dioxide (\(CO_2\)) or water (\(H_2O\)) are well-established. These known quantities allow us to easily compute the enthalpy of combustion for methane, with the prior knowledge that the \(\Delta H_f\) of the reactants and products involved will remain consistent regardless of the reaction route.
Combustion of Methane
The combustion of methane is a common chemical reaction used to demonstrate the practical application of Hess鈥檚 Law. Methane (\(CH_4\)), when burned in oxygen, creates carbon dioxide \((CO_2)\) and water \((H_2O)\), releasing energy in the process. This reaction is exothermic, meaning it emits heat, and this is reflected by the negative enthalpy change.

By utilizing the known \(\Delta H_f\) values for the reactants and products, Hess's Law enables the straightforward calculation of the overall enthalpy change of methane's combustion. This is crucial in energy-related industries, where understanding the energy output from fuel combustion is fundamental for efficiency and environmental impact assessments.

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Most popular questions from this chapter

The standard enthalpy change \(\Delta H^{\circ}\) for the thermal decomposition of silver nitrate according to the following equation is \(+78.67 \mathrm{~kJ}\) : $$ \mathrm{AgNO}_{3}(s) \longrightarrow \mathrm{AgNO}_{2}(s)+\frac{1}{2} \mathrm{O}_{2}(g) $$ The standard enthalpy of formation of \(\mathrm{AgNO}_{3}(s)\) is \(-123.02 \mathrm{~kJ} / \mathrm{mol}\). Calculate the standard enthalpy of formation of \(\mathrm{AgNO}_{2}(s)\).

A quantity of \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.862 \mathrm{M} \mathrm{HCl}\) is mixed with \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.431 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the \(\mathrm{HCl}\) and \(\mathrm{Ba}(\mathrm{OH})_{2}\) solutions is the same at \(20.48^{\circ} \mathrm{C}\). For the process $$ \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) $$ the heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\). What is the final temperature of the mixed solution?

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.

Why are cold, damp air and hot, humid air more uncomfortable than dry air at the same temperatures? (The specific heats of water vapor and air are approximately \(1.9 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), respectively.

A 1.00 -mole sample of ammonia at 14.0 atm and \(25^{\circ} \mathrm{C}\) in a cylinder fitted with a movable piston \(\mathrm{ex}\) pands against a constant external pressure of 1.00 atm. At equilibrium, the pressure and volume of the gas are 1.00 atm and 23.5 L, respectively. (a) Calculate the final temperature of the sample. (b) Calculate \(q, w,\) and \(\Delta E\) for the process. The specific heat of ammonia is \(0.0258 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\)
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