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What is the connection between Hess's law and the fact that \(H\) is a state function?

Short Answer

Expert verified
The connection between Hess's law and enthalpy (H) being a state function lies in the fact that both concepts revolve around the idea that the total enthalpy change only depends on the initial and final states, not on the path taken. Hess's law states that the total enthalpy change of a chemical reaction is the same, regardless of the path taken to complete the reaction. Enthalpy is a state function, meaning that its change (ΔH) only depends on the initial and final states, not on the path taken. This directly results in the conclusion that the overall enthalpy change for a series of reactions leading to the same final state, as stated by Hess's law, will always yield the same overall enthalpy change, regardless of the order or path taken.

Step by step solution

01

Define Hess's Law

Hess's law states that the total enthalpy change of a chemical reaction is the same, regardless of the path taken to complete the reaction. In other words, the total enthalpy change for a series of reactions that leads to the same final state is the same, no matter the order or path taken to reach that state.
02

Define state functions

A state function is a property of a system that depends only on its current state, not on the path taken to reach that state. State functions have the characteristic that any change in the function's value for a process depends only on the initial and final states, not on the specific path taken between the two states.
03

Explain the properties of enthalpy as a state function

Enthalpy (H) is a state function, which means that the change in enthalpy (ΔH) for any process only depends on the initial and final states, not on the path taken. If we consider a chemical reaction that has an initial state A and final state B, we can write the change in enthalpy for this reaction as: ΔH(A→B) = H(B) - H(A) Since enthalpy is a state function, ΔH(A→B) will be the same no matter how we get from state A to state B.
04

Relate Hess's Law to enthalpy being a state function

Now, let's consider a series of reactions that leads from initial state A to final state B. As per Hess's law, the overall enthalpy change of this series of reactions will be the same, regardless of the path taken. This property directly results from the fact that enthalpy is a state function. Let's say the overall reaction can be written as the sum of two separate reactions: Reaction 1: A → C Reaction 2: C → B Then the overall reaction can be written as: Reaction 3: A → B (formed by adding Reaction 1 and Reaction 2) If enthalpy is a state function, then the overall enthalpy change for Reaction 3 will be equal to the sum of the enthalpy changes for Reaction 1 and Reaction 2, regardless of the specific path taken: ΔH(A→B) = ΔH(A→C) + ΔH(C→B) This relationship is a direct consequence of the fact that enthalpy is a state function – since enthalpy only depends on the initial and final states, adding separate reactions that lead to the same final state will always yield the same overall enthalpy change. This is the connection between Hess's law and the fact that enthalpy is a state function.

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Key Concepts

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

State Functions
In the world of science, a state function is like a snapshot of a system. It doesn't care how you got to a certain point; it just matters where you are. This means that state functions are properties that depend only on the present state of a system, not the route taken to arrive there.
Think of it as the difference between taking a scenic route or the highway to reach a mountain top. The altitude of the peak is the same regardless of the path, which is just like a state function. Whether you hike up, drive, or take a helicopter, the altitude—the state function—remains a constant that describes the mountain's height.
In chemistry, state functions include things like pressure, temperature, and enthalpy. These properties are invaluable because they simplify the analysis of chemical reactions, helping scientists focus on where a system starts and ends, rather than the complex path it takes to transition between these states.
Enthalpy
Enthalpy, symbolized as \(H\), is a key player in understanding thermodynamics. It's a state function that relates to the heat content of a system. When we talk about enthalpy, we are usually interested in changes in enthalpy, indicated by \(\Delta H\).
This change in enthalpy tells us the heat absorbed or released during a chemical reaction under constant pressure.
  • For an exothermic reaction, \(\Delta H\) is negative because the system releases heat.
  • For an endothermic reaction, \(\Delta H\) is positive as the system absorbs heat.
Enthalpy can be best understood through the expression:
\[\Delta H = H_{final} - H_{initial} \]This equation highlights how enthalpy, being a state function, only depends on the initial and final states of the system. Regardless of the changes occurring during the reaction, if the initial and final states are the same, \(\Delta H\) remains unchanged.
Chemical Reactions
Chemical reactions are fascinating transformations where substances change into new ones with distinct properties. These reactions involve the breaking and forming of bonds and are usually accompanied by changes in energy, such as heat.
In a chemical reaction, the reactants undergo a transformation to become products, and this shifting involves changes in energy.
Using enthalpy as a guide, chemical reactions can be analyzed for energy changes. According to Hess's Law, which stems from enthalpy being a state function, the total enthalpy change of a reaction remains constant regardless of how it proceeds.
Here's how this concept assists in understanding chemical reactions:
  • We can break complex reactions into a series of simpler steps where the overall enthalpy change is predictable.
  • It makes calculating the energy required or released more manageable, even for reactions that occur in multiple stages.
Hess's Law allows us to mix and match reactions like a puzzle, finding the correct pieces to determine the total energy change—showing elegance in chemical processes and aiding in precise energy calculations.

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Most popular questions from this chapter

You are given \(\Delta H\) for a process that occurs at constant pressure. What additional information do you need to determine \(\Delta E\) for the process?

Calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) reacts with water to form acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\). From the following enthalpy of reaction data and data in Appendix \(C\), calculate \(\Delta H_{f}^{\circ}\) for \(\mathrm{CaC}_{2}(s):\) \(\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g)\) \(\Delta H^{\circ}=-127.2 \mathrm{~kJ}\)

Naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) is a solid aromatic compound often sold as mothballs. The complete combustion of this substance to yield \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(25^{\circ} \mathrm{C}\) yields \(5154 \mathrm{~kJ} / \mathrm{mol}\). (a) Write balanced equations for the formation of naphthalene from the elements and for its combustion. (b) Calculate the standard enthalpy of formation of naphthalene.

When a 9.55-g sample of solid sodium hydroxide dissolves in \(100.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure 5.17), the temperature rises from \(23.6^{\circ} \mathrm{C}\) to \(47.4^{\circ} \mathrm{C}\). Calculate \(\Delta H\) (in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NaOH}\) ) for the solution process $$ \mathrm{NaOH}(s) \stackrel{-\cdots}{\mathrm{Na}}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ Assume that the specific heat of the solution is the same as that of pure water.

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is used as a fuel in race cars. (a) Write a balanced equation for the combustion of liquid methanol in air. (b) Calculate the standard enthalpy change for the reaction, assuming \(\mathrm{H}_{2} \mathrm{O}(g)\) as a product. (c) Calculate the heat produced by combustion per liter of methanol. Methanol has a density of \(0.791 \mathrm{~g} / \mathrm{mL}\) (d) Calculate the mass of \(\mathrm{CO}_{2}\) produced per \(\mathrm{kJ}\) of heat emitted.

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