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Hess's law is really just another statement of the first law of thermodynamics. Explain.

Short Answer

Expert verified
Hess's law and the first law of thermodynamics both focus on conservation principles, with Hess's law conserving enthalpy and the first law conserving energy. Enthalpy is related to internal energy, pressure, and volume, and under constant pressure conditions, the change in enthalpy represents the heat exchanged. Therefore, Hess's law is essentially stating that the total heat exchange in a reaction is constant, which aligns with the first law of thermodynamics' focus on energy conservation, making Hess's law another statement of the first law of thermodynamics.

Step by step solution

01

Understand the First Law of Thermodynamics

The first law of thermodynamics states that energy cannot be created or destroyed, but can only change from one form to another. In other words, the total energy of an isolated system is conserved. Mathematically, it can be expressed as: \(\Delta E = q + w\), where \(\Delta E\) is the change in internal energy, \(q\) is the heat exchanged, and \(w\) is the work done on/by the system.
02

Understand Hess's Law

Hess's law states that the enthalpy change for a given reaction is the same, regardless of the particular way the reaction is carried out. In other words, if a reaction can be expressed as a series of steps, the sum of the enthalpy changes for each step is equal to the enthalpy change for the overall reaction. This is true as long as the initial and final conditions are the same.
03

Compare the Principles of the First Law of Thermodynamics and Hess's Law

Both the first law of thermodynamics and Hess's law are based on the conservation of a certain property of a system. In the case of thermodynamics, it's the conservation of energy, whereas for Hess's law, it's the conservation of enthalpy. These two quantities, though not the same, both relate to the energy changes within a system.
04

Relate Enthalpy to Heat and Internal Energy

Enthalpy (\(H\)) is a state function that represents the total energy of a system. It is related to the internal energy (\(U\)), pressure (\(P\)), and volume (\(V\)) as follows: \(H = U + PV\). For an isobaric (constant pressure) process, the change in enthalpy is equal to the heat exchanged: \(\Delta H = q_p\). This links the concept of enthalpy to the heat term in the first law of thermodynamics equation.
05

Show that Hess's Law is a Reflection of the First Law of Thermodynamics

Since the enthalpy changes in a chemical reaction are equivalent to the heat exchanged in a constant pressure process, Hess's law is essentially stating that the sum of heat exchanges in a series of reactions is equal to the heat exchange in the overall reaction. This aligns with the first law of thermodynamics, which focuses on the conservation of energy in a system, including heat exchanges. Therefore, Hess's law can be considered another statement of the first law of thermodynamics.

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Most popular questions from this chapter

A sample of nickel is heated to \(99.8^{\circ} \mathrm{C}\) and placed in a coffeecup calorimeter containing \(150.0 \mathrm{g}\) water at \(23.5^{\circ} \mathrm{C}\). After the metal cools, the final temperature of metal and water mixture is \(25.0^{\circ} \mathrm{C} .\) If the specific heat capacity of nickel is \(0.444 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) what mass of nickel was originally heated? Assume no heat loss to the surroundings.

Calculate \(\Delta H\) for the reaction $$\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ given the following data: $$\begin{array}{lr}\text { Equation } & \Delta H(\mathrm{kJ}) \\ 2 \mathrm{NH}_{3}(g)+3 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) & -1010 \\ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & -317 \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & -143 \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & -286 \end{array}$$

As a system increases in volume, it absorbs \(52.5 \mathrm{J}\) of energy in the form of heat from the surroundings. The piston is working against a pressure of \(0.500 \mathrm{atm}\). The final volume of the system is \(58.0 \mathrm{L}\). What was the initial volume of the system if the internal energy of the system decreased by \(102.5 \mathrm{J} ?\)

The enthalpy of neutralization for the reaction of a strong acid with a strong base is \(-56 \mathrm{kJ} / \mathrm{mol}\) water produced. How much energy will be released when \(200.0 \mathrm{mL}\) of \(0.400 \mathrm{M} \mathrm{HNO}_{3}\) is mixed with \(150.0 \mathrm{mL}\) of \(0.500 \mathrm{M}\) KOH?

For the process \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) at \(298 \mathrm{K}\) and 1.0 atm, \(\Delta H\) is more positive than \(\Delta E\) by 2.5 kJ/mol. What does the \(2.5 \mathrm{kJ} / \mathrm{mol}\) quantity represent?

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