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The enthalpy change for a reaction is a state function and it is an extensive property. Explain.

Short Answer

Expert verified
The enthalpy change (\(ΔH\)) for a reaction is a state function because it depends only on the initial and final states of the system and is independent of the path taken during the reaction process. It can be demonstrated by Hess's Law, which states that the overall enthalpy change of a reaction is the same, regardless of the individual steps or reactions involved. Additionally, enthalpy change is an extensive property because it depends on the quantity of matter in the system. When the amount of reactants changes, the overall enthalpy change will also change, making it directly proportional to the quantity of matter involved.

Step by step solution

01

Definition of State Function

A state function is a property of a system that depends only on its current state and is independent of the path taken to reach that state. This means that the value of a state function is determined by the initial and final states of a chemical reaction and not on the specific steps taken during the reaction process.
02

Definition of Extensive Property

An extensive property is a property of a system that depends on the quantity of matter in the system. This means that an extensive property is directly proportional to the size or amount of the substances involved in a reaction. Examples of extensive properties include mass, volume, and enthalpy.
03

Enthalpy Change as a State Function

Enthalpy change (\(ΔH\)) is the amount of heat absorbed or released during a chemical reaction under constant pressure conditions. \(ΔH\) is a state function because it only depends on the initial and final states of the system and not on the specific path taken during the reaction. This can be demonstrated by using Hess's Law, which states that the overall enthalpy change of a reaction is the same, regardless of the individual steps or reactions involved in the process. The enthalpy change only depends on the enthalpy values of the reactants and products and not on the intermediate steps of the reaction.
04

Enthalpy Change as an Extensive Property

Enthalpy change is an extensive property because it depends on the amount of reactants and products in the system. When the quantity of reactants changes, the overall enthalpy change of the reaction will also change. For example, if the amount of reactants in a reaction is doubled, the enthalpy change (\(ΔH\)) of the reaction will also double, making it directly proportional to the quantity of matter in the system. To summarize, the enthalpy change for a reaction is a state function because it depends only on the initial and final states of the system, and it is an extensive property because it depends on the quantity of matter in the system.

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Most popular questions from this chapter

The enthalpy of combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) when \(\mathrm{H}_{2} \mathrm{O}(l)\) is formed is \(-891 \mathrm{~kJ} / \mathrm{mol}\) and the enthalpy of combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) when \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) is formed is \(-803 \mathrm{~kJ} / \mathrm{mol} .\) Use these data and Hess's law to determine the enthalpy of vaporization for water.

Given the following data $$ \begin{aligned} \mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) & \longrightarrow 4 \mathrm{PCl}_{3}(g) & & \Delta H=-1225.6 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & & \Delta H=-2967.3 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) & \longrightarrow \mathrm{PCl}_{5}(g) & & \Delta H=-84.2 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \mathrm{Cl}_{3} \mathrm{PO}(g) & & \Delta H=-285.7 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g) $$

Water gas is produced from the reaction of steam with coal: $$ \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}(g) $$ Assuming that coal is pure graphite, calculate \(\Delta H^{\circ}\) for this reaction.

Calculate the internal energy change for each of the following. a. One hundred (100.) joules of work is required to compress a gas. At the same time, the gas releases \(23 \mathrm{~J}\) of heat. b. A piston is compressed from a volume of \(8.30 \mathrm{~L}\) to \(2.80 \mathrm{~L}\) against a constant pressure of \(1.90 \mathrm{~atm} .\) In the process, there is a heat gain by the system of \(350 . \mathrm{J}\). c. A piston expands against \(1.00 \mathrm{~atm}\) of pressure from \(11.2 \mathrm{~L}\) to \(29.1 \mathrm{~L}\). In the process, \(1037 \mathrm{~J}\) of heat is absorbed.

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.

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