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Ch.9 - Thermochemistry: Chemical Energy

McMurry 8th Edition

Ch.9 - Thermochemistry: Chemical Energy

Problem 92

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Chapter 9, Problem 92

What is Hess's law, and why does it 'work'?

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hess's Law

Hess's Law states that the total enthalpy change for a chemical reaction is the same, regardless of the number of steps or the pathway taken. This principle is based on the first law of thermodynamics, which asserts that energy cannot be created or destroyed, only transformed. Therefore, if a reaction can be expressed as the sum of multiple steps, the overall enthalpy change is simply the sum of the enthalpy changes for each individual step.

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Hess's Law

Enthalpy (ΔH)

Enthalpy is a thermodynamic quantity that represents the total heat content of a system at constant pressure. It is a state function, meaning its value depends only on the current state of the system, not on how it reached that state. In the context of Hess's Law, enthalpy changes (ΔH) are crucial for calculating the heat absorbed or released during chemical reactions, allowing for the determination of reaction energetics through indirect means.

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Enthalpy of Formation

State Functions

State functions are properties of a system that depend only on its current state, not on the path taken to reach that state. Examples include enthalpy, pressure, volume, and temperature. In Hess's Law, the concept of state functions is essential because it allows the enthalpy change of a reaction to be calculated from the sum of changes in enthalpy for individual steps, reinforcing the idea that the total change is independent of the reaction pathway.

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Logarithmic Functions

Previous problem

Related Practice

Textbook Question

When 1.50 g of magnesium metal is allowed to react with 200 mL of 6.00 M aqueous HCl, the temperature rises from 25.0 °C to 42.9 °C. Calculate ΔH in kilojoules for the reaction, assumign that the heat capacity of the calorimeter is 776 J/°C, that the specific heat of the final soltuion is the same as that of water [4.18 J(g·°C)] and that the density of the solution is 1.00 g/mL

Textbook Question

A 110.0 g piece of molybdenum metal is heated to 100.0 °C and placed in a calorimeter that contains 150.0 g of water at 24.6 °C. The system reaches equilibirum at a final temeprature of 28.0 °C. Calcualte the specific heat of molybdenum metal in J/g·°C. The specific heat of water is 4.18 J/g·°C

Textbook Question

Citric acid has three dissociable hydrogens. When 5.00 mL of 0.64 M citric acid and 45.00 mL of 0.77 M NaOH are mixed at an initial temperature of 26.0 °C, the temperature rises to 27.9 °C as the citric acid is neutralized. The combined mixture ahs a mass of 51.6 g and a specific heat of 4.0 J/(g·°C). Assuming that no heat is transferred to the surroundings, cal- culate the enthalpy change for the reaction of 1.00 mol of cit- ric acid in kJ. Is the reaction exothermic or endothermic?

Textbook Question

The following steps occur in the reaction of ethyl alcohol (CH3CH2OH) wiht oxygen to yield acetic acid (CH3CO2H). Show that equations 1 and 2 sum to give the net equation and calculate ΔH° for the net equation. (1) CH3CH2OH(l) + 1/2 O2(g) → CH3CHO (g) + H2O(l) ΔH° = -174.2 kJ (2) CH3CHO(g) + 1/2 O2(g) → CH3CO2H(l) ΔH° = -318.4 kJ (Net) CH3CH2OH(l) + O2(g) → CH3CO2H(l) + H2O(l) ΔH° = ?

Textbook Question

The industrial degreasing solvent methylene chloride, CH2Cl2, is prepared from methane by reaction with chlorine: CH4(g) + 2 Cl2(g) → CH2Cl2(g) + 2 HCl(g) Use the following data to calcualte ΔH° in kilojoules for the reaction: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) ΔH° = -98.3 kJ CH3Cl(g) + Cl2(g) → CH2Cl2(g) + HCl(g) ΔH° = -104 kJ

Textbook Question

Hess's law can be used to calculate reaction enthalpies for hypothetical processes that can't be carried out in the labo- ratory. Set up a Hess's law cycle that will let you calculate ∆H° for the conversion of methane to ethylene: 2 CH4(g) → C2H4(g) + 2 H2(g) You can use the following information: 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l) ∆H° = -3120.8 kJ CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ∆H° = -890.3 kJ C2H4(g) + H2(g) → C2H6(g) ∆H° = -136.3 kJ H2O(l) ∆H°f = -285.8 kJ/mol