Can you use an approach similar to Hess’s law to calculate the change in internal energy, _x001F_E, for an overall reaction by summing the _x001F_E values of individual reactions that add up to give the desired overall reaction?

Verified step by step guidance

Identify the target overall reaction for which you want to calculate the change in internal energy, \( \Delta E \).

List the individual reactions that, when combined, will yield the overall reaction. Ensure that these reactions are balanced.

For each individual reaction, note the given \( \Delta E \) values. These are typically provided or can be found in tables of thermodynamic data.

Adjust the \( \Delta E \) values of the individual reactions as necessary. If you reverse a reaction, change the sign of \( \Delta E \). If you multiply a reaction by a coefficient, multiply the \( \Delta E \) by the same coefficient.

Sum the adjusted \( \Delta E \) values of the individual reactions to find the \( \Delta E \) for the overall reaction. This is analogous to Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step.

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 reaction is the sum of the enthalpy changes for the individual steps of the reaction, regardless of the pathway taken. This principle allows chemists to calculate the enthalpy change of a complex reaction by breaking it down into simpler reactions with known enthalpy changes.

Internal Energy

Internal energy is the total energy contained within a system, including kinetic and potential energy of the particles. It is a state function, meaning its change depends only on the initial and final states of the system, not on the path taken. Understanding internal energy is crucial for analyzing thermodynamic processes and reactions.

State Functions

State functions are properties of a system that depend only on its current state, not on how it reached that state. Examples include internal energy, enthalpy, and entropy. In thermodynamics, state functions allow for the simplification of calculations, as changes in these properties can be determined from initial and final conditions without needing to consider the specific process.

Recommended video:

Guided course

03:37

Logarithmic Functions

Related Practice

Textbook Question

Under constant-volume conditions, the heat of combustion of benzoic acid (C₆H₅O₆) is 15.57 kJ/g. A 3.500-g sample of sucrose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 20.94 to 24.72 °C. (a) What is the total heat capacity of the calorimeter?

1939

views

Textbook Question

Under constant-volume conditions, the heat of combustion of benzoic acid (C₆H₅O₆) is 15.57 kJ/g. A 3.500-g sample of sucrose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 20.94 to 24.72 °C. (b) If the size of the sucrose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?

1039

views

Textbook Question

Under constant-volume conditions, the heat of combustion of naphthalene (C₁₀H₈) is 40.18 kJ/g. A 2.50-g sample of naphthalene is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.50 to 28.83 °C. (c) Suppose that in changing samples, a portion of the water in the calorimeter were lost. In what way, if any, would this change the heat capacity of the calorimeter?

1165

views

Textbook Question

Consider the following hypothetical reactions: A → B ΔH_I = +60 kJ B → C ΔH_II = -90 kJ (a) Use Hess’s law to calculate the enthalpy change for the reaction A → C.

Textbook Question

Consider the following hypothetical reactions: A → B ΔH_I = +60 kJ B → C ΔH_II = -90 kJ (b) Construct an enthalpy diagram for substances A, B, and C, and show how Hess's law applies.

1310

views

Textbook Question

Calculate the enthalpy change for the reaction P₄O₆(s) + 2 O₂(g) → P₄O₁₀(s) given the following enthalpies of reaction: P₄(s) + 3 O₂(g) → P₄O₆(s) ΔH = -1640.1 kJ P₄(s) + 5 O₂(g) → P₄O₁₀(s) ΔH = -2940.1 kJ

1297

views