Hess’ Law of Constant Heat Summation

Hess’s Law of Constant Heat Summation states that the total enthalpy change for a chemical reaction is independent of the pathway taken. In other words, the heat released or absorbed in a chemical reaction is the same whether the reaction occurs in one step or in a series of steps.

This law is based on the principle of conservation of energy, which states that energy cannot be created or destroyed. In a chemical reaction, the total amount of energy released or absorbed is the same, regardless of the pathway taken.

Example

The following example shows how Hess’s Law can be used to calculate the enthalpy change for a reaction that cannot be measured directly.

Consider the reaction:

$$\ce{2CO(g) + O_2(g) -> 2CO_2(g)}$$

The enthalpy change for this reaction can be calculated using the following steps:

1. Find the enthalpy change for the following reaction:

$$\ce{CO(g) + 1/2O_2(g) -> CO_2(g)}$$

The enthalpy change for this reaction is -283 kJ/mol.

2. Multiply the enthalpy change for the reaction in step 1 by 2.

This gives us -566 kJ/mol.

3. The enthalpy change for the reaction in step 1 is the same as the enthalpy change for the reaction in step 2.

Therefore, the enthalpy change for the reaction $\ce{2CO(g) + O2(g) -> 2CO2(g)}$ is -566 kJ/mol.

Hess’s Law of Constant Heat Summation is a powerful tool that can be used to calculate the enthalpy change for a chemical reaction. This law is based on the principle of conservation of energy, which states that energy cannot be created or destroyed.

Example-Based on Hess’s Law of Constant Heat Summation

Hess’s Law of Constant Heat Summation states that the total heat change in a chemical reaction is independent of the pathway taken. This means that the heat change for a reaction can be calculated by adding up the heat changes for the individual steps in the reaction.

For example, consider the following reaction:

$$\ce{2H2(g) + O2(g) -> 2H2O(l)}$$

The heat change for this reaction can be calculated by adding up the heat changes for the following steps:

$$\ce{H2(g) + 1/2O2(g) -> H2O(l) ΔH = -285.8 kJ}$$

The total heat change for the reaction is:

$$\ce{ΔH = -285.8 kJ + (-285.8 kJ) = -571.6 kJ}$$

This is the same heat change that would be obtained if the reaction were carried out in a single step.

Hess’s Law can be used to calculate the heat change for any chemical reaction, regardless of the complexity of the reaction. This makes it a very useful tool for thermochemistry.

Applications of Hess’s Law

Hess’s Law has a number of applications in thermochemistry. Some of the most common applications include:

• Calculating the heat change for a reaction that cannot be carried out in a single step.
• Determining the enthalpy of formation of a compound.
• Calculating the heat of combustion of a fuel.
• Predicting the products of a chemical reaction.

Hess’s Law is a powerful tool that can be used to solve a variety of thermochemical problems. It is a fundamental principle of thermodynamics and is used extensively in the fields of chemistry, engineering, and materials science.

Calculation of Enthalpy of Formation

Enthalpy of formation is a measure of the energy change that occurs when a compound is formed from its constituent elements. It is an important thermodynamic property that is used in a variety of chemical calculations.

Standard Enthalpy of Formation

The standard enthalpy of formation of a compound is the enthalpy change that occurs when one mole of the compound is formed from its constituent elements in their standard states. The standard state of an element is the most stable form of the element at a pressure of 1 atm and a temperature of 25°C.

Calculating Enthalpy of Formation

The enthalpy of formation of a compound can be calculated using the following equation:

$$\ce{ΔHf° = ΣΔHf°(products) - ΣΔHf°(reactants)}$$

where:

• ΔHf° is the standard enthalpy of formation of the compound
• ΔHf°(products) is the sum of the standard enthalpies of formation of the products
• ΔHf°(reactants) is the sum of the standard enthalpies of formation of the reactants

Example

To calculate the standard enthalpy of formation of water, we need to know the standard enthalpies of formation of hydrogen and oxygen. The standard enthalpy of formation of hydrogen is 0 kJ/mol, and the standard enthalpy of formation of oxygen is 0 kJ/mol. Therefore, the standard enthalpy of formation of water is:

$$\ce{ΔHf°(H2O) = [2ΔHf°(H2) + ΔHf°(O2)] - [0 kJ/mol + 0 kJ/mol] = 0 kJ/mol}$$

This means that the formation of water from hydrogen and oxygen is a thermoneutral process.

Applications of Enthalpy of Formation

The enthalpy of formation is a useful property for a variety of chemical calculations. For example, it can be used to:

• Predict the products of a chemical reaction
• Calculate the heat released or absorbed by a chemical reaction
• Design chemical processes

The enthalpy of formation is an important thermodynamic property that is used in a variety of chemical calculations. It is a measure of the energy change that occurs when a compound is formed from its constituent elements.

Problem Based on Hess’ Law

Hess’s Law states that the enthalpy change of a chemical reaction is independent of the pathway taken. This means that the enthalpy change of a reaction can be calculated by adding up the enthalpy changes of the individual steps in the reaction.

Problem

Consider the following reaction:

$$\ce{CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)}$$

The enthalpy change of this reaction can be calculated using Hess’s Law by adding up the enthalpy changes of the following steps:

$$\ce{CH4(g) + O2(g) → CO(g) + 2H2O(g) ΔH = -890 kJ}$$

$$\ce{CO(g) + O2(g) → CO2(g) ΔH = -283 kJ}$$

The overall enthalpy change of the reaction is:

$$\ce{ΔH = ΔH1 + ΔH2 = -890 kJ + (-283 kJ) = -1173 kJ}$$

The enthalpy change of the reaction between CH4 and O2 to form CO2 and H2O is -1173 kJ. This value was calculated using Hess’s Law by adding up the enthalpy changes of the individual steps in the reaction.

Hess Law FAQs

Q: What is Hess’s Law?

A: Hess’s Law states that the total enthalpy change for a chemical reaction is independent of the pathway taken. In other words, the enthalpy change for a reaction is the same whether it occurs in one step or multiple steps.

Q: How is Hess’s Law used?

A: Hess’s Law can be used to calculate the enthalpy change for a reaction that cannot be measured directly. This is done by adding up the enthalpy changes for the individual steps of the reaction.

Q: What are some examples of Hess’s Law calculations?

A: Here are a few examples of how Hess’s Law can be used to calculate enthalpy changes:

• The enthalpy change for the combustion of methane can be calculated by adding up the enthalpy changes for the following reactions:

$$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) ΔH = -890 kJ}$$

$$\ce{C(s) + O2(g) -> CO2(g) ΔH = -393 kJ}$$

$$\ce{2H2(g) + O2(g) -> 2H2O(g) ΔH = -572 kJ}$$

The total enthalpy change for the combustion of methane is:

$$\ce{ΔH = -890 kJ + (-393 kJ) + (-572 kJ) = -1855 kJ}$$

• The enthalpy change for the formation of water can be calculated by adding up the enthalpy changes for the following reactions:

$$\ce{H2(g) + 1/2O2(g) -> H2O(g) ΔH = -286 kJ}$$

$$\ce{C(s) + O2(g) -> CO2(g) ΔH = -393 kJ}$$

$$\ce{CO2(g) + H2O(g) -> H2CO3(aq) ΔH = -20 kJ}$$

The total enthalpy change for the formation of water is:

$$\ce{ΔH = -286 kJ + (-393 kJ) + (-20 kJ) = -699 kJ}$$

Q: What are the limitations of Hess’s Law?

A: Hess’s Law only applies to reactions that occur at constant temperature and pressure. It also does not apply to reactions that involve a change in the number of moles of gas.

Q: Is Hess’s Law still used today?

A: Yes, Hess’s Law is still used today by chemists to calculate enthalpy changes for reactions. It is a valuable tool for understanding the thermodynamics of chemical reactions.