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Understanding Energetics and Thermochemistry in IB Chemistry


For You Education students studying energetics and thermochemistry in IB Chemistry

As a student doing IB chemistry, you’ve probably studied the units on thermodynamics and thermochemistry. The discussed topics constitute basic knowledge on the involvement of energy in chemical reactions, its influence upon substance properties, etc.


Energetics is defined as chemical reactions involving heat and other forms that lead to energy changes. Thermochemistry is a different concept that deals with the quantitative measurement of the chemical heat absorbed during a reaction.


Here are some key concepts and examples you should know to succeed in these subtopics:


1.Enthalpy


Enthalpy refers to the heat energy in a chemical reaction. It is usually employed in determining the heat of reaction, which is the difference in enthalpy between a chemically reactive substance and its product. Understanding this concept is vital to appreciating the spontaneity of such reactions as well as determining their feasibility.


For example, let's consider the combustion of methane:


CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)


The heat of reaction can be calculated by subtracting the enthalpy of the reactants from the enthalpy of the products:


ΔH = [ΔHf(CO2) + 2ΔHf(H2O)] - [ΔHf(CH4) + 2ΔHf(O2)]


where ΔHf is the standard enthalpy of formation of each compound.


2.Entropy


Entropy is an indicator of disorder or randomness in any system. The entropy change in chemical reactions is a comparison of the entropy for products against the entropy for reactants.


For example, let's consider the reaction of solid sodium bicarbonate with hydrochloric acid:


NaHCO3 (s) + HCl (aq) → NaCl (aq) + CO2 (g) + H2O (l)


The entropy change can be calculated using the formula:


ΔS = S(products) - S(reactants)


where S is the molar entropy of each substance. In this case, the entropy change is positive, indicating an increase in disorder.


Tips for success in Energetics and Thermochemistry:


1. Understand the difference between enthalpy and entropy, and how they are related.


        2. Practice calculating enthalpy changes and using them to predict the feasibility of reactions.


3. Pay attention to units and standard conditions when working with thermochemical equations.


        4. Use visual aids such as diagrams and charts to help understand complex concepts.


By mastering these key concepts and practicing with examples, you can succeed in the challenging subtopics of energetics and thermochemistry in IB Chemistry.


Worked Example:


Example 1:

Calculate the enthalpy change for the reaction:


2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)


The following bond enthalpies are given:


C-C: 348 kJ/mol


C-H: 413 kJ/mol


O=O: 498 kJ/mol


C=O: 805 kJ/mol


O-H: 463 kJ/mol


Solution:

This problem requires us to calculate enthalpy changes based on bond enthalpies. The reaction can be broken down into the bond breaking and bond making steps:


Bond breaking: 4 C-C bonds x 348 kJ/mol = 1392 kJ/mol 12 C-H bonds x 413 kJ/mol = 4956 kJ/mol 7 O=O bonds x 498 kJ/mol = 3486 kJ/mol


Bond making: 8 C=O bonds x 805 kJ/mol = 6440 kJ/mol 12 O-H bonds x 463 kJ/mol = 5556 kJ/mol


Enthalpy change = (bond making energy) - (bond breaking energy)


Enthalpy change = 6440 kJ/mol + 5556 kJ/mol - 1392 kJ/mol - 4956 kJ/mol - 3486 kJ/mol


Enthalpy change = -122 kJ/mol


Therefore, the enthalpy change for the reaction is -122 kJ/mol.


Example 2:

Calculate the standard enthalpy change of the following reaction at 298 K:


N2(g) + 3H2(g) → 2NH3(g)


The following standard enthalpies of formation are given:


N2(g): 0 kJ/mol


H2(g): 0 kJ/mol


NH3(g): -46 kJ/mol


Solution:

In this problem, we need to use the standard enthalpy of formation to calculate the standard enthalpy change of the reaction. Standard enthalpy changes can be calculated using the following formula.


ΔH°rxn = ΣnΔH°f(products) - ΣnΔH°f(reactants)


Where ΔH°f is the standard enthalpy of reaction, n is the stoichiometric coefficient of each component in the equilibrium chemical equation.


Substituting the values given in the problem, we get:


ΔH°rxn = [2 x ΔH°f(NH3)] - [ΔH°f(N2) + 3 x ΔH°f(H2)]


ΔH°rxn = [2 x (-46 kJ/mol)] - [0 kJ/mol + 3 x 0 kJ/mol]


ΔH°rxn = -92 kJ/mol


Therefore, the standard enthalpy change of the reaction is -92 kJ/mol.

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