Exothermic and Endothermic Reactions

Every chemical reaction involves energy changes. Enthalpy (H) is the total heat content of a system at constant pressure. We measure the enthalpy change (ΔH) -- the difference in enthalpy between the products and reactants.

Exothermic (ΔH < 0)

Energy is released to the surroundings. Products have less energy than reactants.

E

Reactants

↓ ΔH < 0

Products

Examples: Combustion, neutralisation, respiration

Endothermic (ΔH > 0)

Energy is absorbed from the surroundings. Products have more energy than reactants.

E

Reactants

↑ ΔH > 0

Products

Examples: Photosynthesis, thermal decomposition, dissolving NH₄NO₃

Key convention: ΔH = H_products − H_reactants. A negative value means the system has lost energy (exothermic); a positive value means the system has gained energy (endothermic).

Enthalpy Diagrams and Activation Energy

An enthalpy diagram (energy profile diagram) shows the energy changes during a reaction. The activation energy (E_a) is the minimum energy required for the reaction to proceed -- it represents the energy barrier that must be overcome.

Energy Profile: Exothermic Reaction

Reactants

Starting energy level

↑

Transition State (E_a)

Peak energy -- activation energy barrier

↓

Products

Lower energy level (ΔH < 0)

E_a = Reactants → Peak ΔH = Reactants − Products

Catalyst effect: A catalyst lowers the activation energy (E_a) by providing an alternative reaction pathway. It does not change ΔH -- the overall enthalpy change remains the same.

Hess's Law

Hess's law states that the total enthalpy change for a reaction is the same regardless of the route taken, provided the initial and final conditions are the same. This is a consequence of enthalpy being a state function.

Hess's Law Cycle

Reactants (A)

ΔH (direct) →

Products (B)

ΔH₁ ↓

Route 1 = Route 2
ΔH = ΔH₁ + ΔH₂

↑ ΔH₂

Intermediate (C)

Applying Hess's Law with Standard Enthalpies of Formation

Formula: ΔH_rxn = ΣΔH_f(products) − ΣΔH_f(reactants)

The standard enthalpy of formation (ΔH_f°) is the enthalpy change when one mole of a compound forms from its elements in their standard states (298 K, 100 kPa).

By definition, ΔH_f° of any element in its standard state is zero.

Key Vocabulary

Enthalpy (H)

The total heat content of a system at constant pressure. We measure changes in enthalpy (ΔH) rather than absolute values.

Exothermic Reaction

A reaction that releases heat to the surroundings, resulting in a negative ΔH value.

Activation Energy (E_a)

The minimum energy required for reactant molecules to undergo a successful collision and form products.

Hess's Law

The total enthalpy change for a reaction is independent of the route taken, depending only on the initial and final states.

Worked Examples

1

The combustion of methane has ΔH = −890 kJ/mol. Is this exothermic or endothermic? Explain.

Step 1: Check the sign of ΔH. Here, ΔH = −890 kJ/mol, which is negative.

Step 2: A negative ΔH means the products have less energy than the reactants.

Answer: This is an exothermic reaction. Energy (890 kJ per mole of methane) is released to the surroundings as heat.

2

Using Hess's law, calculate ΔH for: C(s) + O₂(g) → CO₂(g), given: (1) C(s) + ½O₂(g) → CO(g), ΔH₁ = −110 kJ/mol; (2) CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283 kJ/mol.

Step 1: By Hess's law, ΔH = ΔH₁ + ΔH₂ (since reactions 1 and 2 can be added to give the target reaction).

Step 2: ΔH = (−110) + (−283) = −393 kJ/mol.

Answer: ΔH = −393 kJ/mol. The combustion of carbon to carbon dioxide releases 393 kJ per mole.

3

Calculate ΔH_rxn for 2NO(g) + O₂(g) → 2NO₂(g) using ΔH_f°: NO(g) = +90.3 kJ/mol, NO₂(g) = +33.2 kJ/mol.

Step 1: Apply ΔH_rxn = ΣΔH_f(products) − ΣΔH_f(reactants).

Step 2: Products: 2 × (+33.2) = +66.4 kJ. Reactants: 2 × (+90.3) + 0 = +180.6 kJ.

Step 3: ΔH_rxn = +66.4 − (+180.6) = −114.2 kJ/mol.

Answer: ΔH_rxn = −114.2 kJ/mol. The reaction is exothermic.

Knowledge Check

Select the correct answer for each question. Click "Check Answer" to see if you are right.

Question 1

A reaction has ΔH = +56 kJ/mol. This reaction is:

Exothermic -- it releases energy Endothermic -- it absorbs energy Neither -- the sign tells us nothing Exothermic -- a positive value means heat is given out

Question 2

On an enthalpy diagram for an exothermic reaction, the products are:

Higher than the reactants At the same level as the reactants Lower than the reactants At the transition state

Question 3

A catalyst affects a reaction by:

Making ΔH more negative Increasing the activation energy Changing the enthalpy of the products Lowering the activation energy without changing ΔH

Question 4

Hess's law states that the total enthalpy change for a reaction depends only on:

The initial and final states, not the route taken The number of steps in the reaction The activation energy of each step Whether a catalyst is present

Question 5

Given: ΔH_f°(H₂O) = −286 kJ/mol and ΔH_f°(H₂) = 0 kJ/mol. What is ΔH for 2H₂(g) + O₂(g) → 2H₂O(l)?

−286 kJ −572 kJ +572 kJ +286 kJ

Enthalpy Changes

Exothermic and Endothermic Reactions

Exothermic (ΔH < 0)

Endothermic (ΔH > 0)

Enthalpy Diagrams and Activation Energy

Energy Profile: Exothermic Reaction

Hess's Law

Hess's Law Cycle

Applying Hess's Law with Standard Enthalpies of Formation

Key Vocabulary

Enthalpy (H)

Exothermic Reaction

Activation Energy (E_a)

Hess's Law

Worked Examples

The combustion of methane has ΔH = −890 kJ/mol. Is this exothermic or endothermic? Explain.

Using Hess's law, calculate ΔH for: C(s) + O₂(g) → CO₂(g), given: (1) C(s) + ½O₂(g) → CO(g), ΔH₁ = −110 kJ/mol; (2) CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283 kJ/mol.

Calculate ΔH_rxn for 2NO(g) + O₂(g) → 2NO₂(g) using ΔH_f°: NO(g) = +90.3 kJ/mol, NO₂(g) = +33.2 kJ/mol.

Knowledge Check

Key Concepts Summary

Enthalpy Changes

Exothermic and Endothermic Reactions

Exothermic (ΔH < 0)

Endothermic (ΔH > 0)

Enthalpy Diagrams and Activation Energy

Energy Profile: Exothermic Reaction

Hess's Law

Hess's Law Cycle

Applying Hess's Law with Standard Enthalpies of Formation

Key Vocabulary

Enthalpy (H)

Exothermic Reaction

Activation Energy (Ea)

Hess's Law

Worked Examples

The combustion of methane has ΔH = −890 kJ/mol. Is this exothermic or endothermic? Explain.

Using Hess's law, calculate ΔH for: C(s) + O2(g) → CO2(g), given: (1) C(s) + ½O2(g) → CO(g), ΔH1 = −110 kJ/mol; (2) CO(g) + ½O2(g) → CO2(g), ΔH2 = −283 kJ/mol.

Calculate ΔHrxn for 2NO(g) + O2(g) → 2NO2(g) using ΔHf°: NO(g) = +90.3 kJ/mol, NO2(g) = +33.2 kJ/mol.

Knowledge Check

Key Concepts Summary

Activation Energy (E_a)

Using Hess's law, calculate ΔH for: C(s) + O₂(g) → CO₂(g), given: (1) C(s) + ½O₂(g) → CO(g), ΔH₁ = −110 kJ/mol; (2) CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283 kJ/mol.

Calculate ΔH_rxn for 2NO(g) + O₂(g) → 2NO₂(g) using ΔH_f°: NO(g) = +90.3 kJ/mol, NO₂(g) = +33.2 kJ/mol.