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Energetics of Reactions - Exothermic and Endothermic Reactions

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Energy Conservation: The Law of Conservation of Energy states that energy cannot be created or destroyed, only transferred. In chemical reactions, energy is transferred between the system (reactants and products) and the surroundings.

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Exothermic Reactions: These reactions release thermal energy to the surroundings. As a result, the temperature of the surroundings increases. The enthalpy change ΔH\Delta H is negative (ΔH<0\Delta H < 0) because the products have less chemical energy than the reactants. Common examples include combustion, neutralization, and respiration.

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Endothermic Reactions: These reactions absorb thermal energy from the surroundings. As a result, the temperature of the surroundings decreases. The enthalpy change ΔH\Delta H is positive (ΔH>0\Delta H > 0) because the products have more chemical energy than the reactants. Common examples include photosynthesis and thermal decomposition (e.g., CaCO3→CaO+CO2CaCO_{3} \rightarrow CaO + CO_{2}).

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Enthalpy Change (ΔH\Delta H): This represents the heat content change of a system at constant pressure, typically measured in kJ⋅mol−1kJ \cdot mol^{-1}.

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Activation Energy (EaE_{a}): The minimum energy required for reactant particles to collide with enough force to break bonds and start a reaction. Both exothermic and endothermic reactions require activation energy.

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Bond Energetics: Breaking chemical bonds is an endothermic process (requires energy input). Forming chemical bonds is an exothermic process (releases energy). A reaction is exothermic overall if more energy is released when forming bonds than was taken in to break them.

📐Formulae

ΔH=Hproducts−Hreactants\Delta H = H_{products} - H_{reactants}

ΔH=Energy absorbed to break bonds−Energy released to form bonds\Delta H = \text{Energy absorbed to break bonds} - \text{Energy released to form bonds}

Q=m⋅c⋅ΔTQ = m \cdot c \cdot \Delta T

💡Examples

Problem 1:

In a reaction between Hydrochloric Acid and Sodium Hydroxide, the initial temperature was 22∘C22^{\circ}C. After the reaction, the temperature rose to 28∘C28^{\circ}C. Classify the reaction and state the sign of ΔH\Delta H.

Solution:

The temperature of the surroundings increased by ΔT=28∘C−22∘C=6∘C\Delta T = 28^{\circ}C - 22^{\circ}C = 6^{\circ}C. This indicates that energy was released into the surroundings. Therefore, the reaction is exothermic, and the enthalpy change ΔH\Delta H is negative (ΔH<0\Delta H < 0).

Explanation:

In exothermic reactions, chemical potential energy is converted into thermal energy, causing the temperature of the mixture to rise.

Problem 2:

Calculate the enthalpy change (ΔH\Delta H) for the reaction: H2(g)+Cl2(g)→2HCl(g)H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}. Given bond energies: H−H=436 kJ/molH-H = 436 \text{ kJ/mol}, Cl−Cl=242 kJ/molCl-Cl = 242 \text{ kJ/mol}, H−Cl=431 kJ/molH-Cl = 431 \text{ kJ/mol}.

Solution:

  1. Energy to break bonds (Reactants): (1×436)+(1×242)=678 kJ/mol(1 \times 436) + (1 \times 242) = 678 \text{ kJ/mol}
  2. Energy released forming bonds (Products): 2×431=862 kJ/mol2 \times 431 = 862 \text{ kJ/mol}
  3. ΔH=Energy to break−Energy to form\Delta H = \text{Energy to break} - \text{Energy to form}: ΔH=678−862=−184 kJ/mol\Delta H = 678 - 862 = -184 \text{ kJ/mol}

Explanation:

Since the total energy released during bond formation (862 kJ/mol862 \text{ kJ/mol}) is greater than the energy required to break the reactant bonds (678 kJ/mol678 \text{ kJ/mol}), the reaction is exothermic, resulting in a negative ΔH\Delta H.