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Chemistry - Chemical Energetics

Grade 10IGCSE

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

🔑Concepts

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Exothermic Reactions: Reactions that release thermal energy to the surroundings. The temperature of the surroundings increases. The enthalpy change ΔH\Delta H is negative (ΔH<0\Delta H < 0). Examples include combustion and neutralization.

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Endothermic Reactions: Reactions that absorb thermal energy from the surroundings. The temperature of the surroundings decreases. The enthalpy change ΔH\Delta H is positive (ΔH>0\Delta H > 0). Examples include photosynthesis and thermal decomposition.

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Activation Energy (EaE_a): The minimum amount of energy that colliding particles must possess to react. On an energy profile diagram, it is the energy gap between the reactants and the peak of the curve.

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Bond Energies: Breaking chemical bonds is an endothermic process (requires energy intake), while forming chemical bonds is an exothermic process (releases energy).

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Enthalpy Change (ΔH\Delta H): The overall energy change in a reaction, calculated as the difference between the energy required to break reactant bonds and the energy released when product bonds are formed.

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Energy Profile Diagrams: Graphical representations of energy changes. In exothermic reactions, the products are at a lower energy level than the reactants. In endothermic reactions, the products are at a higher energy level.

📐Formulae

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=mcΔTQ = mc\Delta T

ΔH=Qn\Delta H = \frac{Q}{n}

Bond Energy Calculation: ΔH=∑Bond energies (reactants)−∑Bond energies (products)\text{Bond Energy Calculation: } \Delta H = \sum \text{Bond energies (reactants)} - \sum \text{Bond energies (products)}

💡Examples

Problem 1:

Calculate the enthalpy change (ΔH\Delta H) for the combustion of methane: CH4+2O2→CO2+2H2OCH_4 + 2O_2 \rightarrow CO_2 + 2H_2O. Given bond energies: C−H=413 kJ/molC-H = 413 \text{ kJ/mol}, O=O=495 kJ/molO=O = 495 \text{ kJ/mol}, C=O=799 kJ/molC=O = 799 \text{ kJ/mol}, and O−H=463 kJ/molO-H = 463 \text{ kJ/mol}.

Solution:

  1. Energy to break bonds (reactants): (4×413)+(2×495)=1652+990=2642 kJ/mol(4 \times 413) + (2 \times 495) = 1652 + 990 = 2642 \text{ kJ/mol}. \n2. Energy released forming bonds (products): (2×799)+(4×463)=1598+1852=3450 kJ/mol(2 \times 799) + (4 \times 463) = 1598 + 1852 = 3450 \text{ kJ/mol}. \n3. ΔH=2642−3450=−808 kJ/mol\Delta H = 2642 - 3450 = -808 \text{ kJ/mol}.

Explanation:

Since ΔH\Delta H is negative (−808 kJ/mol-808 \text{ kJ/mol}), the reaction is exothermic. More energy is released during the formation of the C=OC=O and O−HO-H bonds in the products than is absorbed to break the C−HC-H and O=OO=O bonds in the reactants.

Problem 2:

During a neutralization reaction between NaOHNaOH and HClHCl, the temperature of 50 g50 \text{ g} of water increased by 10∘C10^{\circ}C. Calculate the heat energy QQ released. (Specific heat capacity of water c=4.18 J/g∘Cc = 4.18 \text{ J/g}^{\circ}C)

Solution:

Q=m×c×ΔTQ = m \times c \times \Delta T \n Q=50 g×4.18 J/g∘C×10∘C=2090 JQ = 50 \text{ g} \times 4.18 \text{ J/g}^{\circ}C \times 10^{\circ}C = 2090 \text{ J}

Explanation:

The formula Q=mcΔTQ = mc\Delta T is used to calculate the thermal energy transferred to the solution. Here, the mass mm is the mass of the reaction mixture, and ΔT\Delta T is the change in temperature.

Chemical Energetics - Revision Notes & Key Formulas | IGCSE Grade 10 Science