krit.club logo

Chemistry - Chemical energetics (Exothermic and Endothermic reactions)

Grade 9IGCSE

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

🔑Concepts

•

Chemical reactions involve an energy change, usually in the form of heat. This change is measured as the enthalpy change, denoted as ΔH\Delta H.

•

In an Exothermic reaction, energy is released to the surroundings. The temperature of the surroundings increases. The energy of the products is lower than the energy of the reactants, so ΔH\Delta H is negative (ΔH<0\Delta H < 0).

•

In an Endothermic reaction, energy is absorbed from the surroundings. The temperature of the surroundings decreases. The energy of the products is higher than the energy of the reactants, so ΔH\Delta H is positive (ΔH>0\Delta H > 0).

•

Bond Breaking is an endothermic process because energy must be supplied to overcome the forces of attraction between atoms.

•

Bond Making is an exothermic process because energy is released when new chemical bonds are formed.

•

The Activation Energy (EaE_a) is the minimum amount of energy that colliding particles must possess for a chemical reaction to occur.

•

Common exothermic processes: Combustion of fuels (e.g., CH4+2O2→CO2+2H2OCH_4 + 2O_2 \rightarrow CO_2 + 2H_2O), Neutralization reactions, and Respiration.

•

Common endothermic processes: Thermal decomposition (e.g., CaCO3→CaO+CO2CaCO_3 \rightarrow CaO + CO_2), Photosynthesis, and dissolving NH4NO3NH_4NO_3 in water.

📐Formulae

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

Q=mcΔTQ = mc\Delta T

ΔH=−Qn\Delta H = -\frac{Q}{n}

Energy change per mole=Total Energy Change (kJ)Number of moles (mol)\text{Energy change per mole} = \frac{\text{Total Energy Change (kJ)}}{\text{Number of moles (mol)}}

💡Examples

Problem 1:

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=243 kJ/molCl-Cl = 243\text{ kJ/mol}, and H−Cl=432 kJ/molH-Cl = 432\text{ kJ/mol}.

Solution:

  1. Energy to break bonds (Reactants): (1×436)+(1×243)=679 kJ/mol(1 \times 436) + (1 \times 243) = 679\text{ kJ/mol}.
  2. Energy released forming bonds (Products): 2×432=864 kJ/mol2 \times 432 = 864\text{ kJ/mol}.
  3. ΔH=679−864=−185 kJ/mol\Delta H = 679 - 864 = -185\text{ kJ/mol}.

Explanation:

Since the total energy released during bond making (864 kJ/mol864\text{ kJ/mol}) is greater than the energy required for bond breaking (679 kJ/mol679\text{ kJ/mol}), the overall reaction is exothermic, resulting in a negative ΔH\Delta H.

Problem 2:

In a calorimetry experiment, 0.05 mol0.05\text{ mol} of HClHCl reacts with 0.05 mol0.05\text{ mol} of NaOHNaOH. The temperature of 100 g100\text{ g} of water increases by 3∘C3^{\circ}C. Calculate the heat energy released (QQ). (Specific heat capacity of water c=4.18 J/g∘Cc = 4.18\text{ J/g}^{\circ}C)

Solution:

Q=mcΔTQ = mc\Delta T Q=100 g×4.18 J/g∘C×3∘C=1254 JQ = 100\text{ g} \times 4.18\text{ J/g}^{\circ}C \times 3^{\circ}C = 1254\text{ J}.

Explanation:

The heat energy released is calculated using the mass of the solution, the specific heat capacity, and the observed temperature rise. This is an exothermic neutralization reaction.