Unit 4: Energy Changes

Understanding the flow of heat in chemical reactions and how it is measured.

4.1 Exothermic & Endothermic Reactions

Every chemical substance possesses stored chemical potential energy, known as enthalpy (H). During a chemical reaction, this enthalpy changes as reactants are converted into products. The overall enthalpy change ($\Delta H$) is the heat energy exchanged with the surroundings at constant pressure.

  • Exothermic Reaction: A reaction that releases heat into the surroundings. The products are more stable (lower enthalpy) than the reactants. This results in a negative $\Delta H$ and a rise in temperature.
  • Endothermic Reaction: A reaction that absorbs heat from the surroundings. The products are less stable (higher enthalpy) than the reactants. This results in a positive $\Delta H$ and a drop in temperature.

These changes can be visualized with enthalpy level diagrams, which plot enthalpy against the process of the reaction.

Solved Examples:
  1. Is the combustion of wood exothermic or endothermic? Explain.
    Solution: Exothermic. It releases a large amount of energy as heat and light. The products (CO₂, H₂O) have lower enthalpy than the reactants (wood and O₂).
  2. When solid ammonium nitrate is dissolved in water, the beaker feels cold. What does this indicate?
    Solution: This indicates an endothermic process. The dissolving process absorbs heat from the surroundings (the water and the beaker), causing the temperature to drop. $\Delta H$ is positive.
  3. Draw an enthalpy level diagram for an exothermic reaction.
    Solution: The diagram should show the reactants at a higher enthalpy level than the products, with a downward arrow indicating a negative $\Delta H$.
  4. What is the sign of $\Delta H$ for melting ice?
    Solution: Positive ($\Delta H > 0$). Energy must be absorbed from the surroundings to break the bonds holding the water molecules in a fixed lattice structure.
  5. A chemical reaction has $\Delta H = -250 \, kJ/mol$. Is the reaction exothermic or endothermic?
    Solution: The negative sign indicates that the reaction is exothermic.
  6. What happens to the enthalpy of the surroundings during an endothermic reaction?
    Solution: The enthalpy (heat content) of the surroundings decreases because the reaction absorbs heat from it.
  7. Give an example of a common exothermic process other than combustion.
    Solution: The neutralization of a strong acid with a strong base (e.g., HCl + NaOH) is a common and highly exothermic reaction.
  8. Photosynthesis requires sunlight to proceed. Is it exothermic or endothermic?
    Solution: Endothermic. It absorbs light energy from the sun to convert CO₂ and H₂O into glucose, a higher-energy molecule.
  9. If a reaction is highly exothermic, are the products more or less stable than the reactants?
    Solution: The products are more stable. They have lower chemical potential energy (enthalpy) because energy has been released.
  10. A cold pack for injuries often contains ammonium nitrate and water. Explain how it works.
    Solution: When the inner pouch is broken, the ammonium nitrate dissolves in the water. This dissolution is a highly endothermic process, absorbing heat from the surroundings (the injury) and making the pack feel cold.

4.2 Molar Enthalpy Changes

Because the amount of heat exchanged depends on the amount of substance reacting, enthalpy changes are standardized as molar enthalpy changes, measured in kilojoules per mole (kJ/mol). This allows for fair comparison between reactions. Several specific types of molar enthalpy changes are defined under standard conditions (101 kPa, 298 K).

  • Standard Enthalpy of Formation ($\Delta H_f^\circ$): The enthalpy change when one mole of a compound is formed from its elements in their standard states. The $\Delta H_f^\circ$ of an element in its standard state is zero.
  • Standard Enthalpy of Combustion ($\Delta H_c^\circ$): The enthalpy change when one mole of a substance is completely burned in excess oxygen.
  • Standard Enthalpy of Neutralization ($\Delta H_{neut}^\circ$): The enthalpy change when one mole of water is formed from the reaction of an acid and a base.
  • Standard Enthalpy of Solution ($\Delta H_{sol}^\circ$): The enthalpy change when one mole of a substance dissolves completely in a large excess of solvent.
Solved Examples:
  1. Write the equation for the standard enthalpy of formation of liquid water ($H_2O$).
    Solution: $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$. Note the formation of exactly one mole of the product.
  2. The combustion of 2 moles of methane releases 1780 kJ of heat. What is the molar enthalpy of combustion?
    Solution: $\Delta H_c = \frac{-1780 \, kJ}{2 \, mol} = -890 \, kJ/mol$. The sign is negative because heat is released.
  3. Write the equation for the standard enthalpy of combustion of ethanol ($C_2H_5OH$).
    Solution: $C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$. Note the combustion of exactly one mole of the substance.
  4. The reaction of HCl with NaOH is $HCl + NaOH \rightarrow NaCl + H_2O, \Delta H = -57.3 \, kJ$. Why is this the molar enthalpy of neutralization?
    Solution: Because the reaction as written produces exactly one mole of water from the acid and base.
  5. The $\Delta H_f^\circ$ of $O_2(g)$ is 0 kJ/mol, but the $\Delta H_f^\circ$ of $O_3(g)$ (ozone) is +142.7 kJ/mol. Why?
    Solution: $O_2(g)$ is the most stable allotrope of oxygen in its standard state, so its formation from itself requires zero energy. Ozone ($O_3$) is a less stable allotrope, so energy is required to form it from $O_2$.
  6. When 0.1 mol of KCl is dissolved, 1.72 kJ of heat is absorbed. Calculate $\Delta H_{sol}^\circ$.
    Solution: $\Delta H_{sol}^\circ = \frac{+1.72 \, kJ}{0.1 \, mol} = +17.2 \, kJ/mol$. The sign is positive because heat is absorbed (endothermic).
  7. Write the equation representing the enthalpy of formation of solid sodium chloride (NaCl).
    Solution: $Na(s) + \frac{1}{2}Cl_2(g) \rightarrow NaCl(s)$. The elements must be in their standard states (solid sodium, gaseous chlorine).
  8. The combustion of 10 g of ethane ($C_2H_6$, M=30 g/mol) releases 519 kJ. Find $\Delta H_c^\circ$.
    Solution: Moles of ethane = $10 \, g / 30 \, g/mol = 1/3$ mol. $\Delta H_c^\circ = \frac{-519 \, kJ}{1/3 \, mol} = -1557 \, kJ/mol$.
  9. Which of the following represents an enthalpy of formation? (a) $2H_2+O_2 \rightarrow 2H_2O$ (b) $C(diamond) \rightarrow C(graphite)$
    Solution: Neither. (a) forms two moles of product. For (b), diamond is not the standard state of carbon. The correct formation equation for graphite would be from itself, with $\Delta H_f^\circ = 0$.
  10. The $\Delta H_{neut}^\circ$ for a weak acid and strong base is less exothermic than for a strong acid/strong base. Why?
    Solution: Some energy is absorbed to fully ionize the weak acid before neutralization can occur. This absorbed energy reduces the total amount of heat released.

4.3 Measuring Enthalpy Changes (Calorimetry)

Calorimetry is the experimental technique used to measure heat changes in chemical reactions. For reactions in solution, a simple calorimeter (like a polystyrene cup) is used. The heat absorbed or released by the reaction causes a temperature change ($\Delta T$) in the solution.

The heat change (q) is calculated using the formula $q = mc\Delta T$, where 'm' is the mass of the solution, 'c' is its specific heat capacity, and '$\Delta T$' is the temperature change. The molar enthalpy change ($\Delta H$) is then found by dividing the heat change by the number of moles (n) of the limiting reactant.

Key Formula: $$ q = mc\Delta T $$ $$ \Delta H = \frac{q}{n} $$
  • $q$ = heat change (in Joules, J)
  • $m$ = mass of the substance being heated (usually water/solution in grams, g)
  • $c$ = specific heat capacity of the substance (for water, $c = 4.18 \, J \, g^{-1} K^{-1}$)
  • $\Delta T$ = change in temperature (in K or °C)
  • $n$ = moles of reactant
Note: If temperature increases, the reaction is exothermic ($\Delta H$ is negative). If temperature decreases, it's endothermic ($\Delta H$ is positive).
Solved Examples:
  1. When 50 cm³ of 1.0 M HCl is mixed with 50 cm³ of 1.0 M NaOH, the temperature rises by 6.8°C. Calculate $\Delta H_{neut}^\circ$. (Assume density = 1.0 g/cm³, c = 4.18 J/g/K).
    Solution: Total volume = 100 cm³, so mass = 100 g. $q = 100 \times 4.18 \times 6.8 = 2842.4$ J. Moles H₂O formed = $1.0 \times 0.050 = 0.05$ mol. $\Delta H = -2842.4 / 0.05 = -56848$ J/mol $\approx -56.8$ kJ/mol.
  2. Dissolving 4.0 g of NaOH (40 g/mol) in 100 cm³ of water raises the temperature by 10.5°C. Calculate $\Delta H_{sol}^\circ$.
    Solution: Mass of solution = 104 g. $q = 104 \times 4.18 \times 10.5 = 4563$ J. Moles NaOH = $4.0/40 = 0.1$ mol. $\Delta H = -4563 / 0.1 = -45630$ J/mol $\approx -45.6$ kJ/mol.
  3. How much heat energy is needed to raise the temperature of 250 g of water from 20°C to 100°C?
    Solution: $\Delta T = 100 - 20 = 80$°C. $q = 250 \times 4.18 \times 80 = 83600$ J or 83.6 kJ.
  4. An excess of zinc is added to 25 cm³ of 0.5 M $CuSO_4$. The temperature rises by 12.5°C. Find $\Delta H$ for the reaction.
    Solution: Mass = 25 g. $q = 25 \times 4.18 \times 12.5 = 1306.25$ J. Moles $CuSO_4 = 0.5 \times 0.025 = 0.0125$ mol. $\Delta H = -1306.25 / 0.0125 = -104500$ J/mol = -104.5 kJ/mol.
  5. If a reaction releases 5000 J of heat into 200 g of water, what is the expected temperature rise?
    Solution: $\Delta T = q / (mc) = 5000 / (200 \times 4.18) \approx 6.0$°C.
  6. Why is a polystyrene cup used for calorimetry experiments?
    Solution: Polystyrene is a good insulator, which minimizes heat loss to or gain from the surroundings, leading to a more accurate measurement of the temperature change.
  7. What is the main source of error in simple calorimetry?
    Solution: Heat loss to the surroundings. Other sources include incomplete reaction and assuming the solution's density and specific heat capacity are the same as pure water.
  8. When 8.0 g of $NH_4NO_3$ (80 g/mol) dissolves in 100 cm³ of water, the temperature drops from 22.0°C to 15.5°C. Calculate $\Delta H_{sol}^\circ$.
    Solution: Mass = 108 g. $\Delta T = 15.5 - 22.0 = -6.5$°C. $q = 108 \times 4.18 \times (-6.5) = -2933$ J. Moles = $8.0/80 = 0.1$ mol. $\Delta H = (-2933 / 0.1) = +29330$ J/mol $\approx +29.3$ kJ/mol (sign is flipped as temp drop is endothermic).
  9. How many moles of HCl reacted if a neutralization reaction with NaOH in 100 cm³ of water released 2.85 kJ of heat? ($\Delta H_{neut}^\circ = -57$ kJ/mol).
    Solution: n = q / $\Delta H = (-2.85 \, kJ) / (-57 \, kJ/mol) = 0.05$ mol.
  10. Calculate the final temperature if 10 kJ of heat is added to 400 g of water initially at 20°C.
    Solution: $\Delta T = q / (mc) = 10000 / (400 \times 4.18) \approx 6.0$°C. Final Temp = $20.0 + 6.0 = 26.0$°C.

4.4 Enthalpies of Combustion & Fuels

The enthalpy of combustion ($\Delta H_c^\circ$) is a critical measure for evaluating fuels. It is the heat released when one mole of a substance is burned completely in oxygen. This value helps determine the energy content of fuels, often measured in kJ/g. A higher energy content means more energy is released per gram of fuel burned.

Fuels like hydrocarbons (petrol, natural gas) and carbohydrates (in food) release energy through exothermic combustion reactions. This energy can be harnessed for transportation, electricity generation, or metabolic processes in living organisms (respiration). Calorimetry experiments using a spirit burner and a copper can are a common way to estimate the enthalpy of combustion of liquid fuels like ethanol.

Solved Examples:
  1. The $\Delta H_c^\circ$ of methane (CH₄, 16 g/mol) is -890 kJ/mol. Calculate its energy content in kJ/g.
    Solution: Energy content = $890 \, kJ / 16 \, g = 55.6$ kJ/g.
  2. Burning 0.62 g of ethanol (C₂H₅OH, 46 g/mol) heats 100 cm³ of water by 30°C. Calculate the experimental $\Delta H_c^\circ$.
    Solution: $q = 100 \times 4.18 \times 30 = 12540$ J. Moles ethanol = $0.62 / 46 = 0.0135$ mol. $\Delta H_c = -12.54 \, kJ / 0.0135 \, mol \approx -929$ kJ/mol.
  3. Why is the experimental value for $\Delta H_c^\circ$ usually less exothermic than the theoretical value?
    Solution: Significant heat loss to the surroundings, incomplete combustion (producing soot/CO), and evaporation of the fuel are major sources of error that lead to a smaller measured temperature change.
  4. A food label says a 50 g snack contains 1000 kJ of energy. What is the energy content in kJ/g?
    Solution: Energy content = $1000 \, kJ / 50 \, g = 20$ kJ/g.
  5. Which fuel releases more energy per gram: Hydrogen (142 kJ/g) or Coal (31 kJ/g)?
    Solution: Hydrogen releases significantly more energy per gram ($142 > 31$), making it a very efficient fuel by mass.
  6. If burning 10 g of a fuel produces 450 kJ of heat, what is its energy content?
    Solution: Energy content = $450 \, kJ / 10 \, g = 45$ kJ/g.
  7. Why is a copper can used in simple calorimetry for combustion?
    Solution: Copper is an excellent thermal conductor, ensuring that the heat released from the burning fuel is efficiently transferred to the water.
  8. The $\Delta H_c^\circ$ of propane (C₃H₈, 44 g/mol) is -2220 kJ/mol. How much heat is released when 11 g is burned?
    Solution: Moles = $11/44 = 0.25$ mol. Heat released = $0.25 \, mol \times 2220 \, kJ/mol = 555$ kJ.
  9. What is the primary product of incomplete combustion of a hydrocarbon fuel?
    Solution: Carbon monoxide (CO) and/or solid carbon (soot), which are produced when there is insufficient oxygen.
  10. How many grams of fat (energy content ~39 kJ/g) would you need to consume to get 10,000 kJ of energy?
    Solution: Mass = Total Energy / Energy Content = $10000 \, kJ / 39 \, kJ/g \approx 256$ g.

End of Topic Questions

Answer: A reaction that releases heat energy into the surroundings, resulting in a negative $\Delta H$.

Answer: The enthalpy change when one mole of a compound is formed from its elements in their standard states.

Answer: $q = mc\Delta T$.

Answer: Endothermic, as it requires energy input to turn liquid into gas.

Answer: Less stable, because they have higher enthalpy (energy has been absorbed).

Answer: $C(s, graphite) + O_2(g) \rightarrow CO_2(g)$.

Answer: 4.18 J g⁻¹ K⁻¹ or 4.18 J g⁻¹ °C⁻¹.

Answer: Negative, because the process is exothermic.

Answer: Zero.

Answer: $q = 50 \times 4.18 \times 20 = 4180$ J or 4.18 kJ.

Answer: The amount of energy released per unit mass (usually kJ/g) when the fuel is burned.

Answer: The enthalpy change when one mole of water is formed from the reaction of an acid and a base under standard conditions.

Answer: $\Delta H_c = -700 \, kJ / 0.5 \, mol = -1400$ kJ/mol.

Answer: The reactants are at a lower enthalpy level than the products.

Answer: The most stable form of a substance at 101 kPa (1 atm) and a specified temperature, usually 298 K (25°C).

Answer: $C(s, graphite) + 2H_2(g) \rightarrow CH_4(g)$.

Answer: $q = 200 \times 4.18 \times 5 = 4180$ J. Since it's a temperature drop, the reaction absorbed 4.18 kJ.

Answer: Exothermic combustion.

Answer: It absorbs heat.

Answer: Exothermic, indicated by the negative sign.
⬅ Back to Unit 4 Overview Next: Rates of Reaction ➡