Thermochemical equations Knowledge-Based Study

Study notes

1. What is a Thermochemical Equation?

   A thermochemical equation is a balanced chemical equation that explicitly states the physical states of all chemical species and the associated enthalpy change ($\Delta H$) of the reaction. Unlike conventional chemical equations that only track mass and atom conservation, thermochemical equations also quantify the energy exchanged between the system and its surroundings under constant pressure.

   The enthalpy change is typically reported in units of kilojoules per mole ($\text{kJ/mol}$) or kilojoules ($\text{kJ}$) relative to the stoichiometric coefficients written in the equation.

   Sign Conventions for Enthalpy Change

   • Exothermic Reactions: Heat is released into the surroundings. Enthalpy decreases, meaning $\Delta H < 0$ (negative value). Energy may also be written as a product on the right side of the equation.
   • Endothermic Reactions: Heat is absorbed from the surroundings. Enthalpy increases, meaning $\Delta H > 0$ (positive value). Energy may also be written as a reactant on the left side of the equation.

   Enthalpy as an Extensive Property

   Enthalpy is an extensive property, which means its value depends directly on the quantity of matter undergoing transformation. Consequently, the value of $\Delta H$ corresponds strictly to the exact molar quantities represented by the stoichiometric coefficients.

   Rules for Manipulating Thermochemical Equations

   1. Scaling Coefficients: If all stoichiometric coefficients in a balanced equation are multiplied or divided by an integer or fraction $n$, the enthalpy change must also be multiplied or divided by that same factor $n$.
      For example, if:
      $$\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \quad \Delta H = -92.2\text{ kJ}$$
      Then halving the coefficients yields:
      $$\frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \rightarrow \text{NH}_3(g) \quad \Delta H = -46.1\text{ kJ}$$
   2. Reversing the Direction: When a reaction is reversed, the magnitude of $\Delta H$ remains identical, but its sign is inverted.
      $$2\text{NH}_3(g) \rightarrow \text{N}_2(g) + 3\text{H}_2(g) \quad \Delta H = +92.2\text{ kJ}$$

   Why Physical States Must Be Specified

   Enthalpy changes depend significantly on the physical states (phase) of all reactants and products. Phase transformations—such as melting, boiling, and sublimation—involve heat exchange. Therefore, writing different physical states changes the overall $\Delta H$ of the reaction.

   Consider the formation of liquid water versus water vapor:

   • Formation of liquid water:
     $$2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l) \quad \Delta H = -571.6\text{ kJ}$$
   • Formation of gaseous water:
     $$2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(g) \quad \Delta H = -483.6\text{ kJ}$$

   The release of less heat during the production of gaseous water reflects the fact that vapor retains the latent heat of vaporization ($\Delta H_{\text{vap}}$). Omitting the state symbols $(s, l, g, aq)$ makes a thermochemical equation ambiguous and quantitatively inaccurate.

   Using $\Delta H$ as a Conversion Factor

   In quantitative stoichiometry, the enthalpy change can serve as a conversion factor between moles of substances consumed or produced and the thermal energy transferred.

   Worked Example

   Problem: Calculate the amount of heat released when 8.00 g of methane ($\text{CH}_4$) is completely combusted in excess oxygen according to the thermochemical equation:

   $$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H = -890.4\text{ kJ}$$

   Solution:

   1. Determine the molar mass of $\text{CH}_4$:
      $\text{Molar mass} = 12.01 + 4(1.008) = 16.04\text{ g/mol}$
   2. Convert the mass of $\text{CH}_4$ to moles:
      $n = \frac{8.00\text{ g}}{16.04\text{ g/mol}} = 0.4988\text{ mol}$
   3. Use the stoichiometric ratio between $\text{CH}_4$ and $\Delta H$ (1 mol $\text{CH}_4$ corresponds to $-890.4\text{ kJ}$):
      $q = 0.4988\text{ mol} \times (-890.4\text{ kJ/mol}) = -444\text{ kJ}$

   Thus, 444 kJ of heat is released into the surroundings.