6.3 Hess's Law

HESS'S LAW (6.3)

A) Enthalpy (H)

  • Definition: Enthalpy is a state function which means it is a property that depends solely on the initial and final states of a system, irrespective of the path taken.

Key Equation:

  • Change in enthalpy, represented as \Delta H, is defined as the heat absorbed or released during a reaction.
    • Example: \Delta H = 68 \text{ kJ} (indicating heat is absorbed, denoted as +).
    • Example Reaction:
    • \text{N}2 + 2\text{O}2 \rightarrow 2\text{NO}_2
    • Alternative Reaction: \text{N}2 + \text{O}2 \rightarrow 2\text{NO}
    • Another Reaction: 2\text{NO} + \text{O}2 \rightarrow 2\text{NO}2
Path Function vs State Function:

  • State Function:

    • Depends only on the start and end points.
    • Example: Elevation.

  • Path Function:

    • Depends on the process or the way the system transitions from one state to another.
    • Example: Distance traveled in a specific path.

B) Characteristics of Enthalpy Changes

  1. It is acceptable to reverse a reaction and change the sign of \Delta H accordingly.
    • Example: If \text{H}2(g) + \text{O}2(g) \rightarrow \text{H}_2\text{O}(l) has \Delta H = -286 \text{ kJ}, then its reverse reaction has \Delta H = +286 \text{ kJ}.
  2. Magnitude of \Delta H: The magnitude of the change in enthalpy is proportional to the quantities of the reactants and products involved in a reaction.
    • Example: For the reaction \text{H}2(g) + \text{O}2(g) \rightarrow \text{H}2\text{O}(l), the enthalpy change is \Delta H = -286 \text{ kJ} per mole of \text{O}2 consumed.

C) Sample Problem #1: Calculation of Entropy Change for Carbon Forms

  • Problem: Calculate \Delta H for the conversion of graphite to diamond: \text{C(graphite)} \rightarrow \text{C(diamond)}

Given Information:

  1. Reaction: \text{C(graphite)} + \text{O}2 \rightarrow \text{CO}2

    • Enthalpy Change: \Delta H = -394 \text{ kJ}

  2. Reaction: \text{C(diamond)} + \text{O}2 \rightarrow \text{CO}2

    • Enthalpy Change: \Delta H = -396 \text{ kJ}

Steps to Calculate:

  1. Flip the equation for diamond to align with the desired reaction.
    • This means taking the equation for diamond and changing the sign in front of its enthalpy:
      \Delta H = +396 \text{ kJ}
  2. Combine equations:
    • \text{C(graphite)} + \frac{9}{10} \text{O}_2
      ightarrow ext{C(diamond)} will have its enthalpy added to that for graphite.

Final Calculation:

  • \Delta H = -394 \text{ kJ} + 396 \text{ kJ} = +2 \text{ kJ}

C) Sample Problem #2: Calculation of Enthalpy for Boranes Formation

  • Problem: Calculate \Delta H for the formation of \text{B}2\text{H}6(g) from elemental boron and hydrogen: 2\text{B}(s) + 3\text{H}2(g) \rightarrow \text{B}2\text{H}_6(g)

Given Information:

  1. Reaction: 2\text{B}(s) + \frac{3}{2} \text{O}2(g) \rightarrow \text{B}2\text{O}_3(s)

    • Enthalpy Change: \Delta H = -1273 \text{ kJ}

  2. Reaction: \text{B}2\text{H}6(g) + 3 \text{O}2(g) \rightarrow \text{B}2\text{O}3(s) + 3\text{H}2\text{O}(g)

    • Enthalpy Change: \Delta H = -2035 \text{ kJ}

  3. Reaction: \text{H}2(g) + \text{O}2(g) \rightarrow \text{H}_2\text{O}(l)

    • Enthalpy Change: \Delta H = -286 \text{ kJ}

  4. Reaction for water formation, decomposing water into gas:

    • \text{H}2\text{O}(l) \rightarrow \text{H}2\text{O}(g)

Final Calculation Steps:

  1. Total heat for the reactions needed, adjustments for stoichiometry:

    • Multiply the needed reactions for proper substance formation and apply the corresponding enthalpy changes.
    • Upon calculation through rearranging, arrive at total enthalpy for full reaction.

  2. Example Calculation (illustrative):

    • \Delta H = 3(-286 \text{ kJ}) - 1273 \text{ kJ} + - (-2035 \text{ kJ}) + \text{others}

  3. Compute the specific enthalpy changes involved, ensuring to adhere to signs and states of matter in calculations.

    • Empirical values require all terms to be collected accordingly, for example:
  • Final Result including calculations = \text{Final } \Delta H = !Final Value!