Questions 22-25 refer to the following information. N2O4(g) ⇄ 2 NO2(g) Kp = 3.0 at 70°C colorless br — Thermodynamics Chemistry Question

This question wraps up our multi-part analysis of the dinitrogen tetroxide and nitrogen dioxide equilibrium system by connecting the molecular-level process of bond-breaking to its thermodynamic consequences.

STEPS:

1. Identify the molecular process occurring in the forward reaction:
* Look at the Lewis structure of the reactant, dinitrogen tetroxide ($\text{N}_2\text{O}_4$). It consists of two symmetric $\text{NO}_2$ groups held together by a single nitrogen–nitrogen ($\text{N–N}$) covalent bond.
* In the forward reaction:
$$\text{N}_2\text{O}_4(g) \rightarrow 2\ \text{NO}_2(g)$$
* The single $\text{N–N}$ bond is broken to produce two separate $\text{NO}_2$ molecules, and no new chemical bonds are formed during this process.
2. Apply the fundamental energetic principle of chemical bonds:
* A covalent bond is a stable, low-energy state resulting from electrostatic attractions between the nuclei and shared electrons.
* To disrupt this stable state and separate the atoms, energy must always be absorbed/input from the surroundings.
* Therefore, bond breaking is always an endothermic process (energy is required), whereas bond formation is always exothermic (energy is released).
3. Relate the energy change to the sign of $\Delta H^\circ$:
* By thermodynamic convention, when a system absorbs heat/energy from its surroundings (an endothermic process), its enthalpy increases.
* This is represented by a positive enthalpy change: $\Delta H^\circ > 0$.
4. Cross-reference with experimental observations:
* This molecular-level reasoning is perfectly consistent with our observation from Question 23, where placing the system in an ice bath shifted the equilibrium to the left.
* A shift toward reactants upon cooling is macroscopic, experimental proof that the forward reaction is indeed endothermic ($\Delta H^\circ > 0$).
5. Select the matching option:
* Putting these concepts together: $\Delta H^\circ > 0$ because energy is required to break the $\text{N–N}$ bond. This corresponds exactly to Option D.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: This option states that $\Delta H^\circ < 0$ (exothermic) and makes the fundamental conceptual error of claiming that "energy is released when a bond breaks." Breaking a bond is never a source of energy release; it always requires an energy input.
• Option B is incorrect: Although it correctly states that "energy is required to break the $\text{N–N}$ bond," it makes a thermodynamic error by associating this endothermic process with a negative enthalpy sign ($\Delta H^\circ < 0$), which represents an exothermic process.
• Option C is incorrect: While it correctly identifies that $\Delta H^\circ > 0$, it pairs this with the incorrect physical explanation that "energy is released when the $\text{N–N}$ bond breaks."