Question

Does \(Q\) for the formation of \(1 \mathrm{~mol}\) of NO from its elements differ from \(Q\) for the decomposition of \(1 \mathrm{~mol}\) of \(\mathrm{NO}\) to its elements? Explain and give the relationship between the two \(Q\) 's.

Short answer

Yes, they differ by sign. \( Q_f = -Q_d \)

Step-by-step solution

- Understand Formation and Decomposition

First, understand that formation involves creating a compound from its elements, while decomposition involves breaking a compound down into its elements.

- Write the Chemical Reactions

For formation of NO: \[ \frac{1}{2} \text{N}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{NO} \] For decomposition of NO: \[ \text{NO} \rightarrow \frac{1}{2} \text{N}_2 + \frac{1}{2} \text{O}_2 \]

- Define the Heat of Reaction (Q)

Let \( Q_f \) be the heat of formation for the formation reaction and \( Q_d \) be the heat of decomposition for the decomposition reaction.

- Relationship Between Q_f and Q_d

Heats of formation and decomposition are related by being equal in magnitude but opposite in sign. Thus, \( Q_f = -Q_d \).

- Conclusion

Yes, \( Q \) for the formation of NO differs from \( Q \) for the decomposition of NO by sign. The formation of NO releases or absorbs an amount of energy, while the decomposition of NO absorbs or releases the same amount of energy respectively, but in the opposite direction.

Key concepts

Heat of Formation

The heat of formation, often denoted as \( Q_f \), is the energy change when one mole of a compound is formed from its constituent elements in their standard states. For example, when NO forms from nitrogen (\text{N}_2) and oxygen (\text{O}_2), this reaction occurs under standard conditions.

The chemical equation for the formation of NO is: \[ \frac{1}{2} \text{N}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{NO} \] The energy change in this process is the heat of formation.

When we talk about the heat of formation, we usually refer to this energy change at 25°C and 1 atmosphere pressure (standard conditions). If the process releases heat, the value of \( Q_f \) is negative, indicating an exothermic reaction. If it absorbs heat, \( Q_f \) is positive, indicating an endothermic reaction.

This concept helps us understand the stability of compounds: a large negative \( Q_f \) often means a very stable compound.

Heat of Decomposition

The heat of decomposition, denoted as \( Q_d \), is the energy change when one mole of a compound breaks down into its elements. This is the reverse process of formation.

For NO, the decomposition reaction is: \[ \text{NO} \rightarrow \frac{1}{2} \text{N}_2 + \frac{1}{2} \text{O}_2 \] This reaction also happens under standard conditions. The energy change in this process is the heat of decomposition.

Similar to formation, the heat of decomposition has a sign indicating whether the reaction releases or absorbs energy. If \( Q_d \) is negative, the decomposition releases energy (exothermic); if it's positive, it absorbs energy (endothermic). Usually, decomposition reactions require energy input (endothermic) because breaking chemical bonds typically requires energy.

Enthalpy Sign Relationship

Heats of formation and decomposition are directly related by their signs. This means the heat of formation for a compound is equal in magnitude but opposite in sign to the heat of decomposition.

Mathematically, this relationship is expressed as: \[ Q_f = -Q_d \] If forming a compound releases a certain amount of energy, decomposing that compound must absorb the same amount of energy, and vice versa.

For example, if the formation of NO from nitrogen and oxygen releases energy (egative Q_fegative), the decomposition of NO into nitrogen and oxygen will absorb the same amount of energy (egative Q_degative).

Understanding this relationship helps in predicting how much energy will be required or released in chemical reactions, which is crucial in practical applications like energy production, material synthesis, and understanding reaction mechanisms.