Chapter 17: Problem 64

Consider the reaction $$ \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g) \quad K=4.4 \times 10^{-19} $$ (a) Calculate \(\Delta G^{\circ}\) for the reaction at \(25^{\circ} \mathrm{C}\). (b) Calculate \(\Delta G_{i}^{\circ}\) for \(\mathrm{N}_{2} \mathrm{O}\) at \(25^{\circ} \mathrm{C}\).

### Short Answer

## Step by step solution

## Part (a) - Calculate \(\Delta G^{\circ}\) for the reaction at 25°C

## Part (b) - Calculate \(\Delta G_{i}^{\circ}\) for N2O at 25°C

## Key Concepts

These are the key concepts you need to understand to accurately answer the question.

###### Understanding Equilibrium Constant

In the expression for equilibrium constant:

- For a reaction \( aA + bB \leftrightarrow cC + dD \), the equilibrium constant \( K \) is given by \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \).
- The brackets \([ ]\) denote concentration in terms of molarity (moles per liter).

###### Thermodynamics and Gibbs Free Energy

The relationship can be summarized as follows:

- If \( \Delta G^{\circ} < 0 \), the reaction is spontaneous and will proceed in the forward direction.
- If \( \Delta G^{\circ} = 0 \), the system is at equilibrium.
- If \( \Delta G^{\circ} > 0 \), the reaction is non-spontaneous and tends to go in the reverse direction.

###### The Dynamics of Chemical Reactions

Our exercise involves a seemingly simple reaction: \( \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \rightarrow 3 \mathrm{NO}(g) \). Though it looks straightforward, reactions often occur in complex steps that influence how and when equilibrium is reached.

- The rates at which reactants are converted to products depend on factors like temperature, pressure, and catalysts.
- The energy changes associated with reactions are explained using Gibbs free energy, which considers both enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)).
- The ability to predict reaction spontaneity and direction heavily relies on thermodynamics, as well as the calculated \( \Delta G \).