Chapter 3, Problem 3.24P

Interpretation Introduction

(a)

Interpretation:

The equilibrium constant for the reaction at $\text{25°C}$ is to be calculated.

Concept introduction:

The change in standard free energy is given by an expression shown below.

$\text{Δ}G°=G_{\text{p}}^{°}-G_{\text{r}}^{°}=-2.3RT\log \left(K_{\text{eq}}\right)$

Where,

• $K_{\text{eq}}$ is the equilibrium constant.

• $R$ is the Gas constant.

• $T$ is the absolute temperature in Kelvin.

• $G_{\text{p}}^{°}$ is the standard free energy for product.

• $G_{\text{r}}^{°}$ is the standard free energy for reactant.

Interpretation Introduction

(b)

Interpretation:

The standard free-energy difference between starting materials and products for a reaction that has an equilibrium constant of $305$ is to be calculated.

Concept introduction:

The change in standard free energy is given by an expression shown below.

$\text{Δ}G°=G_{\text{p}}^{°}-G_{\text{r}}^{°}=-2.3RT\log \left(K_{\text{eq}}\right)$

Where,

• $K_{\text{eq}}$ is the equilibrium constant.

• $R$ is the Gas constant.

• $T$ is the absolute temperature in Kelvin.

• $G_{\text{p}}^{°}$ is the standard free energy for product.

• $G_{\text{r}}^{°}$ is the standard free energy for reactant.