Chapter 5, Problem 5G.16E

(a)

Interpretation Introduction

Interpretation:

The Gibbs free energy for the given reaction has to be calculated when the partial pressures of ${\text{H}}_2$, ${\text{I}}_{\text{2}}$ and $\text{HI}$ are $\text{0}\text{.35}\text{bar}$, $\text{0}\text{.18}\text{bar}$ and $\text{2}\text{.85}\text{bar}$, respectively.

  ${\text{H}}_{\text{2}}\text{(g)}{\text{+I}}_{\text{2}}\text{(g)}\to \text{2HI(g)}$

Concept Introduction:

The Gibb’s free energy of a reaction can be calculated using the following expression,

  $\begin{array}{l}{\text{ΔG}}_{\text{r}}\text{=}{\text{ΔG}}_{\text{r}}^{\text{°}}\text{+}\text{RT}\text{lnQ}\\ \\ \text{where,}\\ {\text{ΔG}}_{\text{r}}\text{=}\text{Gibb's}\text{free}\text{energy}\\ {\text{ΔG}}_{\text{r}}^{\text{°}}\text{=}\text{Standard}\text{Gibb's}\text{free}\text{energy}\\ \text{Q}\text{=}\text{reaction}\text{quotient}\end{array}$

The value of reaction quotient can be calculated by the expression,

  $\text{Q}\text{=}\frac{{\left({\text{a}}_{\text{C}}\right)}^{\text{c}}{\left({\text{a}}_{\text{D}}\right)}^{\text{d}}}{{\left({\text{a}}_{\text{A}}\right)}^{\text{a}}{\left({\text{a}}_{\text{B}}\right)}^{\text{b}}}$

(b)

Interpretation Introduction

Interpretation:

  The above reaction mixture whether form reactants, products or at equilibrium has to be determined.

Concept introduction:

Refer to part (a).