Chapter 5, Problem 5G.14E

Interpretation Introduction

Interpretation:

The Gibbs free energy for the given reaction has to be calculated using the given partial pressures of ${\text{PCl}}_{\text{3}}$, ${\text{Cl}}_{\text{2}}$ and ${\text{PCl}}_5$ are $\text{0}\text{.35}\text{bar}$, $\text{0}\text{.45}\text{bar}$ and $\text{1}\text{.02}\text{bar}$, respectively. Also the spontaneous direction of change has to be given.

  ${\text{PCl}}_3\text{(g)}+{\text{Cl}}_{\text{2}}\text{(g)}\to {\text{PCl}}_{\text{5}}\text{(g)}$

Concept Introduction:

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

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

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

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