Chapter 20, Problem 42QRT

Interpretation Introduction

Interpretation:

The reaction given below is product-favored or not has to be determined.

    $\text{2Al(s)}\text{+}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\stackrel{}{\to }\text{2Cr(s)}\text{+}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}$

Concept Introduction:

In order to find whether the reaction is product-favored or not, ${\Delta }_{r}G°$ value can be used.  If the value of ${\Delta }_{r}G°$ is negative, then the reaction is product-favored and if the value of ${\Delta }_{r}G°$ is positive, then the reaction is not product-favored.

  $\begin{array}{l}{\Delta }_{r}G°={\displaystyle \sum \left[(\text{coefficient}\text{of}\text{product})\times {\Delta }_{\text{f}}G°(\text{product})\right]}\\ -{\displaystyle \sum \left[(\text{coefficient}\text{of}\text{reactant})\times {\Delta }_{\text{f}}G°(reac\tan t)\right]}\end{array}$

Explanation of Solution

Given reaction is,

    $\text{2Al(s)}\text{+}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\stackrel{}{\to }\text{2Cr(s)}\text{+}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}$

${\Delta }_{r}G°$ can be calculated as shown below,

$\begin{array}{l}{\Delta }_{r}G°={\displaystyle \sum \left[(\text{coefficient}\text{of}\text{product})\times {\Delta }_{\text{f}}G°(\text{product})\right]}\\ -{\displaystyle \sum \left[(\text{coefficient}\text{of}\text{reactant})\times {\Delta }_{\text{f}}G°(reac\tan t)\right]}\\ =2{\Delta }_{\text{f}}G°\left\{\text{Cr(s)}\right\}+{\Delta }_{\text{f}}G°\left\{{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\right\}-2{\Delta }_{\text{f}}G°\left\{Al\text{(s)}\right\}-{\Delta }_{\text{f}}G°\left\{{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\right\}\\ =2(0)+(-1528.3kJ/mol)-2(0)-(-1058.1\text{kJ/mol})\\ =-1528.3\text{kJ/mol}+1058.1\text{kJ/mol}\\ \text{=}-470.2\text{kJ/mol}\end{array}$

As the value of ${\Delta }_{r}G°$ is negative, the given reaction is product-favored.