Chapter 17, Problem 17.105P

(a)

Interpretation Introduction

Interpretation:

For the given reaction ${\text{K}}_{\text{c}}\text{ = 1}{\text{.26×10}}^{\text{-3}}$ at $\text{298 K}$, ${\text{K}}_{\text{P}}$ has to be calculated.

(b)

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{c}}$ for the formation of $\text{HI}$ at 298K has to be identified.

Concept Introduction:

Equilibrium constant:

The relationship between the concentration of products and concentration of reactants in a chemical reaction at equilibrium is said to be equilibrium constant.  It is denoted by $\text{K}$.

For a reaction,

  $\text{xX + yY }\rightleftharpoons \text{ zZ}$

The expression of $\text{K}$ can be given as

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = }\frac{{\text{[Z]}}^{\text{z}}}{{\text{[X]}}^{\text{x}}{\text{[Y]}}^{\text{y}}}\\ \\ \text{where,}\text{[X] = equilibrium concentration of X}\\ \\ \text{[Y] = equilibrium concentration of Y}\\ \\ \text{[Z] = equilibrium concentration of Z}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The ${\text{ΔH}}_{\text{reaction}}^{\text{o}}$ for ${\text{HI decomposition for the ΔH}}_{\text{f}}^{\text{o}}\text{ has to be calculated}$

Concept Introduction:

Equilibrium constant:

The relationship between the concentration of products and concentration of reactants in a chemical reaction at equilibrium is said to be equilibrium constant.  It is denoted by $\text{K}$.

For a reaction,

  $\text{xX + yY }\rightleftharpoons \text{ zZ}$

The expression of $\text{K}$ can be given as

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = }\frac{{\text{[Z]}}^{\text{z}}}{{\text{[X]}}^{\text{x}}{\text{[Y]}}^{\text{y}}}\\ \\ \text{where,}\text{[X] = equilibrium concentration of X}\\ \\ \text{[Y] = equilibrium concentration of Y}\\ \\ \text{[Z] = equilibrium concentration of Z}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The ${\text{ΔH}}_{\text{reaction}}^{\text{o}}$ for ${\text{HI decomposition for the ΔH}}_{\text{f}}^{\text{o}}\text{ has to be calculated}$ from the van’t off equation.