Chapter 6, Problem 42E

Interpretation Introduction

Interpretation: Partial pressure of ${\text{H}}_2$ , ${\text{I}}_2$ , and $\text{HI}$ at equilibrium is to be calculated.

Concept introduction: Consider a hypothetical reversible reaction as follows:

  $\text{aA}+\text{bB}\rightleftharpoons \text{cC}+\text{dD}$

Here proper quotient of pressure at particular temperature can be written as follows:

  ${\text{Q}}_{\text{p}}=\frac{{\left(\frac{{\text{P}}_{\text{C}}}{{\text{P}}_{\text{0}}}\right)}^{\text{c}}{\left(\frac{{\text{P}}_{\text{D}}}{{\text{P}}_{\text{0}}}\right)}^{\text{d}}}{{\left(\frac{{\text{P}}_{\text{A}}}{{\text{P}}_{\text{0}}}\right)}^{\text{a}}{\left(\frac{{\text{P}}_{\text{B}}}{{\text{P}}_{\text{0}}}\right)}^{\text{d}}}$

At equilibrium, proper quotient of pressure can be represented by ${\text{K}}_{\text{P}}$ .

So at equilibrium, the relation can be written as follows:

  ${\text{K}}_{\text{p}}=\frac{{\left(\frac{{\text{P}}_{\text{C}}}{{\text{P}}_{\text{0}}}\right)}_{\text{eq}}^{\text{c}}{\left(\frac{{\text{P}}_{\text{D}}}{{\text{P}}_{\text{0}}}\right)}_{\text{eq}}^{\text{d}}}{{\left(\frac{{\text{P}}_{\text{A}}}{{\text{P}}_{\text{0}}}\right)}_{\text{eq}}^{\text{a}}{\left(\frac{{\text{P}}_{\text{B}}}{{\text{P}}_{\text{0}}}\right)}_{\text{eq}}^{\text{b}}}$

Where,

• ${\text{K}}_{\text{P}}$is the equilibrium constant.
• ${\left[{\text{P}}_{\text{A}}\right]}_{\text{eq}}$ and ${\left[{\text{P}}_{\text{B}}\right]}_{\text{eq}}$ are the partial pressure of reactant ‘A’ and ‘B’ respectively at equilibrium.
• ${\left[{\text{P}}_{\text{C}}\right]}_{\text{eq}}$ and ${\left[{\text{P}}_{\text{D}}\right]}_{\text{eq}}$ are the partial pressure of product ‘C’ and ‘D’ respectively at equilibrium.
• ‘a’ and ‘b’ are the stoichiometric coefficient of reactant ‘A’ and ‘B’ respectively.
• c’ and ‘d’ are the stoichiometric coefficient of product ‘C’ and ‘D’ respectively.