Chapter 14, Problem 14.61QE

Interpretation Introduction

Interpretation:

The expression for the equilibrium constant and the partial pressure of ${\text{CO}}_{\text{2}}$ for the following reaction has to be determined.

  ${\text{CaCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons \text{CaO}\left(\text{s}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)$

Concept Introduction:

The condition of equilibrium is a state of balance of processes that runs in opposite directions. At equilibrium, the formation of a product from the reactant balances the formation of reactant from the product. Also, the change in concentration of reaction and product seems to be negligible at equilibrium state.

The general equilibrium reaction is as follows:

  $a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$

Here,

$\text{A}$ and $\text{B}$ are the reactants.

$\text{C}$ and $\text{D}$ are products.

$a$ and $b$ are the stoichiometric coefficients of reactants.

$c$ and $d$ are the stoichiometric coefficients of products.

The expression of the equilibrium constant for the above reaction is as follows:

  $K_{\text{p}}=\frac{P_{\text{C}}^c{P}_{\text{D}}^d}{P_{\text{A}}^a{P}_{\text{B}}^b}$

Here,

$K_{\text{p}}$ is the equilibrium constant.

$P_{\text{C}}$ is the partial pressure of $\text{C}$.

$P_{\text{D}}$ is the partial pressure of $\text{D}$.

$P_{\text{A}}$ is the partial pressure of $\text{A}$.

$P_{\text{B}}$ is the partial pressure of $\text{B}$.