Chapter 23, Problem 38QP

Interpretation Introduction

Interpretation:

The value of $\Delta {\text{H}}^{\circ }$ for the decomposition reaction is to be determined.

Concept Introduction:

The enthalpy of reaction is calculated by the expression as

$\Delta {\text{H}}^{\circ }=\sum \Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{products}\right)-\sum \Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{reactants}\right)$.

Here, $\Delta {\text{H}}^{\circ }$ is the standard change in enthalpy of reaction, $\sum \Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{reactants}\right)$ is the sum of standard change in enthalpy of reactants, $\sum \Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{products}\right)$ is the standard change in enthalpy of products.

Answer to Problem 38QP

Solution:

The $\Delta {\text{H}}^{\circ }$ value for the decomposition reaction is equal to $117.6\text{ kJ/mol}$.

The $\Delta {\text{H}}^{\circ }$ value for the decomposition reaction is equal to $177.8\text{ kJ/mol}$.

Explanation of Solution

a) ${\text{MgCO}}_3\left(s\right)\to \text{MgO}\left(s\right)+{\text{CO}}_2\left(g\right)$.

The value of $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{MgCO}}_3\right)$ from appendix $2$ is $\text{-1112}\text{.9}\text{kJ/mol}$.

The value of $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)$ from appendix $2$ is $-393.5\text{kJ/mol}$.

The value of $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{MgO}\right)$ from appendix $2$ is $\text{-601}\text{.8}\text{kJ/mol}$.

The value of $\Delta {\text{H}}^{\circ }$ for the given reaction is calculated by using the relation given below:

$\Delta {\text{H}}^{\circ }=\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{MgO}\right)+\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)-\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{MgCO}}_3\right)$

Here, $\Delta {\text{H}}^{\circ }$ is the standard change in enthalpy, $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{MgO}\right)$ is the standard change in enthalpy of MgO, $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)$ is the standard change in enthalpy of ${\text{CO}}_2$ and $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{MgCO}}_3\right)$ is the standard change in enthalpy of ${\text{MgCO}}_3$.

Substitute $\text{-601}\text{.8}\text{kJ/mol}$ for $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{MgO}\right)$, $-393.5\text{kJ/mol}$ for $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right),$ and $\text{-1112}\text{.9}\text{kJ/mol}$ for $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{MgCO}}_3\right)$ in the equation as:

$\begin{array}{c}\Delta {\text{H}}^{\circ }=\left(1\right)\left(\text{-601}\text{.8}\text{kJ/mol}\right)+\left(1\right)\left(-393.5\text{kJ/mol}\right)-\left(1\right)\left(\text{-1112}\text{.9}\text{kJ/mol}\right)\\ =117.6\text{ kJ/mol}\text{.}\end{array}$

b) ${\text{CaCO}}_3\left(s\right)\to \text{CaO}\left(s\right)+{\text{CO}}_2\left(g\right)$.

The value of $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CaCO}}_3\right)$ from appendix $2$ is $-1206.9\text{kJ/mol}$.

The value of $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)$ from appendix $2$ is $-393.5\text{kJ/mol}$.

The value of $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{CaO}\right)$ from appendix $2$ is $\text{-635}\text{.6}\text{kJ/mol}$.

The value of $\Delta {\text{H}}^{\circ }$ for the given reaction is calculated by using the relation given below:

$\Delta {\text{H}}^{\circ }=\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{CaO}\right)+\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)-\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CaCO}}_3\right)$.

Here, $\Delta {\text{H}}^{\circ }$ is the standard change in enthalpy, $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{CaO}\right)$ is the standard change in enthalpy of CaO, $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)$ is the standard change in enthalpy of ${\text{CO}}_2$ and $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CaCO}}_3\right)$ is the standard change in enthalpy of ${\text{CaCO}}_3$.

Substitute $\text{-635}\text{.6}\text{kJ/mol}$ for $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{CaO}\right)$, $-393.5\text{kJ/mol}$ for $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)$, and $-1206.9\text{kJ/mol}$ for $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CaCO}}_3\right)$ in the equation as

$\begin{array}{c}\Delta {\text{H}}^{\circ }=\left(1\right)\left(\text{-635}\text{.6}\text{kJ/mol}\right)+\left(1\right)\left(-393.5\text{kJ/mol}\right)-\left(1\right)\left(-1206.9\text{kJ/mol}\right)\\ =177.8\text{ kJ/mol}\text{.}\end{array}$

The value of $\Delta {\text{H}}^{\circ }$ for ${\text{MgCO}}_3$ is less as compared to the value of $\Delta {\text{H}}^{\circ }$ for ${\text{CaCO}}_3$, and thus on heating, ${\text{MgCO}}_3$ is more easily decomposed.