Chapter 21, Problem 21.33QP

(a)

Interpretation Introduction

Interpretation:

The $\Delta H^{\circ }$ values for the given decomposition reaction has to be determined.

Concept introduction:

To calculate $\Delta H^{\circ }$ value we can use the following formula

$\Delta H^o{}_{\text{reaction}}=\sum \Delta H^o{}_f(\text{products})-\sum \Delta H^o{}_f(\text{reactants})$

Where, $\Delta H^{\circ }$= standard enthalpy change formation

• $\Delta H^{\circ }$ is inversely proportional to the decomposition of reaction.
• When $\Delta H^{\circ }$ increases decomposition of reaction decreases.

Answer to Problem 21.33QP

The $\Delta H^{\circ }$ values for the given decomposition is

$\Delta H^{\circ }$ of ${\text{MgCO}}_{\text{3}}$ 117.6 kJ/mol

Explanation of Solution

To determine $\Delta H^{\circ }$ values for decomposition of ${\text{MgCO}}_{\text{3(s)}}\to {\text{MgO}}_{\text{(s)}}{\text{+CO}}_{\text{2(g)}}$

Substitute the suitable values in the following equation we get the answer as follows

$\Delta H^o{}_{{\text{MgCO}}_{\text{3}}}=\Delta H^o{}_f\left(\text{MgO}\right)+\Delta H^o{}_f\left({\text{CO}}_{\text{2}}\right)-\Delta H^o{}_f\left({\text{MgCO}}_{\text{3}}\right)$

$\Delta H^o{}_{{\text{MgCO}}_{\text{3}}}=\left(\text{1}\right)\left(\text{-601}\text{.8 kJ/mol}\right)\text{ + }\left(\text{1}\right)\left(\text{-393}\text{.5 kJ/mol}\right)\text{ - }\left(\text{1}\right)\left(\text{-1112}\text{.9 kJ/mol}\right)\text{ = 117}\text{.6 kJ/mol}$

Conclusion

The $\Delta H^{\circ }$ values for the given decomposition reaction was determined

$\Delta H^{\circ }$ of ${\text{MgCO}}_{\text{3}}$ 117.6 kJ/mol

(b)

Interpretation Introduction

Interpretation:

The $\Delta H^{\circ }$ values for the given decomposition reaction and easy decomposed by heat has to be explained.

Concept introduction:

To calculate $\Delta H^{\circ }$ value we can use the following formula

$\Delta H^o{}_{\text{reaction}}=\sum \Delta H^o{}_f(\text{products})-\sum \Delta H^o{}_f(\text{reactants})$

Where, $\Delta H^{\circ }$= standard enthalpy change formation

• $\Delta H^{\circ }$ is inversely proportional to the decomposition of reaction.
• When $\Delta H^{\circ }$ increases decomposition of reaction decreases.

Answer to Problem 21.33QP

The $\Delta H^{\circ }$ values for the given decomposition is

$\Delta H^{\circ }$ of ${\text{CaCO}}_{\text{3}}$177.8 kJ/mol

• ${\text{MgCO}}_{\text{3}}$ easily decomposes than ${\text{CaCO}}_{\text{3}}$.

Explanation of Solution

To determine $\Delta H^{\circ }$ values for decomposition of ${\text{CaCO}}_{\text{3(s)}}\to {\text{CaO}}_{\text{(s)}}{\text{+CO}}_{\text{2(g)}}$

Substitute the suitable values in the following equation we get the answer as follows

$\Delta H^o{}_{{\text{CaCO}}_{\text{3}}}=\Delta H^o{}_f\left(\text{CaO}\right)+\Delta H^o{}_f\left({\text{CO}}_{\text{2}}\right)-\Delta H^o{}_f\left({\text{CaCO}}_{\text{3}}\right)$

$\Delta H^o{}_{{\text{CaCO}}_{\text{3}}}=\left(\text{1}\right)\left(\text{-635}\text{.6 kJ/mol}\right)\text{ + }\left(\text{1}\right)\left(\text{-393}\text{.5 kJ/mol}\right)\text{ - }\left(\text{1}\right)\left(\text{-1206}\text{.9 kJ/mol}\right)\text{ = 177}\text{.8 kJ/mol}$

To explain which one easily decomposes.

• $\Delta H^{\circ }$ is inversely proportional to the decomposition of reaction.
• When $\Delta H^{\circ }$ increases decomposition of reaction decreases.
• So, ${\text{MgCO}}_{\text{3}}$ easily decomposes than ${\text{CaCO}}_{\text{3}}$.

Conclusion

The $\Delta H^{\circ }$ values for the given decomposition reaction and which one gets easy decomposed was explained.

$\Delta H^{\circ }$ of ${\text{CaCO}}_{\text{3}}$177.8 kJ/mol

• ${\text{MgCO}}_{\text{3}}$ is easily decomposes than ${\text{CaCO}}_{\text{3}}$