Chapter 4, Problem 81QRT

(a)

Interpretation Introduction

Interpretation:

The reaction enthalpy value for given reaction has to be calculated.

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

Concept Introduction:

Enthalpy is the amount energy absorbed or released in a process.

The reaction enthalpy change can be calculated by using the following equation.

${\text{ΔH}}^{\circ }{}_{\text{rxn}}\text{ =}\sum {\text{ ΔH}}^{\text{°}}{}_{\text{produdcts}}\text{- }\sum {\text{ΔH}}^{\text{°}}{}_{\text{reactants}}$

Where,

  $\sum {\text{ΔH}}^{\text{°}}{}_{\text{reactants}}$ is the summation of standard formation enthalpy of all reactants.

  $\sum {\text{ΔH}}^{\text{°}}{}_{\text{produdcts}}$ is the summation of standard enthalpy of all products

Answer to Problem 81QRT

The reaction enthalpy value is $\text{178}\text{.32 kJ/mol}$.

Explanation of Solution

$\text{CaO}\left(\text{s}\right)$ and ${\text{CO}}_{\text{2}}\left(\text{g}\right)$ are products. ${\text{CaCO}}_{\text{3}}\left(\text{s}\right)$ is reactant.

The standard reaction enthalpy is determined as follows by considering the given balanced equation.

  ${\Delta }_r{H}^{\circ }\text{ = }{\Delta }_f{H}^{\circ }\left[\text{CaO}\left(\text{s}\right)\right]+{\Delta }_f{H}^{\circ }\left[{\text{CO}}_{\text{2}}\left(\text{g}\right)\right]-\left\{{\Delta }_f{H}^{\circ }\left[{\text{CaCO}}_{\text{3}}\left(\text{s}\right)\right]\right\}$

The standard formation enthalpy values, ${\Delta }_f{H}^{\circ }$ are taken from Appendix J and substituted into above equation in order to get the reaction enthalpy value.

  $\begin{array}{c}{\Delta }_r{H}^{\circ }\text{ =}\left[-635.09\text{ kJ/mol}\right]+\left[\text{-393}\text{.509 kJ/mol}\right]-\left[-1206.92\text{ kJ/mol}\right]\\ \\ \text{= + }178.32\text{ kJ/mol}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The reaction enthalpy value for given reaction has to be calculated.

    $\text{2HI}\left(\text{g}\right)+{\text{F}}_{\text{2}}\left(\text{g}\right)\to \text{2HF}\left(\text{g}\right){\text{+ I}}_{\text{2}}\left(\text{s}\right)$

Concept Introduction:

Refer part (a).

Answer to Problem 81QRT

The reaction enthalpy value is $\text{-595}\text{.2 kJ/mol}$.

Explanation of Solution

$\text{2HF}\left(\text{g}\right)$ and ${\text{I}}_{\text{2}}\left(\text{s}\right)$ are products. $\text{2HI}\left(\text{g}\right)$ and ${\text{F}}_{\text{2}}\left(\text{g}\right)$ are reactants.

The standard reaction enthalpy is determined as follows by considering the given balanced equation.

  ${\Delta }_r{H}^{\circ }\text{ = 2}{\Delta }_f{H}^{\circ }\left[\text{HF}\left(\text{g}\right)\right]+{\Delta }_f{H}^{\circ }\left[{\text{I}}_{\text{2}}\left(\text{s}\right)\right]-2{\Delta }_f{H}^{\circ }\left[\text{HI}\left(\text{g}\right)\right]-2{\Delta }_f{H}^{\circ }\left[{\text{F}}_{\text{2}}\left(\text{g}\right)\right]$

The standard formation enthalpy values, ${\Delta }_f{H}^{\circ }$ are taken from Appendix J and substituted into above equation in order to get the reaction enthalpy value.

  $\begin{array}{c}{\Delta }_r{H}^{\circ }\text{ = 2}\left[-271.1\text{ kJ/mol}\right]+\left[\text{0 kJ/mol}\right]-2\left[26.48\text{ kJ/mol}\right]-2\left[\text{0 kJ/mol}\right]\\ \\ \text{= }-595.2\text{ kJ/mol}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The reaction enthalpy value for given reaction has to be calculated.

    ${\text{SF}}_{\text{6}}\left(\text{g}\right)+{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)\to \text{6HF}\left(\text{g}\right){\text{+SO}}_3\left(\text{g}\right)$

Concept Introduction:

Refer part (a).

Answer to Problem 81QRT

The reaction enthalpy value is $\text{44 kJ/mol}$.

Explanation of Solution

$\text{HF}\left(\text{g}\right)$ and ${\text{SO}}_3\left(\text{g}\right)$ are products. ${\text{SF}}_{\text{6}}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$ are reactants.

The standard reaction enthalpy is determined as follows by considering the given balanced equation.

  ${\Delta }_r{H}^{\circ }\text{ = 6 }{\Delta }_f{H}^{\circ }\left[\text{HF}\left(\text{g}\right)\right]+{\Delta }_f{H}^{\circ }\left[{\text{SO}}_3\left(\text{g}\right)\right]-{\Delta }_f{H}^{\circ }\left[{\text{SF}}_{\text{6}}\left(\text{g}\right)\right]-3{\Delta }_f{H}^{\circ }\left[{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\right]$

The standard formation enthalpy values, ${\Delta }_f{H}^{\circ }$ are taken from Appendix J and substituted into above equation in order to get the reaction enthalpy value.

  $\begin{array}{c}{\Delta }_r{H}^{\circ }\text{ = 6}\left[-271.1\text{ kJ/mol}\right]+\left[\text{-395}\text{.72 kJ/mol}\right]-\left[-1209\text{ kJ/mol}\right]-3\left[\text{-285}\text{.83 kJ/mol}\right]\\ \\ \text{= }44\text{ kJ/mol}\end{array}$