Chapter 9, Problem 9.76P

(a)

Interpretation Introduction

Interpretation:

The enthalpy of the following reaction is to be calculated.

${\text{H}}_2\left(g\right)+{\text{Cl}}_2\left(g\right)\to 2\text{HCl}\left(g\right)$

Concept introduction:

The heat of the reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is defined as the heat released or absorbed during a chemical reaction as a result of the difference in the bond energies (BE) of reactant and product in the reaction. $\Delta H_{\text{rxn}}^{°}$ is negative for exothermic reaction and $\Delta H_{\text{rxn}}^{°}$ is positive for an endothermic reaction.

The formula to calculate $\Delta H_{\text{rxn}}^{°}$ of reaction is as follows:

$\Delta H_{\text{rxn}}^{°}=\sum \Delta H_{\text{reactant bond broken}}^{°}+\sum \Delta H_{\text{product bond formed}}^{°}$

Or,

$\Delta H_{\text{rxn}}^{°}=\sum {\text{BE}}_{\text{reactant bond broken}}-\sum {\text{BE}}_{\text{product bond formed}}$

The bond energy of reactants is positive and the bond energy of products is negative.

Answer to Problem 9.76P

The enthalpy of the following reaction is $-179\text{kJ}$.

Explanation of Solution

The given chemical equation is as follows:

${\text{H}}_2\left(g\right)+{\text{Cl}}_2\left(g\right)\to 2\text{HCl}\left(g\right)$

The number of broken bonds is $\text{1 H}-\text{H}$ bond and $\text{1 Cl}-\text{Cl}$ bonds.

The number of bonds formed is $2\text{ H}-\text{Cl}$ bonds.

The formula to the enthalpy of the given reaction is as follows:

$\Delta H_{\text{rxn}}^{°}=\left({\text{1BE}}_{\text{H}-\text{H}}+1{\text{BE}}_{\text{Cl}-\text{Cl}}\right)-\left({\text{2BE}}_{\text{H}-\text{Cl}}\right)$ (1)

Substitute $432\text{kJ/mol}$ for ${\text{BE}}_{\text{H}-\text{H}}$, $243\text{kJ/mol}$ for ${\text{BE}}_{\text{Cl}-\text{Cl}}$ and $427\text{kJ/mol}$ for ${\text{BE}}_{\text{H}-\text{Cl}}$ in the equation (1).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\left(\left(\text{1 mol}\right)\left(432\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(243\text{kJ/mol}\right)\right)-\left(\text{2 mol}\right)\left(427\text{kJ/mol}\right)\right]\\ =-179\text{kJ}\end{array}$

Conclusion

The enthalpy of the following reaction is $-179\text{kJ}$.

(b)

Interpretation Introduction

Interpretation:

The enthalpy of the following reaction is to be calculated.

${\text{H}}_2\left(g\right)+{\text{I}}_2\left(g\right)\to 2\text{HI}\left(g\right)$

Concept introduction:

The heat of the reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is defined as the heat released or absorbed during a chemical reaction as a result of the difference in the bond energies (BE) of reactant and product in the reaction. $\Delta H_{\text{rxn}}^{°}$ is negative for exothermic reaction and $\Delta H_{\text{rxn}}^{°}$ is positive for an endothermic reaction.

The formula to calculate $\Delta H_{\text{rxn}}^{°}$ of reaction is as follows:

$\Delta H_{\text{rxn}}^{°}=\sum \Delta H_{\text{reactant bond broken}}^{°}+\sum \Delta H_{\text{product bond formed}}^{°}$

Or,

$\Delta H_{\text{rxn}}^{°}=\sum {\text{BE}}_{\text{reactant bond broken}}-\sum {\text{BE}}_{\text{product bond formed}}$

The bond energy of reactants is positive and the bond energy of products is negative.

Answer to Problem 9.76P

The enthalpy of the following reaction is $-7\text{kJ}$.

Explanation of Solution

The given chemical equation is as follows:

${\text{H}}_2\left(g\right)+{\text{I}}_2\left(g\right)\to 2\text{HI}\left(g\right)$

The number of broken bonds is $\text{1 H}-\text{H}$ bond and $\text{1 I}-\text{I}$ bonds.

The number of bonds formed is $2\text{ H}-\text{I}$ bonds.

The formula to the enthalpy of the given reaction is as follows:

$\Delta H_{\text{rxn}}^{°}=\left({\text{1BE}}_{\text{H}-\text{H}}+1{\text{BE}}_{\text{I}-\text{I}}\right)-\left({\text{2BE}}_{\text{H}-\text{I}}\right)$ (2)

Substitute $432\text{kJ/mol}$ for ${\text{BE}}_{\text{H}-\text{H}}$, $151\text{kJ/mol}$ for ${\text{BE}}_{\text{I}-\text{I}}$ and $295\text{kJ/mol}$ for ${\text{BE}}_{\text{H}-\text{I}}$ in the equation (2).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\left(\left(\text{1 mol}\right)\left(432\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(151\text{kJ/mol}\right)\right)-\left(\text{2 mol}\right)\left(295\text{kJ/mol}\right)\right]\\ =-7\text{kJ}\end{array}$

Conclusion

The enthalpy of the following reaction is $-7\text{kJ}$.

(c)

Interpretation Introduction

Interpretation:

The enthalpy of the following reaction is to be calculated.

${\text{2H}}_2\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{H}}_2\text{O}\left(g\right)$

Concept introduction:

The heat of the reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is defined as the heat released or absorbed during a chemical reaction as a result of the difference in the bond energies (BE) of reactant and product in the reaction. $\Delta H_{\text{rxn}}^{°}$ is negative for exothermic reaction and $\Delta H_{\text{rxn}}^{°}$ is positive for an endothermic reaction.

The formula to calculate $\Delta H_{\text{rxn}}^{°}$ of reaction is as follows:

$\Delta H_{\text{rxn}}^{°}=\sum \Delta H_{\text{reactant bond broken}}^{°}+\sum \Delta H_{\text{product bond formed}}^{°}$

Or,

$\Delta H_{\text{rxn}}^{°}=\sum {\text{BE}}_{\text{reactant bond broken}}-\sum {\text{BE}}_{\text{product bond formed}}$

The bond energy of reactants is positive and the bond energy of products is negative.

Answer to Problem 9.76P

The enthalpy of the following reaction is $-506\text{kJ}$.

Explanation of Solution

The given chemical equation is as follows:

${\text{2H}}_2\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{H}}_2\text{O}\left(g\right)$

The number of broken bonds is $\text{2}\text{H}-\text{H}$ bond and $\text{1 O}-\text{O}$ bonds.

The number of bonds formed is $2\text{ O}-\text{H}$ bonds.

The formula to the enthalpy of the given reaction is as follows:

$\Delta H_{\text{rxn}}^{°}=\left({\text{2BE}}_{\text{H}-\text{H}}+1{\text{BE}}_{\text{O}-\text{O}}\right)-\left({\text{4BE}}_{\text{O}-\text{H}}\right)$ (3)

Substitute $432\text{kJ/mol}$ for ${\text{BE}}_{\text{H}-\text{H}}$, $498\text{kJ/mol}$ for ${\text{BE}}_{\text{O}-\text{O}}$ and $467\text{kJ/mol}$ for ${\text{BE}}_{\text{O}-\text{H}}$ in the equation (3).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\left(\left(\text{2 mol}\right)\left(432\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(498\text{kJ/mol}\right)\right)-\left(\text{4 mol}\right)\left(467\text{kJ/mol}\right)\right]\\ =-506\text{kJ}\end{array}$

Conclusion

The enthalpy of the following reaction is $-506\text{kJ}$.