Chapter 4, Problem 61QRT

For the reactions of molecular hydrogen with fluorine and with chlorine:

1. (a) Calculate the enthalpy change for breaking all the bonds in the reactants.
2. (b) Calculate the enthalpy change for forming all the bonds in the products.
3. (c) From the results in parts (a) and (b), calculate the enthalpy change for the reaction.
4. (d) Which reaction is most exothermic?

Interpretation Introduction

Interpretation:

The enthalpy change for breaking of all bonds present in all reactants of given reaction has to be calculated.

Answer to Problem 61QRT

The enthalpy change value for breaking of bonds for fluorine reaction is $594\text{ kJ}$ and the enthalpy change value for breaking of bonds for chlorine reaction is $\text{678 kJ}$.

Explanation of Solution

The reaction of molecular hydrogen with fluorine and chlorine is as follows,

    $\begin{array}{l}{H}_2+F_2\to \text{ 2HF}\\ \\ H_2+C{l}_2\to \text{ 2H}Cl\end{array}$

Above both reactions, involves breaking of1 $\text{H-H}$ and 1 halogen-halogen bond.

    $\begin{array}{c}For{H}_2+F_2\to \text{ 2HF}\\ \\ {\Delta }_r{H}_{\text{reactants}}\text{ = }\left({\Delta }_b{H}_{\text{H-H}}\right)+\left({\Delta }_b{H}_{\text{halogen-halogen}}\right)\\ \\ {\Delta }_r{H}_{\text{reactants}}\text{ = }\left(436\text{ kJ/mol}\right)+\left(158\text{ kJ/mol}\right)\\ \\ =594\text{ kJ}\end{array}$

    $\begin{array}{c}For{H}_2+C{l}_2\to \text{ 2H}Cl\\ \\ {\Delta }_r{H}_{\text{reactants}}\text{ = }\left({\Delta }_b{H}_{\text{H-H}}\right)+\left({\Delta }_b{H}_{\text{halogen-halogen}}\right)\\ \\ {\Delta }_r{H}_{\text{reactants}}\text{ = }\left(436\text{ kJ/mol}\right)+\left(242\text{ kJ/mol}\right)\\ \\ =678\text{ kJ}\end{array}$

Interpretation Introduction

Interpretation:

The enthalpy change for forming of all bonds present in all products of given reaction has to be calculated.

Answer to Problem 61QRT

The enthalpy change value for forming of bonds in fluorine reaction is $-1132\text{ kJ}$ and the enthalpy change value for forming of bonds in chlorine reaction is $\text{-862 kJ}$.

Explanation of Solution

The reaction of molecular hydrogen with fluorine and chlorine is as follows,

    $\begin{array}{l}{H}_2+F_2\to \text{ 2HF}\\ \\ H_2+C{l}_2\to \text{ 2H}Cl\end{array}$

Above both reactions involve formation of 2 hydrogen-halogen bonds. The ${\text{Δ}}_{\text{r}}{\text{H}}_{\text{products}}$ value should be in negative since enthalpy of bond forming should be in opposite sign of the enthalpy of bond breaking.

    $\begin{array}{c}For{H}_2+F_2\to \text{ 2HF}\\ \\ {\Delta }_r{H}_{\text{products}}\text{ = -2}\left({\Delta }_b{H}_{\text{H-halogen}}\right)\\ \\ \text{= -2}\left(566\text{ kJ/mol}\right)\\ \\ =-1132\text{ kJ}\end{array}$

    $\begin{array}{c}For{H}_2+C{l}_2\to \text{ 2H}Cl\\ \\ {\Delta }_r{H}_{\text{products}}\text{ = -2}\left({\Delta }_b{H}_{\text{H-halogen}}\right)\\ \\ \text{ = -2}\left(431\text{ kJ/mol}\right)\\ \\ =-862\text{ kJ}\end{array}$

Interpretation Introduction

Interpretation:

The enthalpy change given reaction has to be calculated.

Concept Introduction:

The enthalpy change in a system $\text{(Δ}{{\rm H}}_{\text{sys}}\text{)}$ can be calculated by the following equation.

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

Where,

  ${\text{ΔH}}^{\text{°}}{}_{\text{reactants}}$ is the standard enthalpy of the reactants

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

Answer to Problem 61QRT

The enthalpy change value for fluorine reaction is $-538\text{ kJ}$ and the enthalpy change value for chlorine reaction is $\text{-184 kJ}$.

Explanation of Solution

The reaction of molecular hydrogen with fluorine and chlorine is as follows,

    $\begin{array}{l}{H}_2+F_2\to \text{ 2HF}\\ \\ H_2+C{l}_2\to \text{ 2H}Cl\end{array}$

The enthalpy change value for each reaction is determined by considering the formula, ${\Delta }_r{H}_{\text{total}}\text{ =}{\Delta }_r{H}_{\text{reactants}}+{\Delta }_r{H}_{\text{products}}$.

    $\begin{array}{c}For{H}_2+F_2\to \text{ 2HF}\\ \\ {\Delta }_r{H}_{\text{total}}\text{ =}{\Delta }_r{H}_{\text{reactants}}+{\Delta }_r{H}_{\text{products}}\\ \\ \text{ = }594kJ+\left(-1132\text{ kJ}\right)\\ \\ \text{= -}538\text{ kJ}\end{array}$

    $\begin{array}{c}For{H}_2+C{l}_2\to \text{ 2H}Cl\\ \\ {\Delta }_r{H}_{\text{total}}\text{ =}{\Delta }_r{H}_{\text{reactants}}+{\Delta }_r{H}_{\text{products}}\\ \\ \text{ = }678kJ+\left(-862\text{ kJ}\right)\\ \\ \text{= }-184\text{ kJ}\end{array}$

Interpretation Introduction

Interpretation:

From the two given reactions, the exothermic reaction has to be identified.

Concept Introduction:

Enthalpy is the amount energy absorbed or released in a process. Under constant pressure conditions the enthalpy change will be equal to molar q.

Exothermic reaction: Exothermic reactions are those in which evolution of heat takes place during any chemical reaction. They release heat because the reactant molecules require less heat for breakage of bonds than the product molecules.

Endothermic reaction: Endothermic reactions are those in which heat is absorbed during any chemical reaction. In such type of reactions, external energy is needed.

Answer to Problem 61QRT

The reaction between molecular hydrogen and fluorine is more exothermic than the other one.

Explanation of Solution

The reaction of molecular hydrogen with fluorine is more exothermic since $-538\text{ kJ}$ is more negative than the value $-184\text{ kJ}$ obtained from reaction between hydrogen and chlorine.