Chapter 4, Problem 62QRT

(a)

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 62QRT

The enthalpy change value for breaking of bonds for bromine reaction is $629\text{ kJ}$ and the enthalpy change value for breaking of bonds for Iodine reaction is $\text{587 kJ}$.

Explanation of Solution

The reactions of molecular hydrogen with bromine and iodine are as follows,

    $\begin{array}{l}{H}_2+B{r}_2\to \text{ 2HBr}\\ \\ H_2+I_2\to \text{ 2H}I\end{array}$

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

    $\begin{array}{c}For{H}_2+B{r}_2\to \text{ 2HBr}\\ \\ {\Delta }_r{H}_{\text{reactants}}\text{ = }\left({\Delta }_b{H}_{\text{H-H}}\right)+\left({\Delta }_b{H}_{\text{halogen-halogen}}\right)\\ \\ {\Delta }_r{H}_{\text{reactants}}\text{ = }\left(436\text{ kJ/mol}\right)+\left(193\text{ kJ/mol}\right)\\ \\ =629\text{ kJ}\end{array}$

    $\begin{array}{c}For{H}_2+I_2\to \text{ 2H}I\\ \\ {\Delta }_r{H}_{\text{reactants}}\text{ = }\left({\Delta }_b{H}_{\text{H-H}}\right)+\left({\Delta }_b{H}_{\text{halogen-halogen}}\right)\\ \\ {\Delta }_r{H}_{\text{reactants}}\text{ = }\left(436\text{ kJ/mol}\right)+\left(151\text{ kJ/mol}\right)\\ \\ =587\text{ kJ}\end{array}$

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer part (a).

Answer to Problem 62QRT

The enthalpy change value for forming of bonds in bromine reaction is $-732\text{ kJ}$ and the enthalpy change value for forming of bonds in iodine reaction is $\text{-598 kJ}$.

Explanation of Solution

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

    $\begin{array}{l}{H}_2+B{r}_2\to \text{ 2HBr}\\ \\ H_2+I_2\to \text{ 2H}I\end{array}$

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.

    $\begin{array}{l}For{H}_2+B{r}_2\to \text{ 2HBr}\\ \\ {\Delta }_r{H}_{\text{products}}\text{ = -2}\left({\Delta }_b{H}_{\text{H-halogen}}\right)\\ \\ \text{= -2}\left(366\text{ kJ/mol}\right)\\ \\ \text{= -732 kJ}\end{array}$

    $\begin{array}{c}For{H}_2+I_2\to \text{ 2H}I\\ \\ {\Delta }_r{H}_{\text{products}}\text{ = -2}\left({\Delta }_b{H}_{\text{H-halogen}}\right)\\ \\ \text{ = -2}\left(\text{299 kJ/mol}\right)\\ \\ \text{= -598 kJ}\end{array}$

(c)

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 62QRT

The enthalpy change value for bromine reaction is $-103\text{ kJ}$ and the enthalpy change value for iodine reaction is $\text{-11 kJ}$.

Explanation of Solution

The reactions of molecular hydrogen with bromine and iodine are as follows,

    $\begin{array}{l}{H}_2+B{r}_2\to \text{ 2HBr}\\ \\ H_2+I_2\to \text{ 2H}I\end{array}$

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

    $\begin{array}{l}For{H}_2+B{r}_2\to \text{ 2HBr}\\ \\ {\Delta }_r{H}_{\text{total}}\text{ =}{\Delta }_r{H}_{\text{reactants}}+{\Delta }_r{H}_{\text{products}}\\ \\ \text{ = }629kJ+\left(-732\text{ kJ}\right)\\ \\ \text{= - }103\text{ kJ}\end{array}$

    $\begin{array}{c}For{H}_2+I_2\to \text{ 2H}I\\ \\ {\Delta }_r{H}_{\text{total}}\text{ =}{\Delta }_r{H}_{\text{reactants}}+{\Delta }_r{H}_{\text{products}}\\ \\ \text{ = 5}87kJ+\left(-598\text{ kJ}\right)\\ \\ \text{= }-11\text{ kJ}\end{array}$

(d)

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 62QRT

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

Explanation of Solution

The reaction of molecular hydrogen with bromine is more exothermic since $-103\text{ kJ}$ is more negative than the value $-11\text{ kJ}$ obtained from reaction between hydrogen and iodine.