Chapter 9, Problem 107AE

(a)

Interpretation Introduction

Interpretation:

The reaction of the $\Delta H^0{}_f$ for solid aluminum oxide should be written.

Concept Introduction :

Change of enthalpy when the formation of 1mole of a compound from its elements with all substances in their standard states is referred to as the standard enthalpy of formation.

Answer to Problem 107AE

  $2Al(s)+\frac{3}{2}{O}_2(g)\to A{l}_2{O}_3(s)$

Explanation of Solution

  $A{l}_2{O}_3$ is solid compound and $Al(s){\text{ and O}}_2(g)$ are considered as the elements that contribute to the formation of $A{l}_2{O}_3$ . Change of enthalpy when the formation of 1mole of $A{l}_2{O}_3$ from $Al(s){\text{ and O}}_2(g)$ in their standard states is referred to as the standard enthalpy of formation ( $\Delta H^0{}_f$ ).

(b)

Interpretation Introduction

Interpretation:

The reaction of the standard enthalpy of combustion of liquid ethanol should be written.

Concept Introduction :

Standard enthalpy of combustion- change in enthalpy when 1mole of a particular substance is completely burned in $O_2$ with all the reactants and products in their standard state under standard condition.

Answer to Problem 107AE

  $C_2{H}_{5}OH(l)+3O_2(g)\to 2C{O}_2(g)+3H_{2}O(l)$

Explanation of Solution

Change of enthalpy in burning (combine vigorously with oxygen) of 1mol of ethanol liquid under standard conditions in their standard states is the standard enthalpy of combustion of liquid ethanol.

  $C_2{H}_{5}OH(l)+3O_2(g)\to 2C{O}_2(g)+3H_{2}O(l)$

(c)

Interpretation Introduction

Interpretation:The reaction of the standard enthalpy of neutralization of barium hydroxide solution by hydrochloric acid should be written.

Concept Introduction : Standard enthalpy of neutralization-change of enthalpy when solutions of alkali and acid react together to form 1mol of water under standard conditions.

Answer to Problem 107AE

  $Ba{(OH)}_2(aq)+2HCl(aq)\to BaC{l}_2(aq)+2H_{2}O(l)$

Explanation of Solution

Under standard conditions, 1 mol of $Ba{(OH)}_2(aq)$ reacts with 2 mol of HCl (aq) to form water and BaCl₂ salt.

The reaction is represented as follows:

  $Ba{(OH)}_2(aq)+2HCl(aq)\to BaC{l}_2(aq)+2H_{2}O(l)$

(d)

Interpretation Introduction

Interpretation:

The reaction of the $\Delta H^0{}_f$ for gaseous vinyl chloride ( $C_2{H}_{3}Cl(g)$ ) should be written.

Concept Introduction:

Change of enthalpy when the formation of 1mole of a compound from its elements with all substances in their standard states is referred to as the standard enthalpy of formation.

Answer to Problem 107AE

  $2C(s)+\frac{3}{2}{H}_2(g)+\frac{1}{2}C{l}_2(g)\to C{H}_{2}CHCl(g)$

Explanation of Solution

  $C_2{H}_{3}Cl(g)$ is gaseous compound and $C(s),C{l}_2{\text{(g) and H}}_2(g)$ are considered as the elements that contribute to the formation of $C_2{H}_{3}Cl(g)$ . Change of enthalpy when the formation of 1mole of $C_2{H}_{3}Cl(g)$ from $C(s),C{l}_2{\text{(g) and H}}_2(g)$ in their standard states is referred to as the standard enthalpy of formation ( $\Delta H^0{}_f$ ).

(e)

Interpretation Introduction

Interpretation:

The reaction of the standard enthalpy of combustion of liquid benzene( $C_6{H}_6(l)$ )should be written.

Concept Introduction:

Standard enthalpy of combustion- change in enthalpy when 1mole of a particular substance is completely burned in $O_2$ with all the reactants and products in their standard state under standard condition.

Answer to Problem 107AE

  $C_6{H}_6(l)+\frac{15}{2}{O}_2(g)\to 6C{O}_2(g)+3H_{2}O(l)$

Explanation of Solution

Change of enthalpy in burning (combine vigorously with oxygen) of 1mol of ethanol liquid under standard conditions in their standard states is the standard enthalpy of combustion of liquid ethanol.

  $C_6{H}_6(l)+\frac{15}{2}{O}_2(g)\to 6C{O}_2(g)+3H_{2}O(l)$

(f)

Interpretation Introduction

Interpretation:

The reaction of enthalpy of solution of solid ammonium bromide should be written.

Concept Introduction:

Change of enthalpy when dissolution of a substance in a solvent at a constant pressure and a temperature by forming in infinite dilution.

Answer to Problem 107AE

  $N{H}_{4}Br(s)\to N{H}_4{}^{+}(aq)+B{r}^{-}(s)$

Explanation of Solution

Solid $N{H}_{4}Br$ will dissociate into its respective ions as $N{H}_4{}^{+}$ and $B{r}^{-}$ at a constant pressure and a temperature. $N{H}_4{}^{+}$ is a polyatomic ion and it will not further dissociated.