Chapter 6, Problem 6.102QP

(a)

Interpretation Introduction

Interpretation:

The equation for the formation of Acetylene and the maximum amount of heat from combustion of Acetylene has to be calculated.

Concept Introduction:

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation (${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

 The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

Answer to Problem 6.102QP

${\text{CaC}}_{\text{2(s)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{Ca(OH)}}_{\text{2(s)}}{\text{+C}}_{\text{2}}{\text{H}}_{\text{(g)}}$

Explanation of Solution

Calcium carbide upon reaction with Water gives Calcium hydroxide and Acetylene. The chemical reaction can be given as,

${\text{CaC}}_{\text{2(s)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{Ca(OH)}}_{\text{2(s)}}{\text{+C}}_{\text{2}}{\text{H}}_{\text{(g)}}$

(b)

Interpretation Introduction

Interpretation:

The equation for the formation of Acetylene and the maximum amount of heat from combustion of Acetylene has to be calculated.

Concept Introduction:

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation (${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

 The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

Answer to Problem 6.102QP

The maximum amount of heat obtained is $\text{1}{\text{.51×10}}^{\text{6}}\text{J}$

Explanation of Solution

The combustion reaction of Acetylene can be given as,

${\text{2C}}_{\text{2}}{\text{H}}_{\text{2(g)}}{\text{+5O}}_{\text{2(g)}}\to {\text{4CO}}_{\text{2(g)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Standard enthalpy of formation of ${\text{CO}}_2\text{=-393}\text{.5kJ}{\text{mol}}^{-1}$

Standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O=-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{O}}_{\text{2}}\text{=0}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{C}}_2{\text{H}}_5\text{=-226}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}$

$\begin{array}{l}{\text{ΔH°}}_{\text{reaction}}\text{=}\left[{\text{4ΔH°}}_{\text{f}}{\text{(CO}}_{\text{2}}{\text{)+2ΔH°}}_{\text{f}}{\text{(H}}_{\text{2}}\text{O)}\right]\text{-}\left[{\text{2ΔH°}}_{\text{f}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)+5ΔH°}}_{\text{f}}{\text{(O}}_{\text{2}}\text{)}\right]\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=}\left[\text{(4)(-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}\text{)+(2)(-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\right]\text{-}\left[\text{(2)(226}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}\text{)+(5)(0)}\right]\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=-2599}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Enthalpy of reaction= $\text{-2599}\text{kJ}{\text{mol}}^{\text{-1}}$

To calculate the moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$

Moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$= $\text{74}\text{.6}\text{g}{\text{CaC}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}{\text{CaC}}_{\text{2}}}{\text{64}\text{.10}\text{g}{\text{CaC}}_{\text{2}}}\text{×}\frac{\text{1}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}}{\text{1}\text{mol}{\text{CaC}}_{\text{2}}}$

Moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$=$\text{1}\text{.16}\text{mol}$

To calculate the amount in heat for $\text{1}\text{.16}\text{mol}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$

Amount of heat = $\text{1}\text{.16}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\text{×}\frac{\text{2599}\text{kJ}}{\text{2}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}}$

Amount of heat =$\text{1}{\text{.51×10}}^{\text{3}}\text{kJ=1}{\text{.51×10}}^{\text{6}}\text{J}$