Chapter 17, Problem 94A

Interpretation Introduction

Interpretation: The heat released during a given chemical reaction is to be calculated.

Concept Introduction: The standard heat of the reaction is equal to the sum of all the standard heat of the formation of the products minus the sum of all the standard heat of the formation of the reactants.

The standard enthalpy of formation is the amount of energy released or consumed when one mole of a substance is formed under standard conditions from its pure elements. The standard enthalpy of formation of a pure substance in its elemental form or pure state is zero.

Answer to Problem 94A

The heat released during the reaction ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6{\text{(s)+6O}}_2\text{(g)}\to 6{\text{CO}}_2{\text{(g)+6H}}_{\text{2}}\text{O(l)}$ is $\text{613}\text{.56 kJ}$

Explanation of Solution

The chemical reaction for which enthalpy change is to be calculated

  ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6{\text{(s)+6O}}_2\text{(g)}\to 6{\text{CO}}_2{\text{(g)+6H}}_{\text{2}}\text{O(l)}$

To calculate the heat released during the reaction, we need to know the heat of the formation of all reactants and products.

  ${\text{ΔH}}^{\text{0}}{}_{\text{rxn}}{\text{= [H}}^{\text{0}}{}_{\text{f}}{\text{(products)-H}}^{\text{0}}{}_{\text{f}}\text{(reactants)]}$

  ${\text{ ΔH}}^{\text{0}}{}_{\text{rxn}}{\text{= [6×H}}^{\text{0}}{}_{\text{f}}{\text{(CO}}_{\text{2}}{\text{)+6×H}}^{\text{0}}{}_{\text{f}}{\text{(H}}_{\text{2}}{\text{O)]-[6×H}}^{\text{0}}{}_{\text{f}}{\text{(O}}_{\text{2}}{\text{)+1×H}}^{\text{0}}{}_{\text{f}}{\text{(C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{)]}$

Given : ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{)= -1260 kJ/mol}$

  ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(O}}_{\text{2}}\text{)= 0 kJ/mol}$ , ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(CO}}_{\text{2}}\text{)= -393}\text{.5 kJ/mol}$ and ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(H}}_{\text{2}}\text{O)= 285}\text{.76 kJ/mol}$

Putting in the values, we get

  $\begin{array}{l}{\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [6mol×(-393}\text{.5kJ/mol)+6mol×(285}\text{.76kJ/mol)]-[6mol×(0kJ/mol)+1mol×-1260kJ/mol]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [-2361kJ+1714}\text{.56kJ]-[0kJ+(-1260kJ)]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [-646}\text{.44kJ]-[(-1260kJ)]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= 613}\text{.56kJ}\end{array}$