Chapter 10, Problem 62A

Interpretation Introduction

Interpretation: The net equation and overall heat of the reaction needs to be determined, if the given changes will apply on the equation:

  $\begin{array}{l}\text{3A + 6B }\stackrel{}{\to }\text{3D ΔH = -403 kJ/mol}\mathrm{...}\text{(1)}\\ \text{E + 2F }\stackrel{}{\to }\text{A ΔH = -105}\text{.2 kJ/mol}\mathrm{....}\text{(2)}\\ \text{C }\stackrel{}{\to }\text{E + 3 D ΔH = +64}\text{.8 kJ/mol}\mathrm{.....}\text{(3)}\end{array}$

Concept Introduction: The change in enthalpy of reaction is always calculated for the balanced chemical equation and can be used to calculate the energy change for the given moles or masses of reactant or product molecules.

Hess’s Las states that if a reaction completed in one step or multiple steps, the total enthalpy change for the reaction will always remain same.

Answer to Problem 62A

  $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.C+F\stackrel{}{\to }B+A+D\Delta H=\text{-87}\text{.4kJ/mol}$

Explanation of Solution

Given information:

  $\begin{array}{l}\text{3A + 6B }\stackrel{}{\to }\text{3D ΔH = -403 kJ/mol}\mathrm{...}\text{(1)}\\ \text{E + 2F }\stackrel{}{\to }\text{A ΔH = -105}\text{.2 kJ/mol}\mathrm{....}\text{(2)}\\ \text{C }\stackrel{}{\to }\text{E + 3 D ΔH = +64}\text{.8 kJ/mol}\mathrm{.....}\text{(3)}\end{array}$

• Reverse equation (1) and multiply by 1/6
• Equation (2) and (3) divided by 2
• Add all the adjusted equations

Adjust the equation as per given instruction:

  $\begin{array}{l}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\times \text{3D }\stackrel{}{\to }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\times \text{3A + }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\times \text{6B ΔH = }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\times \text{-403 kJ/mol}\mathrm{...}\text{(1)}\\ =\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\text{D }\stackrel{}{\to }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\text{A + B ΔH = }-67.2\text{ kJ/mol }\end{array}$

  $\begin{array}{l}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \text{E +}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \text{ 2F }\stackrel{}{\to }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \text{A ΔH = }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \text{ -105}\text{.2 kJ/mol}\mathrm{....}\text{(2)}\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \text{C }\stackrel{}{\to }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \text{E + }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \text{ 3 D ΔH = }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \text{ +64}\text{.8 kJ/mol}\mathrm{.....}\text{(3)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\text{E + F }\stackrel{}{\to }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \text{A ΔH = }-52.6\text{kJ/mol}\mathrm{....}\text{(2)}\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\text{C }\stackrel{}{\to }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\text{E + }\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\text{D ΔH = }32.4\text{ kJ/mol}\mathrm{.....}\text{(3)}\end{array}$

  $\begin{array}{l}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.D\stackrel{}{\to }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.A+B\Delta H=-67.2kJ/mol\mathrm{...}(1)\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.E+F\stackrel{}{\to }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times A\Delta H=-52.6kJ/mol\mathrm{....}(2)\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.C\stackrel{}{\to }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.E+\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.D\Delta H=32.4kJ/mol\mathrm{.....}(3)\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.C+F\stackrel{}{\to }B+A+D\Delta H=\text{-87}\text{.4kJ/mol}\end{array}$

Conclusion

Thus,

  $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.C+F\stackrel{}{\to }B+A+D\Delta H=\text{-87}\text{.4kJ/mol}$