Chapter 10, Problem 64A

Interpretation Introduction

Interpretation: The net reaction and heat of the reaction needs to be calculated.

Concept introduction: The energy flow between two objects due to difference of temperature is termed as heat. In exothermic process, heat is released from system to surroundings and in endothermic process absorption of heat by the system from surroundings occurs.

Answer to Problem 64A

Net reaction is,

  $\text{2F}+\text{C}\to \text{2A}+2\text{B}+\text{2D}$

And overall heat of the reaction is $\text{93}\text{.94}\text{kJ/mol}$ .

Explanation of Solution

The equations are as follows:

  $3\text{A}+6\text{B}\to \text{3D}\Delta H=-403\text{kJ/mol}$...........(1)

  $\text{E}+2\text{F}\to \text{A}\Delta H=-105.2\text{kJ}\text{mol}$...........(2)

  $\text{C}\to \text{E}+\text{3D}\Delta H=64.8\text{kJ/mol}$...........(3)

Reverse equation (1) and multiply if by $\frac{1}{6}$

  $\begin{array}{l}\frac{\text{3}}{6}\text{D}\to \frac{\text{3}}{6}\text{A}+\frac{6}{6}\text{B}\Delta H=\frac{403}{6}\text{kJ/mol}\\ \frac{1}{2}\text{D}\to \frac{1}{2}\text{A}+\text{B}\Delta H=67.17\text{kJ/mol}\end{array}$...........(4)

Now divide equation (2) by 2.

Equation (2) becomes,

  $\frac{1}{2}\text{E}+\frac{2}{2}\text{F}\to \frac{1}{2}\text{A}\Delta H=\frac{-105.2}{2}\text{kJ/mol}=52.6\text{kJ/mol}$...........(5)

Now divide equation (3) by 2. Equation (3) becomes,

  $\frac{1}{2}\text{C}\to \frac{1}{2}\text{E}+\frac{\text{3}}{2}\text{D}\Delta H=\frac{64.8}{2}\text{kJ/mol}=32.4\text{kJ/mol}$...........(6)

Now add equation (4), (5) and (6).

  $\begin{array}{c}\frac{1}{2}\text{D}+\cancel{\frac{1}{2}\text{E}}+\text{F}+\frac{1}{2}\text{C}\to \frac{1}{2}\text{A}+\text{B}+\frac{1}{2}\text{A}+\cancel{\frac{1}{2}\text{E}}+\frac{\text{3}}{2}\text{D}\Delta H=\left(\begin{array}{c}67.17\text{kJ/mol}-52.6\text{kJ}\\ \text{/mol}+32.4\text{kJ/moL}\end{array}\right)\\ \text{F}+\frac{1}{2}\text{C}\to \text{A}+\text{B}+\text{D }\Delta H=46.97\text{kJ/mol}\\ \text{2F}+\text{C}\to \text{2A}+2\text{B}+\text{2D}\Delta H=\left(\begin{array}{l}2\times 46.97\text{kJ/mol}=\\ \text{93}\text{.94}\text{kJ/mol}\end{array}\right)\end{array}$

So, net reaction is,

  $\text{2F}+\text{C}\to \text{2A}+2\text{B}+\text{2D}$

And overall heat of the reaction is $\text{93}\text{.94}\text{kJ/mol}$ .

Conclusion

Net reaction is, $\text{2F}+\text{C}\to \text{2A}+2\text{B}+\text{2D}$ and overall heat of the reaction is $\text{93}\text{.94}\text{kJ/mol}$ .