Chapter 9, Problem 131CP

When I mole of benzene is vaporized at a constant pressure of 1.00 atm and at its boiling point of 353.0 K, 30.79 kJ of energy (heat) is absorbed and the volume change is +28.90 L. What are ∆E and ∆H for this process?

Interpretation Introduction

Interpretation:

          Change in enthalpy $\text{ΔH}$ and change in internal energy $\text{ΔE}$ for the given process has to be determined.

Concept introduction:

Internal energy of a system is total energy present in the system. In simple words, it is the sum of kinetic and potential energy of the particles in the system. According to First law of Thermodynamics, Energy of a system is conserved. It is only transferred from one state to another that is from system to surroundings and vice versa. So $\text{ΔE}$ can be represented as,

                                            ${\text{ΔE}}_{\text{universe}}\text{=}{\text{ΔE}}_{\text{sys}}\text{+}{\text{ΔE}}_{\text{surroundings}}$

Further, $\text{ΔE}$ is also equivalent to sum of either heat gained or lost and either work done on the system or by the system.

                                            $\text{ΔE}\text{=}\text{q}\text{+}\text{w}$

                                  $\begin{array}{l}\text{where}\text{ΔE}\text{=}\text{change}\text{in}\text{internal}\text{energy}\\ \text{q}\text{=}\text{quantity}\text{of}\text{heat}\text{gained}\text{or}\text{heat}\text{lost}\\ \text{w}\text{=}\text{work}\text{done}\end{array}$

Enthalpy is heat content of the system. The value of enthalpy does not depend on the path of a reaction but depend on state of the system. It has a unique value for each state of the system. Thus, enthalpy is a state function. Enthalpy is represented as,

                                  $\begin{array}{l}\text{ H}\text{=}\text{E+}\text{PV}\\ \text{where}\text{H}\text{=}\text{Enthalpy}\\ \text{E}\text{=}\text{Internal}\text{energy}\\ \text{P}\text{=}\text{Pressure}\\ \text{V}\text{=}\text{Volume}\end{array}$

Enthalpy change, denoted by $\text{ΔH}$, refers to heat evolved or absorbed during a reaction. If heat is evolved in the reaction that is exothermic reaction $\text{ΔH}$ has negative value. For an endothermic reaction, $\text{ΔH}$ has positive value. $\text{ΔH}$ can be represented as,

                                       $\text{ΔH}\text{=}\text{ΔE}\text{+}\text{PΔV}$

                             $\begin{array}{l}\text{where,}\text{ΔH}\text{=}\text{Change}\text{in}\text{enthalpy}\\ \text{ΔE}\text{=}\text{Change}\text{in}\text{Internal}\text{energy}\\ \text{ΔV}\text{=}\text{Change}\text{in}\text{volume}\\ \text{P}\text{=}\text{Pressure}\\ \\ \end{array}$

Further, $\text{ΔH}$ also refers to heat absorbed or lost in the reaction at constant pressure. Thus $\text{ΔH}$ can have the following expression also –

                                           $\begin{array}{l}\text{ΔH}\text{=}{\text{q}}_{\text{P}}\\ \text{where}{\text{q}}_{\text{P}}\text{is}\text{heat}\text{gained}\text{or}\text{lost}\text{at}\text{constant}\text{pressure}\text{.}\\ \end{array}$

Answer to Problem 131CP

Answer

Change in enthalpy $\text{ΔH}$ is 30.79 kJ.

Change in internal energy $\text{ΔE}$ is 27.86 kJ.

Explanation of Solution

Explanation

Determine $\text{ΔH}$ of the reaction.

$\begin{array}{l}\text{given}\text{data}\text{:}\text{heat}\text{absorbed}\text{,}\text{q}\text{=}\text{30}\text{.79}\text{kJ}\\ \end{array}$

$\text{ΔH}$ of the reaction corresponds to heat absorbed in the reaction. Therefore,

                                $\text{ΔH}\text{=}{\text{q}}_{\text{p}}=\text{30}\text{.79}\text{kJ}$

Enthalpy of a system corresponds to amount of heat present in the system. Change in enthalpy is change in the heat content of the system during the course of reaction. That is, heat content of the system either decreases or increases after the completion of the reaction. Endothermic reaction proceeds by absorption of heat that at the end of the reaction, heat content of the system is increased. So the amount of heat absorbed or evolved in the reaction corresponds to the enthalpy change of a reaction.

Calculate the work done, ‘w’.

$\begin{array}{l}\text{given}\text{data}\text{:}\text{Pressure,}\text{P}\text{=}\text{1}\text{atm}\\ \text{volume}\text{change,}\text{ΔV}\text{=}\text{28}\text{.90}\text{L}\end{array}$

           $\begin{array}{l}\text{We}\text{know,}\text{ΔH}\text{=}\text{ΔE}\text{+}\text{PΔV}\mathrm{......}\left(\text{1}\right)\\ \text{ΔH}\text{=}\text{q}\mathrm{......}\left(\text{2}\right)\\ \text{ΔE}\text{=}\text{q}\text{+}\text{w}\mathrm{......}\left(\text{3}\right)\\ \text{Subtracting}\left(\text{2}\right)\text{from}\left(\text{3}\right)\text{,}\\ \text{ΔE}-\text{ΔH}\text{=}\text{q}\text{+}\text{w}\text{-}\text{q}\\ \text{From }\left(\text{1}\right)\text{,}\text{ΔE}-\text{ΔH}\text{=}-\text{PΔV}\\ \text{Therefore,}\text{w}\text{=}-\text{PΔV}\\ \text{ w}\text{=}-\text{1}\text{.00}\text{atm}\text{×}\text{29}\text{.80}\text{L}\text{=}-\text{29}\text{.80}\text{L}\text{atm}\\ \text{1}\text{L}\text{atm}\text{=}\text{101}\text{.33}\text{J}\\ \text{Therefore,}\text{w}=-2\text{9}\text{.80}\text{×}\text{101}\text{.33}\text{J}\\ \text{=}-\text{2928}\text{.4}\text{J}\\ \\ \end{array}$

Using the three equations, $\text{ΔH}\text{=}\text{ΔE}\text{+}\text{PΔV}$, $\text{ΔH}\text{=}\text{q}$ and $\text{ΔE}\text{=}\text{q}\text{+}\text{w}$ work done is calculated by substituting the relevant values.

Calculate $\text{ΔE}$ of the reaction.

           $\begin{array}{l}\text{known}\text{data:}\text{q}\text{=}\text{30}\text{.79}\text{kJ}\\ \text{w}\text{=}-\text{2928}\text{.4}\text{J}\text{=}-2.93\text{kJ}\\ \Delta \text{E}\text{=}\text{q}\text{+}\text{w}\\ \text{Therefore,}\Delta \text{E}\text{=}\text{30}\text{.29kJ}-\text{2}\text{.93}\text{kJ}=27.86\text{kJ}\end{array}$

‘q’and work done ‘w’ values are calculated in the previous steps. By substituting these values in the equation, $\Delta \text{E}\text{=}\text{q}\text{+}\text{w}$, change in internal energy $\text{ΔE}$ is obtained.

Conclusion

Conclusion

Change in enthalpy $\text{ΔH}$ and Change in internal energy $\text{ΔE}$ are calculated for the given reaction.