Question

Calculate the amount of energy in the form of heat that is produced when a volume of 3.43 L of SO2(g) is converted to 3.43 L of SO3(g) according to this process at a constant pressure and temperature of 1.00 bar and 25.0 °C . Assume ideal gas behavior.

Answer 1

Explanation:

The reaction equation will be as follows.

              $2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)$

And, the standard enthalpies of formation for $SO_{2}(g)$ and $SO_{3}(g)$ are as follows.

  $\Delta H_{SO_{2}(g)}$ = -296.8 kJ/mol

 $\Delta H_{SO_{3}(g)}$ = -395.7 kJ/mol

Now, according to the ideal gas equation PV = nRT.

    Number of moles of 3.43 L of $SO_{2}$ = $\frac{PV}{RT}$

             = $\frac{1.00 atm \times 3.43 L}{0.0821 L atm/mol K \times 298}$               (as 1 bar = 1 atm (approximately))

             = 0.140

Now, no. of moles of 3.43 L of $SO_{3}$ are also 0.140.

Hence, calculate the enthalpies of formation for 0.140 moles of $SO_{2}(g)$ and $SO_{3}(g)$ as follows.

        $(n \times \Delta H_{SO_{2}(g)})$ and $(n \times SO_{3}(g))$

    $0.140 mol \times (-296.8 kJ/mol)$ and $0.140 \times (-395.7 kJ/mol)$ respectively.

or,                  $0.140 \times [-296.8 kJ/mol - (-395.7 kJ/mol)]$

                          = $0.140 mol \times 98.9 kJ/mol$

                         = 13.84 kJ

Therefore, we can conclude that the amount of energy in the form of heat that is produced is 13.84 kJ.

Answer 2

The amount of heat energy produced in the chemical reaction at 1 bar pressure and 25 degree Celsius temperature is 13.48 kJ.

The amount of energy produced in the reaction is given as the difference in the energy of the reactant and product molecules.

Computation for the amount of energy produced

The given balanced chemical equation is:

$\rm 2\;SO_2\;+\;O_2\;\to\;2\;SO_3$

The molar enthalpy of the reactant and product is:

$\rm \Delta H \;SO_2=-296.8\;kJ/mol\\\Delta H\;SO_3=-395.7\;kJ/mol$

The volume  of $\rm SO_2$ reacted is 3.43 L. The pressure of 1 bar and 25 degrees Celsius is the STP. The moles of $\rm SO_2$ at STP is given as:

$\rm 22.4\;L=1\;mol\\\\3.43\;L=\dfrac{1}{22.4}\;\times\;3.43\;mol\\\\ 3.43\;L=0.15\;mol$

The moles of $\rm SO_2$ reacted is 0.15 mol.

The moles of $\rm SO_3$ produced is given as:

$\rm 22.4\;L=1\;mol\\\\3.43\;L=\dfrac{1}{22.4} \;\times\;3.43\;mol\\\\3.43\;L=0.15\;mol$

The moles of $\rm SO_3$ produced is 0.15 mol.

The amount of energy ($\Delta H$) for the reaction is given as:

$\rm \Delta \textit H=moles\;SO_2\;\times\;\Delta \textit H_{SO_2}-moles\;SO_3\;\times\;\Delta \textitH_{SO_3}\\\\\Delta \textit H=0.15\;\times\;-298.8\;-\;0.15\;\times\;-395.7\;kJ/mol\\\\\Delta \textit H=13.48\;kJ$

The energy of the reaction produced is 13.48kJ.

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