Question

The enthalpy of reaction changes somewhat with temperature. Suppose we wish to calculate \(\Delta H\) for a reaction at a temperature \(T\) that is different from \(298\;{\rm{K}}\). To do this, we can replace the direct reaction at \(T\) with a three-step process. In the first step, the temperature of the reactants is changed from \(T\)to \(298\;{\rm{K}}\).\(\Delta H\) for this step can be calculated from the molar heat capacities of the reactants, which are assumed to be independent of temperature. In the second step, the reaction is conducted at \(298\;{\rm{K}}\) with an enthalpy change \(\Delta {H^o}\). In the third step, the temperature of the products is changed from \(298\;{\rm{K}}\) to \(T\). The sum of these three enthalpy changes is \(\Delta H\) for the reaction at temperature \(T\).

An important process contributing to air pollution is the following chemical reaction \({\rm{S}}{{\rm{O}}_2}(g) + \frac{1}{2}{{\rm{O}}_2}(g) \to {\rm{S}}{{\rm{O}}_3}(g)\). For \({\rm{S}}{{\rm{O}}_2}(g)\), the heat capacity \({C_{\rm{p}}}\) is\(39.9\), for \({{\rm{O}}_2}(g)\) it is \(29.4\), and for \({\rm{S}}{{\rm{O}}_3}(g)\) it is \(50.7{\rm{J}}{{\rm{K}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\). Calculate \(\Delta H\) for the preceding reaction at \(500\;{\rm{K}}\), using the enthalpies of formation at \(298.15\;{\rm{K}}\) from Appendix D.

Short answer

The enthalpies of formation at \(298.15\;{\rm{K}}\) is \(\Delta H = - 99.69\;{\rm{kJ}}\).

Step-by-step solution

Given data

The given reaction is\({\rm{S}}{{\rm{O}}_2}(g) + 1/2{{\rm{O}}_2}(g) \to {\rm{S}}{{\rm{O}}_3}(g)\).

The given heat capacities:

\(\begin{aligned}{}{C_p}\left( {{\rm{S}}{{\rm{O}}_2}} \right) = 39.9\\{C_p}\left( {{{\rm{O}}_2}} \right) = 29.4\\{C_p}\left( {{\rm{S}}{{\rm{O}}_3}} \right) = 50.7\end{aligned}\)

The values of the standard enthalpies of formation:

\(\begin{aligned}{}\Delta H\left( {{\rm{S}}{{\rm{O}}_3}} \right) = - 395.72k{\rm{Jmo}}{l^{ - 1}}\\\Delta H\left( {{\rm{S}}{{\rm{O}}_2}} \right) = - 296.83{\rm{kJmo}}{{\rm{l}}^{ - 1}}\\\Delta H\left( {{{\rm{O}}_2}} \right) = 0{\rm{kJmo}}{{\rm{l}}^{ - 1}}\end{aligned}\)

The all given steps are as follows:

(a) Temperature of the reactants is changed from T to\(298\;{\rm{K}}.\Delta H\)for this step can be calculated from the molar heat capacities of the reactants, which are assumed to be independent of temperature.

(b) The reaction is conducted at\(298\;{\rm{K}}\)with an enthalpy change,\(\Delta {H^o}\).

(c) The temperature of the products is changed from\(298\;{\rm{K}}\)to T. The sum of these three enthalpy changes is\(\Delta H\)for the reaction at temperature\(T\).

Concept of change in enthalpy of reaction

Enthalpy change is the standard enthalpy of formation, which has been determined for a vast number of substances. In any general chemical reaction, the reactants undergo chemical changes and combine to give products. It can be represented by the following equation:

\({\mathop{\rm Re}\nolimits} ac\tan t \to \Pr oduct\)

For any such reaction, the change in enthalpy is represented as\(\Delta H\)and is termed as the reaction enthalpy.

The reaction enthalpy is calculated as:

Mathematically,\(\Delta H = \Delta {H_1} + \Delta {H_2}\).

Where, \(\Delta {H_1}\) is change in enthalpy of reactant and \(\Delta {H_2}\) is change in enthalpy of product.

Calculation of enthalpy of reaction for first step

The enthalpy change for this process is calculated by the formula:

\(\begin{aligned}{}\Delta H = \Delta H\left( {{\rm{S}}{{\rm{O}}_2}} \right) + \Delta H\left( {{{\rm{O}}_2}} \right)\\\Delta H\left( {{\rm{S}}{{\rm{O}}_2}} \right) = n\left( {{\rm{S}}{{\rm{O}}_2}} \right) \times {C_p}\left( {{\rm{S}}{{\rm{O}}_2}} \right) \times \Delta T\\\Delta H\left( {{{\rm{O}}_2}} \right) = n\left( {{{\rm{O}}_2}} \right) \times {C_p}\left( {{{\rm{O}}_2}} \right) \times \Delta T\end{aligned}\)

The final formula can be written as:

\(\Delta H = \Delta T \times \left( {n\left( {{\rm{S}}{{\rm{O}}_2}} \right) \times {C_p}\left( {{\rm{S}}{{\rm{O}}_2}} \right) + n\left( {{{\rm{O}}_2}} \right) \times {C_p}\left( {{{\rm{O}}_2}} \right)} \right).\)

Put the value of all given data in above equation.

\(\begin{aligned}{}\Delta H = (298 - 500) \times \left( {1 \times 39.9 + \frac{1}{2} \times 29.4} \right)\\\Delta {H_1} = - 11.03\;{\rm{kJ}}\end{aligned}\)

Calculation of enthalpy of reaction for second step

The enthalpy change for this process is calculated by the formula.

\(\Delta H = n\left( {{\rm{S}}{{\rm{O}}_3}} \right)\Delta H\left( {{\rm{S}}{{\rm{O}}_3}} \right) - \left( {n\left( {{\rm{S}}{{\rm{O}}_2}} \right)\Delta H\left( {{\rm{S}}{{\rm{O}}_2}} \right) + n\left( {{{\rm{O}}_2}} \right)\Delta H\left( {{{\rm{O}}_2}} \right)} \right)\)

Put the value of all given data in above equation.

\(\begin{aligned}{}\Delta H = 1 \times ( - 395.72) - (1 \times ( - 296.83) + 0)\\\Delta {H_2} = - 98.9\;{\rm{kJ}}\end{aligned}\)

Calculation of enthalpy of reaction for third step

The enthalpy change for this process is the enthalpy change of the product.

\(\begin{aligned}{}\Delta H = \Delta H\left( {{\rm{S}}{{\rm{O}}_3}} \right)\\\Delta H\left( {{\rm{S}}{{\rm{O}}_3}} \right) = n\left( {{\rm{S}}{{\rm{O}}_3}} \right) \times {C_p}\left( {{\rm{S}}{{\rm{O}}_2}} \right) \times \Delta T\end{aligned}\)

Put the value of all given data in above equation.

\(\begin{aligned}{}\Delta H = 1 \times 50.7 \times (500 - 298)\\\Delta {H_3} = 10.24\;{\rm{kJ}}\end{aligned}\)

Calculation of final enthalpy of reaction

Calculate the final enthalpy change at \(500\;{\rm{K}}\).

\(\begin{aligned}{}\Delta H &= \Delta {H_1} + \Delta {H_2} + \Delta {H_3}\\\Delta H &= - 11.02 - 98.9 + 10.24\\\Delta H &= - 99.69\;{\rm{kJ}}.\end{aligned}\)