Consider again the endothermic process of photosynthesis:

\( \mathrm { 6CO_2(g) + 6H_2O(l) \rightarrow C_6H_{12}O_6(s) + O_2(g) } \)

\( \mathrm { ~~~~~~\Delta_rH = +2802.5 kJ } \)

Would the same amount of energy be absorbed if only 1 mole of carbon dioxide reacted? To calculate this, divide the reaction enthalpy by the number of moles of the stipulated substance:

\( \mathrm { \Delta_rH_m~for~CO_2 = \dfrac{\Delta_rH}{n} = \dfrac{+2802.5~kJ}{6~moles} } \)

\( \mathrm { \Delta_rH_m~for~CO_2 = +467.08 \frac{kJ}{mol} } \)

\( \mathrm { \Delta_rH_m } \) molar enthalpy of reaction \( \mathrm { \Delta_rH } \) enthalpy of reaction \( \mathrm { n } \) number of moles

The molar enthalpy of reaction refers to the enthalpy change per mole of the specified chemical undergoing change in the system. In this case, the specified substance is carbon dioxide. Because 467.08 kJ of energy is absorbed when one mole of carbon dioxide is consumed during photosynthesis, we refer to this amount of energy as the molar enthalpy of reaction for carbon dioxide. This quantity has the units kJ/mol and is symbolized as Δ_rH_m. (Notice the subscript m .) If the reaction occurs at SATP, the standard state symbol is also included in the molar enthalpy symbol (Δ_rH_m°)

Learning Tip When communicating a molar enthalpy of reaction, the balanced reaction and the substance in question must be stipulated. In this case, the balanced reaction is 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g) and the substance in question is the carbon dioxide.

Now, consider the exothermic reaction known as cellular respiration.

What is the molar enthalpy of reaction when one mole of carbon dioxide is produced during the process of cellular respiration?

\( \mathrm { \Delta_rH_m~for~CO_2 = \dfrac{\Delta_rH}{n} = \dfrac{-2802.5~kJ}{6~moles} } \)

\( \mathrm { \Delta_rH_m~for~CO_2 = -467.08 \frac{kJ}{mol} } \)

Molar enthalpies can be very useful quantities. In Lesson 4, you will learn how molar enthalpies can be used to compare fuels.

At the beginning of this section, you saw how a reaction enthalpy can be used to calculate a molar enthalpy of reaction. The reverse is also true. That is, molar enthalpies of reactions can be used to calculate the enthalpy of reaction, Δ_rH.

To do this, a balanced chemical reaction is needed as well as the molar enthalpy of reaction for one of the substances in the reaction.

For example, consider the reaction between sulfur dioxide and oxygen to produce sulfur trioxide. Suppose you are told that the standard molar enthalpy of combustion of sulfur dioxide is 98.9 kJ/mol. What would be the Δ_rH for this reaction? Could you handle a question like that?

The first step is to write a balanced reaction.

\( \mathrm { 2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)   } \)

Next, determine a value for Δ_rH. Because we know the standard molar enthalpy for one of the substances involved in the reaction (sulfur dioxide), we can use this information to arrive at a Δ_rH for the balanced reaction as written above. Because this is a combustion reaction, replace the subscript r with the subscript c.

\( \mathrm { \Delta_cH^{\circ} = n\Delta_r H_m^{\circ} } \)

\( \mathrm { \Delta_cH^{\circ} = 2~moles \times \dfrac{-98.9~kJ}{1 mol} } \)

\( \mathrm { \Delta_cH^{\circ} = -197.8~kJ } \)

(Notice that this enthalpy is not a molar amount; therefore, the units are kJ, not kJ/mol)

\( \mathrm { 2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)     } \)

\( \mathrm { \Delta_cH^{\circ} = -197.8~kJ } \)

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 Read pages 495 and 496 in the textbook.

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