- Identify the balanced chemical equation for the combustion of glucose: \(C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)\).
- Use the standard enthalpies of formation for the products (\(CO_2\) and \(H_2O\)) and reactants (glucose and \(O_2\)).
- Apply the formula \(\Delta H_{\text{rxn}}^\circ = \sum n_p \Delta H_f^\circ(\text{products}) - \sum n_r \Delta H_f^\circ(\text{reactants})\) to find the energy change.
- Calculate the result: \(\Delta H_{\text{rxn}}^\circ = \boxed{-2802.5} \text{ kJ/mol}\).
Explanation
- Problem Analysis and Formula Setup
To calculate the standard enthalpy of reaction (\(\Delta H_{\text{rxn}}^\circ\)) for the combustion of glucose, we use the standard enthalpies of formation (\(\Delta H_f^\circ\)) for all participating substances. The chemical equation is given by:
\[C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)\]
The formula for the enthalpy of reaction is:
\[\Delta H_{\text{rxn}}^\circ = \sum n_p \Delta H_f^\circ(\text{products}) - \sum n_r \Delta H_f^\circ(\text{reactants})\]
where \(n_p\) and \(n_r\) are the stoichiometric coefficients of the products and reactants, respectively.
- Identifying Enthalpy of Formation Values
We identify the standard enthalpies of formation for each compound involved in the reaction:
- For glucose (\(C_6H_{12}O_6(s)\)): \(\Delta H_f^\circ = -1273.3 \text{ kJ/mol}\)
- For carbon dioxide (\(CO_2(g)\)): \(\Delta H_f^\circ = -393.5 \text{ kJ/mol}\)
- For liquid water (\(H_2O(l)\)): \(\Delta H_f^\circ = -285.8 \text{ kJ/mol}\)
- For oxygen gas (\(O_2(g)\)): \(\Delta H_f^\circ = 0 \text{ kJ/mol}\) (as it is an element in its standard state)
- Calculations and Final Result
Now we apply the coefficients from the balanced chemical equation to calculate the total enthalpy of the products and the reactants:
- Total enthalpy of products: \(6 \cdot (-393.5 \text{ kJ/mol}) + 6 \cdot (-285.8 \text{ kJ/mol}) = -2361.0 - 1714.8 = -4075.8 \text{ kJ/mol}\)
- Total enthalpy of reactants: \(1 \cdot (-1273.3 \text{ kJ/mol}) + 6 \cdot (0 \text{ kJ/mol}) = -1273.3 \text{ kJ/mol}\)
Finally, calculating the difference:
\[\Delta H_{\text{rxn}}^\circ = (-4075.8 \text{ kJ/mol}) - (-1273.3 \text{ kJ/mol}) = -4075.8 + 1273.3 = -2802.5 \text{ kJ/mol}\]
Examples
Calculating the enthalpy of reaction is vital in fields like nutrition and biochemistry. For instance, by determining the \(\Delta H_{\text{rxn}}^\circ\) of glucose combustion, scientists can quantify the energy released when our bodies metabolize sugar. This provides the fundamental basis for understanding caloric content in food, helping nutritionists design healthy diets and athletes optimize their fuel intake for peak performance.