Questions: Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2 C2H6(g) + 7 O2(g) -> 4 CO2(g) + 6 H2O(g) ΔHrxn° = □ kJ

Solution Steps

To calculate the standard enthalpy change (\(\Delta H^\circ\)) for the reaction, we need the standard heats of formation (\(\Delta H_f^\circ\)) for each compound involved. The standard heat of formation is the change in enthalpy when one mole of a compound is formed from its elements in their standard states.

• \(\Delta H_f^\circ\) for \(\mathrm{C}_2\mathrm{H}_6(g)\)
• \(\Delta H_f^\circ\) for \(\mathrm{O}_2(g)\) is 0 kJ/mol (since it is an element in its standard state)
• \(\Delta H_f^\circ\) for \(\mathrm{CO}_2(g)\)
• \(\Delta H_f^\circ\) for \(\mathrm{H}_2\mathrm{O}(g)\)

The standard enthalpy change for the reaction is calculated using the formula:

\[ \Delta H_r^\circ = \sum (\Delta H_f^\circ \text{ of products}) - \sum (\Delta H_f^\circ \text{ of reactants}) \]

Assuming the standard heats of formation are as follows (values are typically found in a reference table):

• \(\Delta H_f^\circ\) for \(\mathrm{C}_2\mathrm{H}_6(g) = -84.7 \, \text{kJ/mol}\)
• \(\Delta H_f^\circ\) for \(\mathrm{CO}_2(g) = -393.5 \, \text{kJ/mol}\)
• \(\Delta H_f^\circ\) for \(\mathrm{H}_2\mathrm{O}(g) = -241.8 \, \text{kJ/mol}\)

Substitute these values into the formula:

\[ \Delta H_r^\circ = [4 \times (-393.5) + 6 \times (-241.8)] - [2 \times (-84.7) + 7 \times 0] \]

Calculate the enthalpy change for the products and reactants:

• Products: \(4 \times (-393.5) + 6 \times (-241.8) = -1574.0 - 1450.8 = -3024.8 \, \text{kJ}\)
• Reactants: \(2 \times (-84.7) = -169.4 \, \text{kJ}\)

Now, calculate \(\Delta H_r^\circ\):

\[ \Delta H_r^\circ = -3024.8 - (-169.4) = -3024.8 + 169.4 = -2855.4 \, \text{kJ} \]

Final Answer