Question

Using the table of standard formation enthalpies, calculate the reaction enthalpy for the following reaction under standard conditions: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l). Round your answer to the nearest kJ.

a) -1362 kJ
b) -953 kJ
c) -724 kJ
d) -489 kJ

Answer 1

To calculate the reaction enthalpy for the given reaction under standard conditions, we use the standard enthalpies of formation. The calculated value is approximately -1574 kJ.

Step-by-step explanation:

To calculate the reaction enthalpy for the given reaction under standard conditions, we need to use the standard enthalpies of formation for each compound involved. According to the table of standard formation enthalpies, the standard enthalpy of formation for CO₂(g) is -393.5 kJ/mol.

Using this information, we can calculate the reaction enthalpy as follows:

Reactants: 2 mol C₂H₆(g) + 7 mol O₂(g)

Products: 4 mol CO₂(g) + 6 mol H₂O(l)

∆H = (4 mol CO₂(g) × (-393.5 kJ/mol)) + (6 mol H₂O(l) × 0 kJ/mol) - (2 mol C₂H₆(g) × 0 kJ/mol) - (7 mol O₂(g) × 0 kJ/mol)

∆H = -1574 kJ + 0 kJ - 0 kJ - 0 kJ

∆H ≈ -1574 kJ (rounded to the nearest kJ)