Question

Given the following equations and their respective enthalpy changes:1) 2H2 (g) + O2 (g) → 2H2O (l) ΔH = -484 kJ/mol 2) C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l) ΔH = -1,367 kJ/mol 3) C2H5OH (l) → C2H4 (g) + H2O (g) ΔH = +44 kJ/mol Using Hess’s Law, determine the ΔH for the following reaction: C2H5OH (l) → 2CO2 (g) + 3H2 (g)

Answer 1

To use Hess's Law, we need to manipulate the given equations so that they can be combined to form the target equation: C2H5OH  l   2CO2  g  + 3H2  g .First, we need to reverse equation 1 and multiply it by 3/2 to get 3H2  g  as a product:1'  3H2O  l   3H2  g  + 3/2 O2  g  H = +3/2 * 484 kJ/mol = +726 kJ/molNext, we need to reverse equation 3 to get C2H5OH  l  as a reactant:3'  C2H4  g  + H2O  g   C2H5OH  l  H = -44 kJ/molNow, we can add equations 1', 2, and 3' to get the target equation:1'  3H2O  l   3H2  g  + 3/2 O2  g  H = +726 kJ/mol2  C2H5OH  l  + 3O2  g   2CO2  g  + 3H2O  l  H = -1,367 kJ/mol3'  C2H4  g  + H2O  g   C2H5OH  l  H = -44 kJ/mol-------------------------------------------------------------Target: C2H5OH  l   2CO2  g  + 3H2  g  H = ?Now, we can sum the enthalpy changes:H = +726 kJ/mol - 1,367 kJ/mol - 44 kJ/mol = -685 kJ/molSo, the enthalpy change for the target reaction, C2H5OH  l   2CO2  g  + 3H2  g , is H = -685 kJ/mol.