Question

Consider the reaction: 2H2(g) + O2(g) → 2H2O(g) ΔH= −483.6 kJ/molAt constant temperature and pressure, 4 moles of hydrogen and 2 moles of oxygen are allowed to react. What is the maximum amount of work that can be obtained from this reaction?

Answer 1

To find the maximum amount of work that can be obtained from this reaction, we first need to determine the limiting reactant. In this case, we have:4 moles H2 x  1 mole O2 / 2 moles H2  = 2 moles O22 moles O2 x  2 moles H2 / 1 mole O2  = 4 moles H2Since both reactants are present in the stoichiometric ratio, neither is the limiting reactant. Therefore, the reaction will proceed to completion.Now, we can calculate the total enthalpy change  H  for the reaction:H = -483.6 kJ/mol  for 2 moles of H2O produced Since 4 moles of hydrogen and 2 moles of oxygen react to form 4 moles of water, we can calculate the total enthalpy change for this reaction:H_total = -483.6 kJ/mol   4 moles H2O / 2 moles H2O  = -967.2 kJAt constant temperature and pressure, the maximum work that can be obtained from the reaction is equal to the enthalpy change:W_max = H_total = -967.2 kJThe maximum amount of work that can be obtained from this reaction is 967.2 kJ.