To determine which substances have a standard enthalpy of formation \(\Delta H_f = 0 \, \text{kJ/mol}\) by definition, we need to recall that the standard enthalpy of formation of any element in its standard state is zero. Let us analyze each of the given substances to see if it qualifies:
1. \(O_2(g)\): Oxygen gas (\(O_2\)) is an element in its standard state (gaseous form at 1 atm pressure and 25°C). Therefore, its \(\Delta H_f\) is \(0 \, \text{kJ/mol}\).
2. \(N(g)\): Nitrogen as a single atom (\(N(g)\)) is not in its standard state. The standard state of nitrogen is \(N_2(g)\). Hence, \(\Delta H_f\) of \(N(g)\) is not zero.
3. \(H_2O(l)\): Water (\(H_2O(l)\)) is a compound, not an element. Therefore, its \(\Delta H_f\) is not zero.
4. \(Br_2(l)\): Bromine (\(Br_2\)) is an element in its standard state (liquid form at 1 atm pressure and 25°C). Hence, its \(\Delta H_f\) is \(0 \, \text{kJ/mol}\).
5. \(Fe(s)\): Iron (\(Fe\)) in its solid form is an element in its standard state. Thus, its \(\Delta H_f\) is \(0 \, \text{kJ/mol}\).
6. \(He(g)\): Helium (\(He\)) in its gaseous form is an element in its standard state. Therefore, its \(\Delta H_f\) is \(0 \, \text{kJ/mol}\).
In conclusion, the substances with \(\Delta H_f = 0 \, \text{kJ/mol}\) are:
1. \(O_2(g)\)
4. \(Br_2(l)\)
5. \(Fe(s)\)
6. \(He(g)\)
Thus, the substances with \(\Delta H_f = 0 \, \text{kJ/mol}\) by definition are \(O_2(g)\), \(Br_2(l)\), \(Fe(s)\), and \(He(g)\).
The indices corresponding to these substances in the given list are:
[tex]\[ [1, 4, 5, 6] \][/tex]
