calculate enthalpy of H for the reaction

N2H4(l) + 2H2O(l) -> N2(g) + 4H2)(l)

Given the reactions
N2H4(l) + O2(g) -> N2(g) + 2H2O(l) Enthalpy of H = -6.22.2 kJ

H2(g) + (1/2)O2(g) -> H2O(l) enthalpy of H = -285.8 kJ/mol

H2(g) + O2(g) -> H2O2(l) enthalpy of H = -187.8 kJ

To calculate the enthalpy change (ΔH) for the given reaction, you need to use Hess's Law, which states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps that make up the reaction.

Let's break down the given reaction into smaller steps:

1) N2H4(l) + O2(g) -> N2(g) + 2H2O(l) ΔH = -622.2 kJ/mol (given)
2) 2H2(g) + O2(g) -> 2H2O(l) ΔH = 2 * (-285.8 kJ/mol) = -571.6 kJ/mol (using the given enthalpy change of H2(g) + (1/2)O2(g) -> H2O(l))
3) H2(g) + O2(g) -> H2O2(l) ΔH = -187.8 kJ (given)

Now, let's combine these equations to obtain the desired reaction:

N2H4(l) + 2H2O(l) -> N2(g) + 4H2(l)

By manipulating equations 2 and 3, you can create the desired reaction:

4) 2H2(g) + O2(g) -> 2H2O(l) ΔH = -571.6 kJ/mol
5) N2H4(l) + O2(g) -> N2(g) + 2H2O(l) ΔH = -622.2 kJ/mol

Now, we need to cancel out the common compounds in equations 4 and 5:

6) N2H4(l) + 2H2(g) -> N2(g) + 4H2O(l) ΔH = -622.2 kJ/mol - (-571.6 kJ/mol) = -50.6 kJ/mol

The enthalpy change for the given reaction is -50.6 kJ/mol.