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N2H4(g) + H2(g) 2 NH3(g) H1 = –1876 kJ
3 H2(g) + N2(g) 2 NH3(g) H2 = –922 kJ
The H f for the formation of hydrazine: 2 H2(g) + N2(g) N2H4(g) will be_____kJ/mol.
To determine the enthalpy change of the formation reaction, we need to modify the given equations in order to cancel out the reactants and products that are not involved in the formation of N2H4.
The given equations are:
1) N2H4(g) + H2(g) 2 NH3(g) ΔH1 = –1876 kJ
2) 3 H2(g) + N2(g) 2 NH3(g) ΔH2 = –922 kJ
Now we need to rearrange these equations to eliminate NH3 as it is not involved in the formation of N2H4. By multiplying equation 1 by 2 and equation 2 by 1:
1) 2(N2H4(g) + H2(g)) 4 NH3(g) ΔH1' = -2 * 1876 kJ = -3752 kJ
2) (3 H2(g) + N2(g)) 2 NH3(g) ΔH2' = -922 kJ
Now if we subtract equation 2 from equation 1:
ΔH1' - ΔH2' = -3752 kJ - (-922 kJ)
= -3752 kJ + 922 kJ
= -2830 kJ
This means that the enthalpy change of the formation reaction of N2H4 is -2830 kJ/mol.
Therefore, the Hf (standard enthalpy of formation) for the formation of hydrazine (N2H4) is -2830 kJ/mol.