Question

Calculate the standard reaction enthalpy for the reaction

N₂H₄(ℓ) + H₂(g) → 2 NH₃(g)

given

N₂H₄(ℓ) + O₂(g) → N₂(g) + 2H₂O(g) ∆H ◦ = −543 kJ · mol⁻¹

2 H₂(g) + O₂(g) → 2 H₂O(g) ∆H◦ = −484 kJ · mol⁻¹

N₂(g) + 3 H₂(g) → 2 NH₃(g) ∆H◦ = −92.2 kJ · mol⁻¹

1.) −243 kJ · mol⁻¹

2.) −59 kJ · mol⁻¹

3.) −935 kJ · mol⁻¹

4.) −151 kJ · mol⁻¹

5.) −1119 kJ · mol⁻¹

Answer 1

We have to apply Hess's Law

First rearrange the equations so that the elements not present in the original equation will cancel out. Do this by flipping the second equation, which also changes its enthalpy value from negative to positive.

N2H4(l) + O2(g) --->N2(g) + 2H2O(g) ∆H ◦ = -543 kJ · mol -1

2 H20(g) + 2 H2(g) + O2(g) ∆H ◦=+484kJ · mol -1

N2(g) + 3 H2(g) ---> 2 NH3(g) ∆H ◦ = -92.2 kJ · mol -1

Now cancel reactants that are on opposite sides of the yield sign and you are left with,

N2H4(l) + H2 --> 2 NH3(g)

Now to find the enthalpy, simply add the enthalpies used to create this reaction.

(-543 kJ · mol -1)+(+484kJ · mol -1)+(-92.2 kJ · mol -1)

We get a final enthalpy of ∆H ◦ = -151.2 kJ · mol -1 (-151 with sigfigs)