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. −1119 kJ · mol−1 2. −151 kJ · mol−1 3. −243 kJ · mol−1 4. −935 kJ · mol−1 5. −59 kJ · mol−1?

∆H° = −151.2 kJ/mol.

Explanation:

Applying Hess law:

Overall equation,

N2H4(ℓ) + H2(g) → 2NH3(g)

i. N2H4(ℓ) + O2(g) → N2(g) + 2H2O(g) ∆H° = −543 kJ/mol

Flipping,

ii. 2H2(g) + O2(g) → 2H20(g)

to this:

ii. 2H20(g) → 2H2(g) + O2(g)

∆H° = +484kJ/mol

iii. N2(g) + 3 H2(g) → 2 NH3(g)

∆H° = −92.2 kJ/mol

Cancelling reactants on opposite sides,

N2H4(ℓ) + H2(g) → 2 NH3(g)

Calculating Enthalpy, by adding all the enthalpies

(−543 kJ/mol) + (+484kJ/mol) + (−92.2 kJ/mol)

∆H° = −151.2 kJ/mol.

the answer for sodium is 0.186

the answer for rubidium is 0.251