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.

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

2b2t

explanation:

2b2t