1. Calculate AH for the reaction 2NO2(g) → N2O.(g)
delta H = ?
Given:
N2(g) + 202(g) → 2NO2(g) delta H = 67.7 kJ
N2(g) + 202(g) → N2O4(g) delta H = 9.7 kJ

- 58 kJ

Explanation:

The equation of the reaction is given as;

2NO2(g) → N2O4(g)

N2(g) + 202(g) → 2NO2(g) delta H = 67.7 kJ

Since NO2 is the reactant in the equation, we have to reverse this reaction. We now have;

2NO2(g) → N2(g) + 202(g)     delta H = - 67.7 kJ

N2(g) + 202(g) → N2O4(g) delta H = 9.7 kJ

Since NO4 is the product in the equation, we don't have to do anything to this reaction.

Adding the two reactions;

2NO2(g) → N2(g) + 202(g)     delta H = - 67.7 kJ

N2(g) + 202(g) → N2O4(g)     delta H = 9.7 kJ

N2(g) and 2O2 (g) cancels out

2NO2(g) → N2O4(g)       delta H = (-67.7 + 9.7) = - 58 kJ

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.