Hey bud !
You were close congrats !
the only problem i saw was the units part at the end.
remember volume is 3 dimensional
length width and height
so it is m^3
**Visible confusion** no problem
At the first reaction when 2HBr(g) ⇄ H2(g) + Br2(g)
So Kc = [H2] [Br2] / [HBr]^2
7.04X10^-2 = [H2][Br] / [HBr]^2
at the second reaction when 1/2 H2(g) + 1/2 Br2 (g) ⇄ HBr
Its Kc value will = [HBr] / [H2]^1/2*[Br2]^1/2
we will make the first formula of Kc upside down:
1/7.04X10^-2 = [HBr]^2/[H2][Br2]
and by taking the square root:
∴ √(1/7.04X10^-2)= [HBr] / [H2]^1/2*[Br]^1/2
∴ Kc for the second reaction = √(1/7.04X10^-2) = 3.769
Answer:
The correct answer is CaO > LiBr > KI.
Explanation:
Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.
With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.
The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.
