Question

Part A

Calculate the amount of energy released in the formation of one mole of BeS bonds (not lattice energy). The radius of the beryllium ion is 0.31 Å, and the radius of the sulfide ion is 1.84 Å. Note that 1Å=10−10m.

Express your answer with the appropriate units to three significant figures.

Answer 1

Energy of interaction between two ions can calculated using Coloumb's law

k = 2.31 x 10^-28 J.m

q₁ and q₂ are charges on the ions

q₁ = charge of Be = +2

q₂ = charge of S = -2

r = distance between the two ions = sum of radii = 0.31 + 1.84 = 2.15 Å

r =2.15 x10^-10 m

Now let us plug in the values

Attractive force between ions = 4.30 x10^-18 J

Energy released from one molecule = 4.30 x10^-18 J

This is the energy releases for one molecule BeS

Now we have to find for one mole BeS

=2.59 x 10⁶ J

Energy released for one mole BeS = 2.60 x 10⁶ J