Question

Given that (1) the first and second electron affinities of sulfur are –0.332 aJ and 0.980 aJ,

(2) the ionic radii of Zn2 and S2– are 74 pm and 184 pm, and that

(3) Zn(g) --> Zn2+ (g) + 2e-

calculate the energy change for this reaction (per formula unit of ZnS).

Zn(g) + S(g) ---> ZnS(g)

E rxn= ..... aJ?

Answer 1

Answer:

To determine the energy change for the given reaction

Let us consider the coilombs law which is given as follows

E = k(Q1Q2)/r

Where as

Q1 & Q2 are the numerical charge of cation and anion

k is consistent and

k approaches 2.31 x 10-19 J nm

r is the separation between the particles in nm

Along these lines,

E = [(2.31 x 10-19 J nm) x (+2) x (- 2)]/0.258 nm

E = - 35.8 x 10-19 J

E = - 3.58 aJ

So finally,

Erxn = - 3.58 aJ – 0.332 aJ + 0.980 aJ + 4.38 aJ = 1.448 aJ

Erxn = 1.448 aJ