UPSEE Solved Model Paper 3

Given,
S (s) + 3F2 (g) → SF6 ; ΔH = - 1100 kJ ... (i)

1

2

F2 (g) → F (g) ; ΔH = 80 kJ ... (ii)
S (s) → S (g) ; ΔH = + 275 kJ ... (iii)
Subtracting eq (iii) from eq (ii) with 6 gives
S (g) + 3F2β → 6F6 ; ΔH = - 1100 - 275 = - 1375 kJ ... (iv)
Subtracting eq (ii) with 6 gives
3F2 (g) → 6F (g) ; ΔH = 6 × 80 kJ = 480 kJ ... (v)
Subtracting eq (v) from eq (iv)
S (g) + 6 F (g) → SF6 ; ΔH = - 1375 - 480
= - 1855 kJ or
SF6 → S (g) + 6F (g) ; ΔH = + 1855 kJ
Average BE =

1855

6

= 309.17 kJ