UPSEE Solved Model Paper 3

Given, S (s) + $3\mathrm{F}_2$ (g) → $\mathrm{SF}_6$ ; ΔH = - 1100 kJ ... (i) $\frac{1}{2}\mathrm{F}_2$ (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) + $3\mathrm{F}_2\beta$ → $6\mathrm{F}_6$ ; ΔH = - 1100 - 275 = - 1375 kJ ... (iv) Subtracting eq (ii) with 6 gives $3\mathrm{F}_2$ (g) → 6F (g) ; ΔH = 6 × 80 kJ = 480 kJ ... (v) Subtracting eq (v) from eq (iv) S (g) + 6 F (g) → $\mathrm{SF}_6$ ; ΔH = - 1375 - 480 = - 1855 kJ or $\mathrm{SF}_6$ → S (g) + 6F (g) ; ΔH = + 1855 kJ Average BE = $\frac{1855}{6}$ = 309.17 kJ