Question

36-40. Calculate the ∆H0rxn for the reaction 2S(g) + 2 OF2(g)→SO2(g) + SF4(g) Using the following Thermochemical Reactions:

OF2(g) + H2O(l) → O2(g) + 2HF(g) SF4(g) + 2H20(l) → SO2(g) + 4HF(g) S(g) + O2(g)→SO2(g)

∆ = -277kJ ∆ = -828kJ ∆ = -297kJ

Answer 1

$OF_{2(g)} + H_2O_{(l) }→ O_{2(g)} + 2HF_{(g)}\ \ \ ∆H = -277kJ \\ SF_{4(g)} + 2H_2O_{(l)} → SO_{2(g) } + 4HF_{(g)}\ \ \ ∆H = -828kJ\\ S_{(g)} + O_{2(g)}→SO_{2(g)} \ \ \ ∆H = -297kJ$

Multiply the first and third reaction by 2, reverse the second reaction, then add all of them together.

$2OF_{2(g)} + 2H_2O_{(l) } + SO_{2(g) } + 4HF_{(g)} + 2S_{(g)} + 2O_{2(g)}→ 2O_{2(g)} + 4HF_{(g)} + SF_{4(g)} + 2H_2O_{(l)} + 2SO_{2(g)}\ \ \ \ ∆H = 2(-277) +828+2(-297) kJ$

The final reaction is therefore,

$2S_{(g)} + 2 OF_{2(g)}→SO_{2(g)} + SF_{4(g)} \ \ \ \ ∆H= -320kJ$