Question

ate the standard enthalpy change for the reaction Given standard enthalpy of formations △H; (SO(g))--395.2 kJ/mol Δ} l; (SO2(g))--296.9 kJ/mol al 201.0 kJ 403 J 254.6 kJ - 120.4 k.J - 196.6 kJ dl

Answer 1

$Enthalpy\quad change\quad of\quad reaction\quad =\quad \Delta { H }_{ f }^{ 0 }\quad Products\quad -\quad \Delta { H }_{ f }^{ 0 }\quad Reactants$

Now,

$\Delta { H }_{ f }^{ 0 }\quad S{ O }_{ 3 }\quad =\quad -395.2\quad KJ/mol\\ \Delta { H }_{ f }^{ 0 }\quad S{ O }_{ 2\quad }=\quad -296.9\quad KJ/mol\\ \Delta { H }_{ f }^{ 0 }\quad { O }_{ 2 }\quad =\quad 0\quad$

Therefore,Enthalpy change = 2*(-395.2) - 2*(-296.9)

Enthalpy change = -790.4 + 593.8

Enthalpy change = -196.6 KJ