In: Chemistry
Given the provided enthalpies of formation, calculate ΔH for the reaction shown.
2 H2 (g) + O2 (g) → 2 H2O (g) ΔH = -483.6 kJ
3 O2 (g) → 2 O3 (g) ΔH = +285.4 kJ
6 H2 (g) + 2 O3 (g) → 6 H2O (g) ΔH = ?
2 H2 (g) + O2 (g) → 2 H2O (g) ΔH = -483.6 kJ
multiply above equation with 3
6 H2 (g) +3 O2 (g) → 6 H2O (g) ΔH = -1450.8 kJ ------------(1)
3 O2 (g) → 2 O3 (g) ΔH = +285.4 kJ reverse the equation
2O3(g) --------------> 3O2(g) ΔH = -285.4 KJ -----------(2)
add 1 and 2
6 H2 (g) +3 O2 (g) → 6 H2O (g) ΔH = -1450.8 kJ ------------(1)
2O3(g) --------------> 3O2(g) ΔH = -285.4 KJ -----------(2)
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6 H2 (g) + 2 O3 (g) → 6 H2O (g) ΔH = - 1736.2 kJ