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Given the provided enthalpies of formation, calculate ΔH for the reaction shown. 2 H2 (g) +...

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 = ?

Solutions

Expert Solution

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)

----------------------------------------------------------------------------------------------------------------

6 H2 (g) + 2 O3 (g) → 6 H2O (g)    ΔH = - 1736.2 kJ

queen_honey_blossom answered 1 year ago
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