Question

Consider the following data: C(graphite) + O2(g) → CO2(g) ΔrH° = -394 kJmol-1 H2(g) +½ O2(g)...

Consider the following data: C(graphite) + O2(g) → CO2(g) ΔrH° = -394 kJmol-1 H2(g) +½ O2(g) → H2O(l) ΔrH° = -286 kJmol-1 C2H5OH(l) → 2 C(graphite) + 3 H2(l) + 0.5 O2(g) ΔrH° = +228 kJmol-1 What is the value of ΔrH° for the following reaction? C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)

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Answer #1

C(graphite) + O2(g) → CO2(g) ΔrH° = -394 kJmol-1 -----(1)

H2(g) +½ O2(g) → H2O(l) ΔrH° = -286 kJmol-1 ------(2)

C2H5OH(l) → 2 C(graphite) + 3 H2(l) + 0.5 O2(g) ΔrH° = +228 kJmol-1 ----(3)

C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)  ΔrH° = ?

eqn (1) x 2 2C(graphite) + 2O2(g) → 2CO2(g) ΔrH° = 2 x -394 kJmol-1 => -788 kJmol-1 ----- (4)

eqn (2) x 3 3H2(g) +3/2 O2(g) → 3H2O(l) ΔrH° = 3 x -286 kJmol-1 => - 858 kJmol-1 ------(5)

eqn (3) C2H5OH(l) → 2 C(graphite) + 3 H2(l) + 0.5 O2(g) ΔrH° = +228 kJmol-1 ----(3)

add equation (3) + (4) + (5) -----------------------------------------------------------------------------------------------

C2H5OH(l)  + 3 O2(g) → 2 CO2(g) + 3 H2O(l) ΔrH° =  -788 - 858 +228 => -1418 kJmol-1

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