Resposta :
2x CH4(g) + H2O(g) -> CO(g) + 3 H2(g)Δ H=
206,1kJ
2x 2 H2(g) + CO(g) ->CH3OH(l) ΔH= -128,3kJ
2x 2 H2(g) + O2(g) -> 2 H2O(g) ΔH= - 483,6kJ
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2 CH4 + O2 -> 2 CH3OH ΔH -328 KJ
2x 2 H2(g) + CO(g) ->CH3OH(l) ΔH= -128,3kJ
2x 2 H2(g) + O2(g) -> 2 H2O(g) ΔH= - 483,6kJ
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2 CH4 + O2 -> 2 CH3OH ΔH -328 KJ