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Methanol, CH3OH, has been considered as a possible fuel. Consider its oxidation: 2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(g) ΔG = -1372 kJ/mol. What is the maximum work that can be obtained by oxidizing 0.50 mol of methanol under standard conditions?

Respuesta :

Answer: Maximum work that can be obtained by given amount of methanol is -343kJ.

Explanation:

For the given chemical reaction:

[tex]2CH_3OH(l)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g);\Delta G_{rxn}=-1372kJ/mol[/tex]

By Stoichiometry of the reaction:

2 moles of methanol does a work of 1372 kJ.

So, 0.5 moles of methanol will do a work of = [tex]\frac{1372kJ}{2}\times 0.5=343kJ[/tex]

Hence, maximum work that can be obtained by given amount of methanol is -343kJ.

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