? 發問於 科學及數學化學 · 1 十年前

enthalpy change

Given that

C (s) + 1/2 O (g) > CO (g) Delta H = -110kJ/ mol

C (s) = O2 (g) > CO2 (g) delta H = -393kJ/mol

Find the enthalpy change of the reaction :

C + CO2 > 2 CO

我劃唔到個enthalpy cycle 出黎

唔該咁多位

1 個解答

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  • 1 十年前
    最愛解答

    The energy cycle (Born-Haber cycle) is drawn below:

    C(s) + CO2(g) <---------- 2 C(s) + O2(g)

    |...................-393kJ/mol.............|

    |................................................|

    |................................................|

    | delta H.........2 x (-110)kJ/mol |

    |................................................|

    |................................................|

    |................................................|

    |................................................|

    v...............................................|

    2 CO(g) <----------------------------

    By Hess's law:

    Delta H + (-393) = 2 x (-110) (enthalpy change sum of anti-clockwise path = enthalpy change sum of clockwise path)

    So,

    Delta H = -220 + 393

    = +173 kJ/mol

    Note: Those dots are for alignment only and they have no meanings.

    2007-01-07 13:47:45 補充:

    在 clockwise 的 path 中, 有 two moles of CO formed. 所以要 乘 2.而計到的 Delta H 係 based on per mole of C or CO2 reacted. 如要 based on per mole of CO formed, 咁就要除 2, 即 86.5 kJ/mol.

    資料來源: My chemical knowledge
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