? 發問於 科學及數學化學 · 1 十年前

# enthalpy change

Given that

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

C (s) = O2 (g) &gt; CO2 (g) delta H = -393kJ/mol

Find the enthalpy change of the reaction :

C + CO2 &gt; 2 CO

### 1 個解答

• 1 十年前
最愛解答

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

C(s) + CO2(g) &lt;---------- 2 C(s) + O2(g)

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

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

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

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

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

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

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

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

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

2 CO(g) &lt;----------------------------

By Hess&#39;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