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1 Q about Enthalpy
發問:
The standard enthalpies of formation of CO2(g) = -394kJmol-1 H2O(g) = -242kJmol-1 of combustion of ethane = -1560kJmol-1 of reduction of ethene to ethane by gaseous hydrogen is -138... 顯示更多 The standard enthalpies of formation of CO2(g) = -394kJmol-1 H2O(g) = -242kJmol-1 of combustion of ethane = -1560kJmol-1 of reduction of ethene to ethane by gaseous hydrogen is -138 kJmol-1 find the standard enthalpy of formation of ethene. 唔識計 佢個model ans : 184kJmol-1
最佳解答:
這是一條舊題目，而且是一條陳舊而『錯誤』的題目。這題目流傳至今居然還是繼續沒有更改而繼續存在，可堪一嘆！ 題目錯誤的原因，是題目給予的standard enthalpy change of formation of H2O(g)，但在 standard enthalpy change of combustion of ethene 中，state of water under standard conditions should be liquid, i.e. H2O(l)。但是若果你堅持『正確』的話， H2O(l) 與 H2O(g) 不能在計算中互相消去。計不到答案是題目的錯誤，不是我能力的問題。 以下所列的流傳數十年的計法，是將錯就錯變成答案的計法。若果你老師逼你要計到答案的話，你可以照抄。但紅色的 state symbol (g) 其實應該是 (l)。記著，若果你堅持要『計算正確』的話，你會計算不到答案的。（苦笑） ========== Given : Standard enthalpy change of formation of CO2(g): C(graphite) + O2(g) → CO2(g) aaΔH = -394 kJ mol-1 Standard enthalpy change of formation of H2O(g): H2(g) + 0.5O2(g) → H2O(g) aaΔH = -242 kJ mol-1 Standard enthalpy change of combustion of ethane: CH3-H3(g) + 3.5O2(g) → 2CO2(g) + 3H2O(g)aaΔH = -1560 kJ mol-1 Standard enthalpy change of reduction of ethane to ethane by hydrogen: CH2CH2-(g) + H2(g) → CH3CH3(g)aaΔH = -(-138) = +138 kJ mol-1 The solution : CH3CH3(g) → CH2CH2-(g) + H2(g) aaΔH = -(-138) = +138 kJ mol-1 2CO2(g) + 3H2O(g) → CH3-H3(g) + 3.5O2(g) aaΔH = -(-1560) = +1560 kJ mol-1 3H2(g) + 1.5O2(g) → 3H2O(g) aaΔH = 3(-242) = -726 kJ mol-1 2C(graphite) + 2O2(g) → 2CO2(g) aaΔH = 2(-394) = -788 kJ mol-1 Add the above four thermochemical equations together, we get 2C(graphite) + 2H2(g) → CH2CH2(g) Standard enthalpy change of formation ΔH of ethane, ΔHf[CH2CH2] = 138 + 1560 - 726 - 788 = +184 kJ mol-1
其他解答:
1 Q about Enthalpy
發問:
The standard enthalpies of formation of CO2(g) = -394kJmol-1 H2O(g) = -242kJmol-1 of combustion of ethane = -1560kJmol-1 of reduction of ethene to ethane by gaseous hydrogen is -138... 顯示更多 The standard enthalpies of formation of CO2(g) = -394kJmol-1 H2O(g) = -242kJmol-1 of combustion of ethane = -1560kJmol-1 of reduction of ethene to ethane by gaseous hydrogen is -138 kJmol-1 find the standard enthalpy of formation of ethene. 唔識計 佢個model ans : 184kJmol-1
最佳解答:
這是一條舊題目，而且是一條陳舊而『錯誤』的題目。這題目流傳至今居然還是繼續沒有更改而繼續存在，可堪一嘆！ 題目錯誤的原因，是題目給予的standard enthalpy change of formation of H2O(g)，但在 standard enthalpy change of combustion of ethene 中，state of water under standard conditions should be liquid, i.e. H2O(l)。但是若果你堅持『正確』的話， H2O(l) 與 H2O(g) 不能在計算中互相消去。計不到答案是題目的錯誤，不是我能力的問題。 以下所列的流傳數十年的計法，是將錯就錯變成答案的計法。若果你老師逼你要計到答案的話，你可以照抄。但紅色的 state symbol (g) 其實應該是 (l)。記著，若果你堅持要『計算正確』的話，你會計算不到答案的。（苦笑） ========== Given : Standard enthalpy change of formation of CO2(g): C(graphite) + O2(g) → CO2(g) aaΔH = -394 kJ mol-1 Standard enthalpy change of formation of H2O(g): H2(g) + 0.5O2(g) → H2O(g) aaΔH = -242 kJ mol-1 Standard enthalpy change of combustion of ethane: CH3-H3(g) + 3.5O2(g) → 2CO2(g) + 3H2O(g)aaΔH = -1560 kJ mol-1 Standard enthalpy change of reduction of ethane to ethane by hydrogen: CH2CH2-(g) + H2(g) → CH3CH3(g)aaΔH = -(-138) = +138 kJ mol-1 The solution : CH3CH3(g) → CH2CH2-(g) + H2(g) aaΔH = -(-138) = +138 kJ mol-1 2CO2(g) + 3H2O(g) → CH3-H3(g) + 3.5O2(g) aaΔH = -(-1560) = +1560 kJ mol-1 3H2(g) + 1.5O2(g) → 3H2O(g) aaΔH = 3(-242) = -726 kJ mol-1 2C(graphite) + 2O2(g) → 2CO2(g) aaΔH = 2(-394) = -788 kJ mol-1 Add the above four thermochemical equations together, we get 2C(graphite) + 2H2(g) → CH2CH2(g) Standard enthalpy change of formation ΔH of ethane, ΔHf[CH2CH2] = 138 + 1560 - 726 - 788 = +184 kJ mol-1
其他解答:
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With reference to the following energy cycle: 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Mar08/Crazythermalchem1.jpg By Hess's law: △Hf + △H1 + △H2 = △H3 + △H4 △Hf - 138 - 1560 = -242 x 3 - 394 x 2 △Hf = 184 kJ/mol
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