
One mole of water is formed from hydrogen and oxygen releasing 286 kJ
H2?(g) + ?O2?(g)??→?H2O (l)? ? ? ? ?? ? ΔHr?= -286 kJ mol-1
Calculate ΔHr for the reaction below:2H2?(g) + O2?(g)??→?H2O (l)
Answer:
ΔHr?= 2 mol x (-286 kJ mol-1)= - 572 kJ
Calculate ΔHr?for the reaction below
4Fe (s) +O2?(g)??→? 2Fe2O3?(s)
given that ΔHf???[Fe2O3?(s)]??= - 824?kJ mol-1
Answer:
ΔHf?=? 2 mol x ( -824 kJ mol-1)= -?1648 kJ
Identify each of the following as? ΔHr?, ΔHf?, ΔHc??or ΔH?neut
Answer:
Answer 1:?ΔHr?
Answer 2:?ΔHf??as one mole of CO2?is formed from its elements in standard state and ΔHc??as one mole of carbon is burnt in oxygen
Answer 3:?ΔHneut??as one mole of water is formed from the reaction of an acid and alkali
You need to learn well the Standard Enthalpy change definitions as they are frequently tested in exam papers
轉載自savemyexams
以上就是關于【IB DP Chemistry: SL復習筆記5.1.2 Standard Enthalpy Change】的解答,如需了解學校/賽事/課程動態,可至翰林教育官網獲取更多信息。
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