Calculate the enthalpy of reaction for
2N2(g) + 6H2(g) → 4NH3(g)
Given the data:4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l), ΔH1 = -1530 kJ mol-1
H2(g) + ½O2(g )→ H2O(l), ΔH2 = -288 kJ mol-1
Answer:



ΔHr = +6ΔH2 – ΔH1 = + (-288 x 6) - ( -1530 ) = -198 kJ
What is the enthalpy change, in kJ, for the reaction below?
4FeO(s) + O2(g) → 2Fe2O3(s)
Given the data:2Fe(s) + O2(g) → 2FeO(s) ∆H = – 544 kJ
4Fe(s) + 3O2(g) → 2Fe2O3(s) ∆H = –1648 kJ
Answer:

Follow the alternative route and the process the calculation
ΔHr = - ( - 544 x 2) + (- 1648) = - 560 kJ
It is very important you get the arrows in the right direction and that you separate the mathematical operation from the sign of the enthalpy change. Many students get these problems wrong because they confuse the signs with the operations. To avoid this always put brackets around the values and add the mathematical operator in front
转载自savemyexams
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