The enthalpy changes of the following reactions can bemeasured: C 2 H 4 (g) + 3
ID: 684117 • Letter: T
Question
The enthalpy changes of the following reactions can bemeasured: C2H4(g) + 3 O2(g) ----> 2CO2(g) + 2 H2O(l) Hº= - 1411.1 kJ C2H5OH(l) +3 O2(g)--->2CO2 (g) = 3H2O (l) Hº=1367.6kJ a) Use these values and Hess's law to determine the enthalpychange for the reaction C2H4(g)=H20 (l)---->C2H5OH(l) The enthalpy changes of the following reactions can bemeasured: C2H4(g) + 3 O2(g) ----> 2CO2(g) + 2 H2O(l) Hº= - 1411.1 kJ C2H5OH(l) +3 O2(g)--->2CO2 (g) = 3H2O (l) Hº=1367.6kJ a) Use these values and Hess's law to determine the enthalpychange for the reaction C2H4(g)=H20 (l)---->C2H5OH(l) C2H5OH(l) +3 O2(g)--->2CO2 (g) = 3H2O (l) Hº=1367.6kJ a) Use these values and Hess's law to determine the enthalpychange for the reaction C2H4(g)=H20 (l)---->C2H5OH(l)Explanation / Answer
When determining the enthalpy change for a reaction, and given twochemical equations, it is important to find how the first two formthe third. In this case, all we have to do is reverse the second equation inorder to obtain the 3rd equation: C2H4(g) + 3O2 (g) ---->2CO2 (g) + 2H2O (l) H=-1411.1 kJ 3H2O (l) + 2CO2 (g) ----. 3O2 (g)+ C2H5OH (l) H=-1367.6 kJ the 3O2 will cancel out, and the 2H2O will goaway. However, you will be left with one H2O on thereacttants side. This will give you: C2H4(g) =H20(l)---->C2H5OH(l) **When you reverse a chemical reaction, the sign of the enthalpychanges (the +1367.6 kJ become -1367.6 kJ) Now, just add the two enthalpies together to get a total of -2778.7kJ= H
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