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Answer to Question #51532 in Organic Chemistry for olgamykytsey

Question #51532
Ethanoic acid (HC2H3O2) may be produced from ethyne (C2H2) by the following overall
reaction:
2 C2H2(g) + 2 H2O(l) + O2(g) ---> 2HC2H3O2 (l)

Calculate Ho for the production of ethanoic acid using the following thermochemical
equations:
C2H4(g) ---> H2(g) + C2H2(g) ΔHo = 174.5 kJ
C2H5OH(l) ---> C2H4(g) + H2O(l) ΔHo = 44.07 kJ
C2H5OH(l) + O2(g) ---> HC2H3O2(l) + H2O(l) ΔHo = - 495.2 kJ
2 H2(g) + O2(g) ---> 2 H2O(l) ΔHo = - 571.7 kJ
Expert's answer
Hess's law problem. By adding theequations (or the equation for the reverse reaction) you can arrive at the equation for the target reaction. Then add the enthalpy changes for each reaction to find the enthalpy change for the target reaction. If you reverse a reaction, then reverse the sign on ΔH. If you change the coefficients by some factor, use the same factor for ΔH. 

Target reaction 
2C2H2(g) + 2H2O(l) + O2(g) --> 2HC2H3O2(l).......... ΔH

2H2(g) + 2C2H2(g) --> 2C2H4(g).................................ΔH = 2(-174.5 kJ) 
2C2H4(g) + 2H2O(l) --> 2C2H5OH(l)...........................ΔH = 2(-44.07 kJ) 
2C2H5OH(l) + 2O2(g) ---> 2HC2H3O2(l) +2H2O(l) ....ΔH = 2(-495.2 kJ) 
2H2O(l) --> 2H2(g) + O2(g).................... ......................ΔH = +571.1 kJ 
----------------------- ------------------------------------------- ---------------------------- 
2C2H2(g) + 2H2O(l) + O2(g) --> 2HC2H3O2(l).......... ΔH = -856.44 kJ

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