Calculate H for the reaction below using Hess\'Law.Explain. 2C 2 H 6 (g)-->2C 2

Question

Calculate H for the reaction below using Hess'Law.Explain. 2C2H6(g)-->2C2H2(g)+4H2(g)            H=? C2H2(g)+5/2O2(g)--->2CO2(g)+H2O(l)         H=-1300kJ 2H2(g)+O2(g)-->2H2O(l)                                H=-572kJ 2C2H6(g)+7O2(g)-->4CO2(g)+6H2O(l)          H=-3120kJ Calculate H for the reaction below using Hess'Law.Explain. 2C2H6(g)-->2C2H2(g)+4H2(g)            H=? C2H2(g)+5/2O2(g)--->2CO2(g)+H2O(l)         H=-1300kJ 2H2(g)+O2(g)-->2H2O(l)                                H=-572kJ 2C2H6(g)+7O2(g)-->4CO2(g)+6H2O(l)          H=-3120kJ

Explanation / Answer

Reverse and double equation #1 to give 4CO2(g)+2H2O(l) --->  2C2H2(g)+5/O2(g )      H=+2600kJ Reverse and doubling the second equation to give 4H2O(l) -->  4H2(g) +2O2(g)                              H=+1144kJ When you then add the three equations together as nowstated, you get 4CO2(g ) + 2H2O(l)+ 4H2O(l)+ 2C2H6(g)+7O2(g)--> 2C2H2(g) +5O2(g ) +4H2(g) +2O2(g) + 4CO2(g) +6H2O(l) Eliminating the substances which are the same on both thereactant and the product side gives you 4CO2(g ) +2H2O(l) + 4H2O(l)+ 2C2H6(g)+ 7O2(g)--> 2C2H2(g) +5O2(g ) + 4H2(g)+ 2O2(g)+ 4CO2(g)+6H2O(l) leaving 2C2H6(g) --> 2C2H2(g) + 4H2(g) ! thedesired reaction....to find H for this reaction, simply sumup the individual H's for each step +2600 kJ + 1144 kJ -3120 kJ = +624 kJ leaving 2C2H6(g) --> 2C2H2(g) + 4H2(g) ! thedesired reaction....to find H for this reaction, simply sumup the individual H's for each step +2600 kJ + 1144 kJ -3120 kJ = +624 kJ

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