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Hess's Law Example 8.9

Posted: Thu Jan 12, 2017 9:53 am
I'm looking over how the book uses Hess's Law on 8.9 and I don't understand how the book got H2O2 for Step 2.
Propane, C3H8, is a gas used as a fuel for outdoor grills and alternative-fuel vehicles. The enthalpy change for the synthesis of propane from its elements in their standard states is difficult to measure directly, but if you are interested in assessing the thermodynamic properties of its reactions, you need to know its value. Calculate the standard enthalpy of the reaction 3 C(gr) + 4 H2(g) -> C3H8(g) from the following experimental data:
(a) C3H8(g)+  5 O2(g) -> 3 CO2(g) +  4 H2O (l) delta H° = 2220. kJ
(b) C(gr) +  O2(g) -> CO2(g) delta H° = 394 kJ
(c) H2(g) + (1/2) O2(g) -> H2O(l) delta H° = 286 kJ

Step 1: 3 C(gr) + 3 O2(g) -> 3 CO2(g) delta H =  3 * -394 kJ = -1182 kJ
Step 2: 3 CO2(g) + 4 H2O2(l) -> C3H8(g) + 5O2(g) delta H° = + 2220. kJ

Re: Hess's Law Example 8.9

Posted: Thu Jan 12, 2017 11:44 am
This is an error in the book. It should be H2O