HW 8.67

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HW 8.67

Postby Nikhil_Madan_3C » Thu Jan 26, 2017 5:44 pm

Use the information in Tables 8.3, 8.6, and 8.7 to estimate the enthalpy of formation of each of the following compounds in the liquid state. The standard enthalpy of sublimation of carbon is 717 kJ·mol 1. (a) H2O; (b) methanol, CH3OH; (c) benzene, C6H6 (without resonance); (d) benzene, C6H6 (with resonance).

For letter b the equation would be C + 2H2 + (1/2)O2 -> 2CH3OH
When calculating bond enthalpies how would you factor in the C since it is an atom?

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Re: HW 8.67

Postby Palmquist_Sierra_2N » Thu Jan 26, 2017 6:21 pm

Hey! First off, the product side of the reaction should only be CH3OH.
For carbon(C) you have to atomize/ sublime it. You look up C in the (g) form in the chart and see that the enthalpy of formation is 716.68 kJ/mol

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