Calculate standard enthalpy (?H Solution 6CO2(g) + 6H2O(l) --> C6H12O6(s) + 6O2(

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6CO2(g) + 6H2O(l) --> C6H12O6(s) + 6O2(g) dH reaction = dHf products - dHf reactants dH reaction = [dHf C6H12O6(s) + 6O2(g)] - [6CO2(g) + 6H2O(l)] dH reaction = [(–1273.3) + 6(zero)] - [6(-393.5) + 6(-285.8)] dH reaction = [(–1273.3) ] - [(-2361) + (-1714.8)] dH reaction = [(–1273.3) ] - [- 4075.8] dH reaction = (–1273.3) + 4075.8 dH = +2802.5 kJ your answer, rounded to 4 sig figs is dH reaction = 2802 kJ as dH is positive its endothermic rection. Also I have taken these dHf's from http://bilbo.chm.uri.edu/CHM112/tables/thermtable.htm because different texts rarely agree

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