Answer to Question #174901 in General Chemistry for Sana Qureshi

Question #174901

Using standard enthalpies of formation for glucose (-1273.3 kJ/mol) and the enthalpies of formation, calculate the enthalpy of reaction for this formation of 1 mol of glucose. Show you work for full marks. 6CO2(g) + 6H2O(g) → 6 O2(g) + C6H12O6 delta H=? 






1
Expert's answer
2021-03-24T01:11:42-0400
"\\Delta H^0 = \\Sigma\\Delta H^0_{product} - \\Sigma\\Delta H^0_{reactant}"

"\\Delta H^0 = \\Delta H^0(C_{6}H_{12}O_{6}) - 6\\Delta H^0(CO_2) - 6\\Delta H^0 (H_2O)"

"\\Delta H^0 = -1273.3 kJ\/mol - 6*(-393.5kJ\/mol) - 6*(-241.8)kJ\/mol"

"\\Delta H^0 = 2538.5 kJ"


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