Question

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=? 

Answer 1

$$\Delta H^0 = \Sigma\Delta H^0_{product} - \Sigma\Delta H^0_{reactant}$$

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

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

$$\Delta H^0 = 2538.5 kJ$$