Multiple choice

Calculate the standard enthalpy of combustion per gram of glucose at 25oC, whereas the standard enthalpies of formation of CO2(g), H2O(I) and glucose at 25oC are -500 kJ/mol, -400 kJ/mol and -1500 kJ/mol, respectively.

  1. -21.66 kJ/g

  2. +21.66 kJ/g

  3. -3900 kJ/g

  4. +3900 kJ/g

  5. +5400 kJ/g
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A Correct answer
Explanation

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ∆HoC = Σ ∆Hof (products) - Σ ∆Hof (reactants) = [6 x ∆Hof (CO2) + 6 ∆Hof (H2O)] - [∆Hof (C6H12O6) + 6 x ∆Hof (O2)] = [6 x (-500) + 6 x (-400)] – [-1500 + 6 (0)] = (-3000 - 2400) – (-1500) = -5400 + 1500 = -3900 kJ/mol 
The standard enthalpy of combustion per gram of glucose = -3900/180 = -21.66 kJ/g

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