The combusion of glucose produces carbon dioxide and water vapor according to the chemical equation below:
C6H12O6 (s) + 6O2 (g) ---> 6CO2 (g) + 6H2O (g)
DELTA H = -2540 kj
Calculate the energ involved in making 75.0 g of glucose from carbon dioxide and liquid water (photosynthesis)
To calculate the energy involved in making 75.0 g of glucose from carbon dioxide and liquid water (photosynthesis), we need to determine the energy change for the reverse process of the given combustion reaction.
Given:
Combustion reaction: C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g) ΔH = -2540 kJ
Since we need the reverse reaction for photosynthesis, we simply invert the reactants and products, as well as the sign of the enthalpy change:
Photosynthesis reaction: 6CO2 (g) + 6H2O (g) → C6H12O6 (s) + 6O2 (g) ΔH = +2540 kJ
Now, we can calculate the energy involved in making 75.0 g of glucose. First, we need to calculate the molar mass of glucose (C6H12O6):
Molar mass of glucose (C6H12O6) = 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol
Next, we can use the molar mass of glucose to convert grams of glucose to moles:
75.0 g glucose (C6H12O6) × (1 mol glucose / 180.18 g glucose) = 0.416 mol glucose
Now, we can use the equation ΔH = +2540 kJ to calculate the energy involved in making 0.416 mol of glucose:
ΔH = +2540 kJ/mol × 0.416 mol = +1056.64 kJ
Therefore, the energy involved in making 75.0 g of glucose from carbon dioxide and liquid water (photosynthesis) is approximately +1056.64 kJ.