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Answer:(a) 6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g)
(b) 5.55*10^37photons
Explanation:
(a) Here we have to write a balanced thermochemical equation for formation of 1.00 mol of glucose.
In this question it has been given that Chlorophyll absorbs light in the 600 to 700 nm region,
1st we will write a chemical equation for biochemical process of photosynthesis is that CO2 and H2O form glucose (C6H12O6) and O2.
6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g)
The heat change off the reaction can be calculated as,
={(1 mol)(6 mol) }- {(6 mol) [H2O]}
=[1 1273.3 kJ + 6(0)] - [6 (-39.5 kJ) + 6 (-285.840 kJ)]
= 2802.74 or 2802.7 kJ
Thus the balanced equation can be written as,
6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g) = 2802.7 kJ for 1.00 mol of glucose.
