Be sure to answer all parts. the net change in the multistep biochemical process of photosynthesis is that co2 and h2o form glucose (c6h12o6) and o2. chlorophyll absorbs light in the 600 to 700 nm region. what is the minimum number of photons with 680. nm needed to prepare 1.00 mol of glucose? assume co2 to be a gas and glucose to be a solid.

9.86 × 10²⁴ photons

Explanation:

Let's consider the photosynthesis global reaction.

6 CO₂(g) + 6 H₂O(l) →  C₆H₁₂O₆(s) +  6 O₂(g)

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)

where,

ni are the moles of reactants and products

ΔG°f(p) are the standard free Gibbs energies of formation of reactants and products

For the balanced equation for 1.00 mol of glucose, ΔG° is:

ΔG° = [1 mol × ΔG°f(C₆H₁₂O₆(s)) + 6 mol × ΔG°f(O₂(g))] - [6 mol × ΔG°f(CO₂(g)) + 6 mol × ΔG°f(H₂O(l))]

ΔG° = [1 mol × (-910.6 kJ/mol) + 6 mol × 0 kJ/mol] - [6 mol × (-394.4 kJ/mol) + 6 mol × (-237.1 kJ/mol)]

ΔG° = 2878 kJ

The energy of each photon (E) can be calculated using the Planck-Einstein's equation.

E = h . c. λ⁻¹

where,

h is the Planck's constant

c is the speed of light

λ is the wavelength

E = (6.626 × 10⁻³⁴ J.s) × (3.00 × 10⁸ m/s) × (680 × 10⁻⁹ m)⁻¹ = 2.92 × 10⁻¹⁹ J

The minimum number of photons required is: