Chemistry, 25.11.2019 21:31 asialoggin21

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.

(a) write a balanced thermochemical equation for formation of 1.00 mol of glucose

(b) what is the minimum number of photons with î» = 680. nm needed to prepare 1.00 mol of glucose?

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Answers

(a)The balanced thermochemical equation for formation of 1.00 mol of glucose is: $6 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+6 \mathrm{O}_{2}$

(b) The minimum number of photons with λ = 680 nm needed to prepare 1.00 mol of glucose is $5.55 \times 10^{37} \text { photons }$

Explanation:

(a) Both side of reaction have equal number of elements therefore number of reactant is equal to number of product hence following balanced equation is achieved : $6 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+6 \mathrm{O}_{2}$

(b) According to theory of special relativity which expresses fact in equation about mass and energy is: $\mathbf{E}=\mathbf{m} \mathbf{c}^{2} \text { and here } \mathrm{c}^{2} \text { is speed of light }$

Here λ = 680 nm,

$\mathrm{n}\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)=1 \mathrm{mol}$

Molar mass of glucose = 180.156 g/mol

$\text { Therefore } m=180.156 \mathrm{g}, \mathrm{c}=3 \times 10^{8}$

Substituting values in above equation

$\mathrm{E}=180.156 \times\left(3 \times 10^{8}\right)^{2}=1.62 \times 10^{29} \mathrm{J}$

Hence it is known that

E (1 photon) = h × v

$v=\frac{c}{\lambda}$

$v=\frac{3 \times 10^{8}}{6.5 \times 10^{-7}}$

$v=4.41 \times 10^{14} \mathrm{s}^{-1}$

$\text { Substituting values in } \mathrm{E}(1 \text { photon })=\mathrm{h} \times v, \mathrm{h} \text { is Planck constant }=6.626 \times 10^{-34} \mathrm{J} \mathrm{s}$

$\mathrm{E}(1 \text { photon })=\left(6.626 \times 10^{-34}\right) \times\left(4.41 \times 10^{14}\right)=2.92 \times 10^{-19} \mathrm{J}$

$\text { The minimum number of photons }=\frac{E}{\mathrm{E}(1 \text { photon })}$

$\text { The minimum number of photons }=\frac{1.62 \times 10^{29}}{2.92 \times 10^{-19}}$

$\text { Hence the minimum number of photons }=5.55 \times 10^{37} \text { photons }$

Answer from: Quest

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