Answer:
(a)The balanced thermochemical equation for formation of 1.00 mol of glucose is:[tex]6 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+6 \mathrm{O}_{2}[/tex]
(b) The minimum number of photons with λ = 680 nm needed to prepare 1.00 mol of glucose is [tex]5.55 \times 10^{37} \text { photons }[/tex]
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 : [tex]6 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+6 \mathrm{O}_{2}[/tex]
(b) According to theory of special relativity which expresses fact in equation about mass and energy is: [tex]\mathbf{E}=\mathbf{m} \mathbf{c}^{2} \text { and here } \mathrm{c}^{2} \text { is speed of light }[/tex]
Here λ = 680 nm,
[tex]\mathrm{n}\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)=1 \mathrm{mol}[/tex]
Molar mass of glucose = 180.156 g/mol
[tex]\text { Therefore } m=180.156 \mathrm{g}, \mathrm{c}=3 \times 10^{8}[/tex]
Substituting values in above equation
[tex]\mathrm{E}=180.156 \times\left(3 \times 10^{8}\right)^{2}=1.62 \times 10^{29} \mathrm{J}[/tex]
Hence it is known that
E (1 photon) = h × v
[tex]v=\frac{c}{\lambda}[/tex]
[tex]v=\frac{3 \times 10^{8}}{6.5 \times 10^{-7}}[/tex]
[tex]v=4.41 \times 10^{14} \mathrm{s}^{-1}[/tex]
[tex]\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}[/tex]
[tex]\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}[/tex]
[tex]\text { The minimum number of photons }=\frac{E}{\mathrm{E}(1 \text { photon })}[/tex]
[tex]\text { The minimum number of photons }=\frac{1.62 \times 10^{29}}{2.92 \times 10^{-19}}[/tex]
[tex]\text { Hence the minimum number of photons }=5.55 \times 10^{37} \text { photons }[/tex]
