Chemistry, 29.07.2019 04:00 deena7

Carbon dioxide gas and liquid water form solid glucose (c6h12o6) and oxygen gas during photosynthesis. chlorophyll absorbs light in the 600-700 nm region. a) write a balanced thermochemical equation for the formation of 1.00 mol of glucose. i know that it goes like this: 6co2(g)+6h2o(> c6h12o6(s)+6o2( this is already balanced but not yet considered a "thermochemical equation" b) what is the minumum number of photons with y=680 nm needed to form 1.00 mol of gluscose?

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Answer from: mimi19374

A thermochemical equation is the one which includes heat of the reaction

Heat of reaction is also known as "enthalpy of reaction" ( ΔH rxn) and it is the amount of heat absorbed or evolved during the reaction.

The balanced equation for the reaction can be written as

$6 CO_{2}(g)+ 6 H_{2}O(l)\rightarrow C_{6}H_{12}O_{6}(s)+ 6O_{2}(g)$

Let us find ΔH rxn for the given reaction

We need heat of formation of the reactants and products. We will use standard reference table to get this data.

From standard table of enthalpies, we have

$H_{f}^{0} Glucose (s) = -1273.3kJ/mol$

$H_{f}^{0} CO_{2}(g)= -393.5 kJ/mol$

$H_{f}^{0} H_{2}O(l)= -285.8 kJ/mol$

$H_{f}^{0} O_{2}(g)= 0$

Enthalpy of reaction can be calculated using following formula

$\bigtriangleup H^{0} _{rxn}= \sum H_{f} (products) - \sum H_{f} ( reactants)$

Let us plug in the standard enthalpy of formation values we found out from reference table.

$\bigtriangleup H^{0} _{rxn}= [-1273.3 kJ.mol + 0]- [6\times (-393.5kJ/mol) + 6\times (-285.8 kJ/mol)]$

$\bigtriangleup H^{0} _{rxn}= [-1273.3 kJ.mol]- [-4075.8kJ/mol)]$

$\bigtriangleup H^{0} _{rxn}= -1273.3 kJ.mol+ 4075.8kJ/mol$

$\bigtriangleup H^{0} _{rxn}= 2802.5 kJ/mol$

The thermochemical equation for the given reaction can be written as

$6 CO_{2}(g)+ 6 H_{2}O(l)\rightarrow C_{6}H_{12}O_{6}(s)+ 6O_{2}(g).....\bigtriangleup H_{rxn} = 2802.5 kJ/mol$

The wavelength of the light absorbed by chlorophyll is 680 nm.

Let us find the energy of one photon having wavelength = 680 nm.

The relationship between energy and wavelength is given by the following equation.

$E = \frac{h\times c}{\lambda }$

Where E = energy of the photon

h = Plank's constant = 6.626 x 10⁻³⁴ J.s

c = velocity of light = 3.00 x 10⁸ m/s

λ = wavelength of light = 680 nm.

We need the wavelength in meters.

$680 nm\times \frac{1 m}{10^{9}nm}= 6.80 \times 10^{-7}m$

λ = 6.80 x 10⁻⁷ m

Let us find E now.

$E = \frac{6.626\times 10^{-34}J.s\times 3.00 \times 10^{8}m/s}{6.80 \times 10^{-7}m}$

$E = \frac{1.9878 \times 10^{-25}}{6.80\times 10^{-7}}J$

$E = 2.92 \times 10^{-19}J$

Let us convert this to kJ.

$2.92 \times 10^{-19}J \times \frac{1kJ}{1000 J}= 2.92 \times 10^{-22}kJ$

Energy of 1 photon = 2.92 x 10⁻²² kJ

The total amount of energy absorbed during the photosynthesis reaction is 2802.5 kJ

Number of photons = Total energy absorbed / Energy of 1 photon

Number of photons = $\frac{2802.5 kJ}{2.92 \times 10^{-22}kJ}$

Number of photons = 9.60 x 10²⁴

The minimum number of photons needed to form 1.0 mol of glucose is 9.60 x 10²⁴

Answer from: Quest

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

Answer from: Quest

consider the reaction below.

no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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Carbon dioxide gas and liquid water form solid glucose (c6h12o6) and oxygen gas during photosynthesi...

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