Question

10a. Calculate AH° for each of the following 4 reactions without considering the energy provided by solar radiation. Use the information to explain why light is required to drive Reactions 1 and 3. In addition to standard molar enthalpies of formation for ground state substances , take AH(O*, g) 437 kJ mol1, AH(O2*, g) = 90 kJ mol1, AH°q(OH, g) 38.95 kJ mol1 O(g)NO(g) > 0з(g) O(g) O2(g)* NO2(g) O(g)O2(g) Оз(g) O(g)H20(g) -»20H(g) 14 24 34 44 10b. Determine the maximum wavelength of the photon required to make reactions 1 and 3 go.

Answer 1

Hi,

Hope you are doing well.

__________________________________________________________________________

PART (10.a)

Given that,

AH°f(O*, g) = 437 kJ mol-1 AHO:(02*, g) = 90 kJ mol-1 AH°(OH, g) = 38.95 kJ mol-1

__________________________________________________________________________

Reaction 1:

NO2(g) → O(g) + NO(g)

We know that, $\Delta H^{o}$ of this reaction is given by,

ΔΗΣ ΔΗ(products) –Σ ΔΗ(products)

:. ΔΗ° = ΔΗ(Ο) + ΔΗ(ΝΟ) – ΔΗ(NO2(g)

:. AH° = 437 kJ/mol +90.25 kJ/mol ] - [33.18 kJ/mol

:. AH = +494.07 kJ/mol

__________________________________________________________________________

Reaction 2:

O(g) + O2(g) → O3(g)

We know that, $\Delta H^{o}$ of this reaction is given by,

ΔΗΣ ΔΗ(products) –Σ ΔΗ(products)

:. ΔΗ° = ΔΗ(Ο39) – ΔΗ(Οq) + ΔΗ? (O2(g)

:: AH° = [142.7 kJ/mol – [437 kJ/mol +0 kJ/mol

.:. ΔΗ =-294.3 kJ/mol

__________________________________________________________________________

Reaction 3:

O3(g) → O(g)* + O2(g)*

We know that, $\Delta H^{o}$ of this reaction is given by,

ΔΗΣ ΔΗ(products) –Σ ΔΗ(products)

.:. ΔΗ = ΔΗ? (O) + ΔΗ(Ο39) – ΔΗ(Ο39)

:. AH° = 437 kJ/mol + 90 kJ/mol)] – [142.7 kJ/mol

.:. ΔΗ = +384.3 kJ/mol

__________________________________________________________________________

Reaction 4:

O(g)* + H2O(g) + 2OH(g)

We know that, $\Delta H^{o}$ of this reaction is given by,

ΔΗΣ ΔΗ(products) –Σ ΔΗ(products)

... ΔΗ° = 2.ΔΗ (OH0) - ΔΗ(Ο) + ΔΗ? (Η2Ο)

:. AH° = 2 38.95 kJ/mol – 437 kJ/mol +-241.82 kJ/mol

:. AH = -117.28 kJ/mol

__________________________________________________________________________

From the above calculation, we have $\mathbf{\Delta H^{o}}$ value for the reaction 1 and 3 is positive while  $\mathbf{\Delta H^{o}}$ value for the reaction 2 and 4 is negative.

For a reaction having negative $\mathbf{\Delta H^{o}}$ value will release energy and a reaction having positive $\mathbf{\Delta H^{o}}$ value will absorb energy.

Therefore reaction 1 and 3 requires energy to carry out the reaction. This energy will be provided by light (light energy).

__________________________________________________________________________

PART (10.b)

For Reaction 1:

We have,

AH = +494.07 kJ/mol

For 1 mol, Energy required is,

E=nX AH° = 1 mol x 494.07 kJ/mol = 494.07 kJ

We know that, Energy of the photon is given by,

$E=\frac{hC}{\lambda}$

Where we have,

Planck's constant,

h = 6.626 x 10-34 J.s

Speed of light,

C = 3 x 108 m/s

Therefore, Wavelength of the photon is given by,

d-hc _ 6.626 x 10-62.5 x 3 025 * 3 * 10% m/s 494.07 x 103 J

= 4.0233 x 10-31 m

__________________________________________________________________________

For Reaction 3:

We have,

AH = +384.3 kJ/mol

For 1 mol, Energy required is,

E=nX AH° = 1 mol x 384.3 kJ/mol = 384.3 kJ

We know that, Energy of the photon is given by,

$E=\frac{hC}{\lambda}$

Where we have,

Planck's constant,

h = 6.626 x 10-34 J.s

Speed of light,

C = 3 x 108 m/s

Therefore, Wavelength of the photon is given by,

hc ta 6.626 X 10-34 J.s x 3 x 108 m/s 384.3 x 103 J

X= 5.1725 x 10-31 m ​​​​​​​

__________________________________________________________________________

Hope this helped for your studies. Keep learning. Have a good day.

Feel free to clear any doubts at the comment section.

Please don't forget to give a thumps up.

Thank you. :)