Chapter 24, Problem 29GQ

(a)

Interpretation Introduction

Interpretation:

The value of enthalpy change ${\Delta }_{\text{r}}{H}^{°}$ for the production of one mole of glucose by the process of photosynthesis at $25{\text{}}^{°}\text{C}$ has to be calculated.

  $6{\text{CO}}_{\text{2}}\left(\text{g}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)+6{\text{O}}_{\text{2}}\left(\text{g}\right)$

Concept introduction:

The change in the enthalpy of a reaction when the reactant is converted into product under standard conditions is called standard enthalpy of reaction.

The expression for standard enthalpy of reaction is,

${\text{Δ}}_{\text{r}}{H}^{\text{°}}=\sum n{\Delta }_f{H}^{\text{°}}\left(\text{products}\right)-\sum n{\Delta }_f{H}^{\text{°}}\left(\text{reactants}\right)$ (1)

Here, ${\Delta }_f{H}^{\text{°}}$ is the standard enthalpy of formation and $n$ is the number of moles of reactant and product in the balanced chemical reaction.

Explanation of Solution

The value of ${\Delta }_{\text{r}}{H}^{°}$ for the production of one mole of glucose by the process of photosynthesis at $25{\text{}}^{°}\text{C}$ is calculated below.

Given:

Refer to Appendix L for the values of standard enthalpy of formation.

The standard enthalpy of formation of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)$ is $-\text{1273}\text{.3 kJ/mol}$.

The standard enthalpy of formation of ${\text{O}}_{\text{2}}\left(\text{g}\right)$ is $\text{0 kJ/mol}$.

The standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$ is $-285.8\text{ kJ/mol}$.

The standard enthalpy of formation of ${\text{CO}}_{\text{2}}\left(\text{g}\right)$ is $-\text{393}\text{.5 kJ/mol}$.

The reaction involved in photosynthesis is,

  $6{\text{CO}}_{\text{2}}\left(\text{g}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)+6{\text{O}}_{\text{2}}\left(\text{g}\right)$

The ${\Delta }_{\text{r}}{H}^{°}$ can be calculated by the following expression,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{H}^{\text{°}}=\sum n{\Delta }_f{H}^{\text{°}}\left(\text{products}\right)-\sum n{\Delta }_f{H}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left[\begin{array}{l}\left({\text{1 mol C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{\text{°}}\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\right]+\\ \left({\text{6 mol O}}_2\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{\text{°}}\left[{\text{O}}_2\left(\text{g}\right)\right]\end{array}\right]-\\ \left[\begin{array}{l}\left({\text{6 mol CO}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{\text{°}}\left[{\text{CO}}_{\text{2}}\left(\text{g}\right)\right]+\\ \left({\text{6 mol H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{\text{°}}\left[{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substitute the value of ${\Delta }_f{H}^{\text{°}}$.

$\begin{array}{c}{\text{Δ}}_{\text{r}}{H}^{\text{°}}=\left[\begin{array}{l}\left[\begin{array}{l}\left({\text{1 mol C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\text{/mol-rxn}\right)\left(-\text{1273}\text{.3 kJ/mol}\right)+\\ \left({\text{6 mol O}}_2\left(\text{g}\right)\text{/mol-rxn}\right)\left(0\right)\end{array}\right]-\\ \left[\begin{array}{l}\left({\text{6 mol CO}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(-\text{393}\text{.5 kJ/mol}\right)+\\ \left({\text{6 mol H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{/mol-rxn}\right)\left(-285.8\text{ kJ/mol}\right)\end{array}\right]\end{array}\right]\\ =+2803kJ\text{/}mol\text{-}rxn\end{array}$

The value of ${\Delta }_{\text{r}}{H}^{°}$ for the production of one mole of glucose by the process of photosynthesis at $25{\text{}}^{°}\text{C}$ is $+2803kJ\text{/}mol\text{-}rxn$.

(b)

Interpretation Introduction

Interpretation:

The enthalpy change involved in producing one molecule of glucose by the process of photosynthesis at $25{\text{}}^{°}\text{C}$ has to be calculated.

Concept introduction:

The relationship between the number of moles and the number of molecules is,

$\text{Number of moles, }n=\frac{\text{number of molecules}\left(N\right)}{\text{Avagadro }\text{}\text{number}\left(N_A\right)}$

Also,

$1\text{ mole = 6}\text{.023}\times {\text{10}}^{23}\text{ molecules}$

Explanation of Solution

The enthalpy change involved in producing one molecule of glucose by the process of photosynthesis at $25{\text{}}^{°}\text{C}$ is calculated below.

Given:

The enthalpy change involved in the production of one mole of glucose by the process of photosynthesis at $25{\text{}}^{°}\text{C}$ is $\text{+2803 kJ/mol-rxn}$.

Since $1\text{ mole = 6}\text{.023}\times {\text{10}}^{23}\text{ molecules}$ therefore enthalpy change involved in the production of one molecule of glucose is,

 $\begin{array}{c}{\text{Δ}}_{\text{r}}{H}^{\text{°}}=\left(\frac{\text{2803 kJ}\left(\frac{1000\text{ J}}{1\text{ kJ}}\right)}{\text{mol}}\right)\left(\frac{1\text{ mol}}{6.023\times {10}^{23}\text{ molecule}}\right)\\ =4.654\times 10^{-18}J\text{/}molecule\end{array}$

The enthalpy change involved in producing one molecule of glucose by the process of photosynthesis is $4.654\times 10^{-18}J\text{/}molecule$.

(c)

Interpretation Introduction

Interpretation:

The energy of a photon of light having a wavelength $650\text{ nm}$ has to be calculated.

Concept introduction:

The energy of one photon $\left(E\right)$ is given as,

$\begin{array}{c}E=h\nu \\ =\frac{hc}{\lambda }\end{array}$ (2)

Here, $c$ is the speed of light, $\lambda$ is the wavelength of light, $h$ is Planks constant.

Value of $c=3.0\times {10}^8\text{ m}\cdot {\text{s}}^{-1}$ and $h=6.626\times {10}^{-34}\text{ J}\cdot {\text{s}}^{-1}$.

Explanation of Solution

The energy of a photon of light having wavelength $650\text{ nm}$ is calculated below.

Given:

The wavelength of light is $650\text{ nm}$.

The speed of light is $3.0\times {10}^8\text{ m}\cdot {\text{s}}^{-1}$.

Planks constant is $6.626\times {10}^{-34}\text{ J}\cdot {\text{s}}^{-1}$.

Substitute the values in equation (2).

$\begin{array}{c}E=\frac{\left(6.626\times {10}^{-34}\text{ J}\cdot {\text{s}}^{-1}\right)\left(3.0\times {10}^8\text{ m}\cdot {\text{s}}^{-1}\right)}{\left(650\text{ nm}\right)\left(\frac{1\text{ m}}{{10}^9\text{}\text{nm}}\right)}\\ =3.1\times 10^{-19}J\end{array}$

The energy of a photon of light having wavelength $650\text{ nm}$ is $3.1\times 10^{-19}J$.

(d)

Interpretation Introduction

Interpretation:

The absorption of one photon at $650\text{ nm}$ can lead to the production of one molecule of glucose or multiple photons must be absorbed has to be predicted.

Concept introduction: If the energy of one photon is greater than the energy required for the production of one mole of substance then single photon absorption is sufficient for the reaction to proceed else multiple photons must be absorbed.

Explanation of Solution

The energy of one photon calculated in part(c) is $\text{3}{\text{.1×10}}^{-\text{19}}\text{ J}$ and the energy required for the production of one molecule of glucose calculated in part(b) is $\text{4}{\text{.654×10}}^{-\text{18}}\text{ J/molecule}$. Since the energy required for the production of one molecule of glucose is greater than the energy of one photon thus multiple photons must be absorbed for the reaction to proceed.

Therefore multiple photons must be absorbed for the production of one molecule of glucose by photosynthesis reaction.