Chapter 20, Problem 11P

Summary Introduction

To calculate: The equilibrium constant for the reaction given in question at 25°C.

Introduction:

Chloroplast is a double membrane bound organelle present in green plants and algae. Chloroplast contains thylakoid membrane in which two photosystem units PS I and PS II are present. These photosystem units absorb sunlight and pass it to an antenna molecule to drive the photosynthesis process.

Explanation of Solution

To calculate the equilibrium constant for the reaction, first the standard reduction potential is calculated for two half reactions of the given reaction:

${\text{2H}}_{\text{2}}\text{O}\text{+}{\text{2NADP}}^{\text{+}}\to \text{2NADPH}\text{+}{\text{2H}}^{\text{+}}\text{+}{\text{O}}_{\text{2}}$

The two half reactions of the given equations are given below as:

${\text{NADP}}^{\text{+}}\text{+}{\text{2H}}^{\text{+}}\text{+}{\text{2e}}^{-}\to \text{NADPH;}{\text{E}}_{\text{1}}{{}^{\prime }}^{°}\text{=}\text{-0}\text{.342V}$

$\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{+}{\text{2H}}^{\text{+}}\text{+}{\text{2e}}^{-}\to {\text{H}}_{\text{2}}\text{O;}{\text{E}}_{\text{2}}{{}^{\prime }}^{°}\text{=}\text{0}\text{.816V}$

Now, the change in standard reduction potential can be calculated by using the formula,

$\text{Δ}{G^{\prime }}^{\text{°}}\text{=}-n\text{FΔ}{E^{\prime }}^{\text{°}}$

In the given formula, $\text{Δ}{G^{\prime }}^{\text{°}}$ determines the standard free energy change, ‘n’ is transfer of a four electrons. This transfer of four electrons is necessary for the production of one mole of oxygen molecule. F is Faraday constant ( 96.45 kJ/Vmol) and $\text{Δ}{E^{\prime }}^{\text{°}}$ is the difference between standard reduction potential between two half reactions.

Thus, the value of ${\text{Δ}{\text{E}}^{\prime }}^{\text{°}}$ can be calculated as:

$\begin{array}{c}{\text{Δ}}^{{\text{E}}^{\prime }}=0.816-(-\text{0}\text{.324)}\\ \text{=}-1.14\text{V}\end{array}$

Now, standard free energy change can be calculated as:

$\begin{array}{c}\text{Δ}{G^{\prime }}^{\text{°}}\text{=}-n\text{FΔ}{E^{\prime }}^{\text{°}}\\ =-4(\text{96}\text{.45}\text{kJ/V}\cdot \text{mol)}\text{×}(-1.14\text{V)}\\ \text{=}439.812\text{kJ/mol}\\ \simeq 440\text{kJ/mol}\end{array}$

Thus, standard free energy change is 440 kJ/mol.

Now, the equilibrium constant for the above given reaction is calculated under steady state. Thus, the equilibrium constant of the given reaction can be calculated by the formula given below:

$\text{Δ}G\text{=}-\text{RT}\text{ln}{K}^{\prime }eq$

In the given equation, $\text{Δ}G$ determines the free energy which is calculated as440 kJ/mol, and R and T are constants whose value is $\text{R}\text{=}\text{8}\text{.315}\text{J/molK}$ and $\text{T}\text{=}\text{25}\text{°C}/\text{298}\text{K}$.

Now, from above equation, ΔG can be calculated as:

$\begin{array}{c}\ln K^{\prime }eq=\frac{-\Delta G}{\text{RT}}\\ =\frac{\text{-(440kJ/mol)}}{\text{2}\text{.48kJ/mol}}\\ =-177\end{array}$

The natural logarithm can be calculated as:

$\begin{array}{c}{K}^{\prime }\text{eq}\text{=}{\text{e}}^{\text{-177}}\\ \text{=}\text{1}\text{.348}\text{×}{\text{10}}^{\text{-77}}\\ \text{=1}\text{.35}\text{×}{\text{10}}^{\text{-77}}\end{array}$

The equilibrium constant for the given reaction is 1.35×10^-77.

Conclusion

The equilibrium constant for the reaction at 25°C is $\underset{\_}{1.35\times {10}^{-77}}.$.

Summary Introduction

To determine: The way in which chloroplast overcome the unfavorable equilibrium.

Introduction:

Chloroplast is a double membrane bound organelle present in green plants and algae. Chloroplast contains thylakoid membrane in which two photosystem units PS I and PS II are present. These photosystem units absorb sunlight and pass it to an antenna molecule to drive the photosynthesis process.

Explanation of Solution

The calculated equilibrium constant value is 1.35×10^-77. A large equilibrium constant value is very unfavorable to drive a reaction. To overcome the equilibrium constant barrier, chloroplast would use both photosystem units (PS I, and PS II). The striking of light in both photosystem units would reduce the equilibrium constant, and help the chloroplast to overcome the equilibrium barrier. Thus, in chloroplast light energy input overcomes this barrier.