Chapter 15, Problem 15.133QP

Photosynthesis can be represented by

$\begin{array}{l}{\text{6CO}}_{\text{2}}\left(g\right)+{\text{6H}}_{\text{2}}\text{O}\left(l\right)\rightleftarrows {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)+{\text{6O}}_{\text{2}}\left(g\right)\\ \Delta H^{\circ }=\text{2801kJ/mol}\end{array}$

Explain how the equilibrium would be affected by the following changes: (a) partial pressure of CO₂ is increased, (b) O₂ is removed from the mixture, (c)C₆H₁₂O₆(glucose) is removed from the mixture, (d) more water is added, (e) a catalyst is added, (f) temperature is decreased.

Interpretation Introduction

Interpretation:

The effect of equilibrium changes (or) directions should be identified given the enthalpy of $\text{(ΔH°)}$  photochemical reactions with following statements.

Concept Introduction:

Concept of equilibrium process: The any system at equilibrium subject to change in concentration, temperature, volume or pressure then the system readjusts itself to partly counteract the effect of the applied change and new equilibrium is formed. This equilibrium explain to simple way, a system equilibrium is distributed the system will adjust itself in such a way that the effect of the change will be reduced.

Photo chemical equilibrium: These processes are photostationary state of a reversible reaction is the equilibrium chemical composition under a specific kind of electromagnetic (UV) irradiation.

Thermal reactions: This is one type of chemical reactions it release energy by light or heat it is called exothermic reaction, several chemical reactions are accompanied by the absorption of heat this type of process endothermic reactions.

Forward Reaction: This type of reaction has involved irreversible, if obtained product cannot be converted back in to respective reactants under the same conditions. Backward Reaction: This type of reaction process involved a reversible, if the products can be converted into a back to reactants.

Answer to Problem 15.133QP

The factors effects of temperature, pressure and decreasing of reactant given the statement of photo chemical equilibrium (a-f) are showed below.

$\begin{array}{l}{\text{6CO}}_{\text{2}}\text{(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}\text{+}{\text{6O}}_{\text{2}}\text{(g)}\text{ΔH}°=2801\text{KJ/mol}\\ \text{Given the statements (a-f)}\\ \text{a)}\text{.}{\text{6CO}}_{\text{2}}\text{(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}\text{+}{\text{6O}}_{\text{2}}\text{(g)}\\ [Increa\sin gpressure]\\ b).{\text{6CO}}_{\text{2}}\text{(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}\text{+}{\text{6O}}_{\text{2}}\text{(g)}\\ [RemovedO2formthereaction]\\ c).{\text{6CO}}_{\text{2}}\text{(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}\text{+}{\text{6O}}_{\text{2}}\text{(g)}\\ [Removed\text{glucose}formthereaction]\\ d).{\text{6CO}}_{\text{2}}\text{(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}\text{+}{\text{6O}}_{\text{2}}\text{(g)}\\ [Addedforsomereac\tan t]\\ e).{\text{6CO}}_{\text{2}}\text{(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}\text{+}{\text{6O}}_{\text{2}}\text{(g)}\\ \text{[Added}\text{for}\text{some}\text{amount}\text{of}\text{catalyst]}\\ f).{\text{6CO}}_{\text{2}}\text{(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}\text{+}{\text{6O}}_{\text{2}}\text{(g)}\\ [Temprature\text{decresing]}\end{array}$

Explanation of Solution

To Identify: Given the photo chemical equilibrium reactions (a-c) are the directions should be identified.

Write and analyze the following equilibrium reactions.

$\begin{array}{l}\text{a)}\text{.}{\text{6CO}}_{\text{2}}\text{(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}\text{+}{\text{6O}}_{\text{2}}\text{(g)}----[\text{Equilibrium}\text{shfit}\text{to right}]\\ [Increa\sin gpressureKp]\\ b).{\text{6CO}}_{\text{2}}\text{(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}\text{+}{\text{6O}}_{\text{2}}\text{(g)}----[\text{Equilibrium}\text{shfit}\text{to right}]\\ [Removed{O}_{2}formthereaction]\\ c).{\text{6CO}}_{\text{2}}\text{(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}\text{+}{\text{6O}}_{\text{2}}\text{(g)}----[Equilibriumnochange]\\ [Removed\text{glucose}formthereaction]\\ \end{array}$

Given the photo chemical equilibrium reactions involved some equilibrium steps (a-c) are fallows,

Pressure: Given the photo chemical reaction, if we increased the amount of pressure, the equilibrium directions migrated to left side to right side. Because product this reaction process will decrease the number of each moles of gas. The product glucose formation also increases.

Removed O₂: Given the photochemical reactions are, removed from some amount of (O₂) in product formation side, the total equilibrium process will be migrated to right side.

Removed glucose:  If we removed from the product, given the equilibrium process it can does not affected the equilibrium process.

To Identify: Given the photo chemical equilibrium reactions (d-f) are the directions should be identified.

Write and analyze the following equilibrium reactions.

$\begin{array}{l}d).{\text{6CO}}_{\text{2}}\text{(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}\text{+}{\text{6O}}_{\text{2}}\text{(g)}------[Equilibrium\text{unaffected]}\\ [Addedforsomereac\tan t]\\ e).{\text{6CO}}_{\text{2}}\text{(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}\text{+}{\text{6O}}_{\text{2}}\text{(g)}-----[Equilibrium\text{unaffected}]\\ \text{[Added}\text{for}\text{some}\text{amount}\text{of}\text{catalyst]}\\ f).{\text{6CO}}_{\text{2}}\text{(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}\text{+}{\text{6O}}_{\text{2}}\text{(g)}-----[Equilibrium\text{shifts}\text{to left}]\\ [Temprature\text{decresing]}\end{array}$

Given the photo chemical equilibrium reactions involved some equilibrium steps (d-f) are fallows.

Adding of reactant: Consider the reaction (d), if added a small amount of hemoglobin ( ${\text{CO}}_{\text{2}}{\text{and H}}_{\text{2}}\text{O}$) for product formation, total equilibrium does not affected and also this process also no any changes,  towards the left, right side Particularly reaction successfully can’t proceeds in right side.

Role of catalyst: If added some of catalyst in this reaction, the equilibrium reactions does not affected any changes. Because this reaction undergoes for gas, liquid phase, so reaction successfully can’t proceeds in right side.

Temperature: This reaction conversion reaction is endothermic reaction (heat absorption) and it equilibrium moved into left side, when temperature decreased, so this reaction get negative $\text{(-}\Delta \text{H)}$ values and its equilibrium constant (Kc) become a lesser values. Hence the concentrations of reactant increase while that product glucose concentration decrease.

Conclusion

The different statements are analyzed and discussed given the photochemical equilibrium reactions and respective $\text{(ΔH)}$ values