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A green plant synthesizes glucose by photosynthesis, as shown in the reaction $$6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)$$ Animals use glucose as a source of energy: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ If we were to assume that both these processes occur to the same extent in a cyclic process, what thermodynamic property must have a nonzero value?

Short Answer

Expert verified
The thermodynamic property that must have a nonzero value in a cyclic process of photosynthesis and glucose breakdown is the Gibbs free energy (ΔG), as it determines the spontaneity of a process. If ΔG ≠ 0, it means one of the reactions is spontaneous while the other is non-spontaneous, providing a driving force for the cyclic process to continue.

Step by step solution

01

Enthalpy (ΔH)

Enthalpy measures the heat exchange of a reaction under constant pressure. If the cyclic process has a zero enthalpy change, it would mean that the reactions are going in opposite directions, canceling out the heat exchange. This means both reactions have equal, but opposite, enthalpy changes. However, it doesn't necessarily imply that the process wouldn't be spontaneous.
02

Entropy (ΔS)

Entropy is a measure of the randomness or disorder of a system. For a cyclic process, if the entropy change (ΔS) is zero, it would imply that the reactions have equal, but opposite, changes in randomness or disorder. However, it also doesn't ensure if the process would be spontaneous or not.
03

Gibbs Free Energy (ΔG)

Finally, we have Gibbs free energy (ΔG), which determines the spontaneity of a process. If the Gibbs free energy change (ΔG) is positive, the process is non-spontaneous, while if it's negative, the process is spontaneous. In a cyclic process like the one given, if ΔG ≠ 0, then it means one of the reactions is spontaneous while the other is non-spontaneous, providing a driving force for the process to continue as a cycle. Therefore, the thermodynamic property that must have a nonzero value in a cyclic process of photosynthesis and glucose breakdown is the Gibbs free energy (ΔG).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Photosynthesis
Photosynthesis is the process by which green plants, algae, and some bacteria convert sunlight, water, and carbon dioxide into glucose and oxygen. This conversion is represented by the chemical equation: \[6 \mathrm{CO}_{2}(g) + 6 \mathrm{H}_{2}O(l) \longrightarrow \mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}(s) + 6 \mathrm{O}_{2}(g)\].

This process is the cornerstone of life on Earth as it provides the organic compounds and oxygen necessary for the survival of most organisms. In the context of thermodynamics, photosynthesis is an endothermic reaction, meaning it absorbs energy, specifically from the sun, to proceed. While it appears to defy the tendency of systems to maximize entropy, it's driven by the constant influx of solar energy.
Glucose Metabolism
Glucose metabolism includes various biochemical processes that break down glucose to generate energy for cellular activities. Animals metabolize glucose in a series of reactions known as cellular respiration, particularly glycolysis, the citric acid cycle, and oxidative phosphorylation. The equation \[\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}(s) + 6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g) + 6 \mathrm{H}_{2}O(l)\]
represents the catabolism of glucose where it combines with oxygen, leading to carbon dioxide, water, and the release of energy in the form of ATP. This exothermic reaction is essential for maintaining the life processes of organisms that are not capable of photosynthesis.
Thermodynamics
Thermodynamics is the study of heat and energy transfer, conversion, and the work done by or on systems during these processes. It's summed up by its laws, with the first law indicating the conservation of energy and the second law stating that the entropy of an isolated system always increases over time. In the context of our reactions, the first law relates to the conservation of energy between photosynthesis and glucose metabolism, while the second law guides the spontaneous flow of the reactions based on the increase of total entropy, albeit locally decreased during photosynthesis.
Entropy
Entropy measures the disorder or randomness of a system and is a fundamental aspect of the second law of thermodynamics. An increase in entropy reflects an increase in disorder and energy distribution. While both photosynthesis and glucose metabolism influence the system's entropy, the overall entropy change of the universe will always be positive, according to the second law. This principle allows us to understand why there is a continuous requirement for energy input, such as from sunlight, to maintain the cyclic order of processes like photosynthesis.

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Most popular questions from this chapter

When most biologic enzymes are heated, they lose their catalytic activity. The change Original enzyme \(\longrightarrow\) new form that occurs on heating is endothermic and spontaneous. Is the structure of the original enzyme or its new form more ordered (has the smaller positional probability)? Explain.

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2}\). Consider the following reactions and approximate standard free energy changes: $$\begin{array}{clr}\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array}$$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$

Consider the reaction $$2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\). b. Would the reaction be more spontaneous at high or low temperatures?

Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\) using the following data: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6}\) at \(600 . \mathrm{K}\) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37}\) at 600. \(\mathrm{K}\)

At 1 atm, liquid water is heated above \(100^{\circ} \mathrm{C}\). For this process, which of the following choices (i-iv) is correct for \(\Delta S_{\text {sur }}\) ? \(\Delta S\) ? \(\Delta S_{\text {univ }} ?\) Explain each answer. i. greater than zero ii. less than zero iii. equal to zero iv. cannot be determined

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