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Chapter 13: Problem 72

The formation of glucose from water and carbon dioxide is one of the more important chemical reactions in the world. Plants perform this reaction through the process of photosynthesis, creating the base of the food chain: $$ 6 \mathrm{H}_{2} \mathrm{O}(g)+6 \mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) $$ At a particular temperature, the following equilibrium concentrations were found: \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=7.91 \times 10^{-2} M,\left[\mathrm{CO}_{2}(g)\right]=\) \(9.3 \times 10^{-1} M\), and \(\left[\mathrm{O}_{2}(g)\right]=2.4 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

Short Answer

Expert verified
The value of K for the photosynthesis reaction at this temperature is approximately \(2.74\times10^{-24}\).

Step by step solution

01

Write the expression for the equilibrium constant (K)

The equilibrium constant (K) is the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of its stoichiometric coefficient in the balanced chemical equation. Using the given equation for photosynthesis, we can write the expression for K as: \[K = \frac{[\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}]^1[\mathrm{O}_{2}]^6}{[\mathrm{H}_{2}\mathrm{O}]^6 [\mathrm{CO}_{2}]^6}\]
02

Plug in the given equilibrium concentrations

Now, substitute the equilibrium concentrations given in the problem (\([\mathrm{H}_{2}\mathrm{O}] = 7.91\times10^{-2} M\), \([\mathrm{CO}_{2}] = 9.3\times10^{-1} M\), and \([\mathrm{O}_{2}] = 2.4\times10^{-3} M\)) into the expression for K. Note that we are not given the concentration of glucose, so we can assume it remains constant and does not affect the calculation of K. \[K = \frac{[\mathrm{O}_{2}]^6}{[\mathrm{H}_{2}\mathrm{O}]^6 [\mathrm{CO}_{2}]^6} = \frac{(2.4\times10^{-3})^6}{(7.91\times10^{-2})^6 (9.3\times10^{-1})^6}\]
03

Calculate K

Now, we can calculate the equilibrium constant using the substituted values: \[ K = \frac{(2.4\times10^{-3})^6}{(7.91\times10^{-2})^6 (9.3\times10^{-1})^6} \approx 2.74\times10^{-24} \] Thus, the value of K for the photosynthesis reaction at this temperature is approximately \(2.74\times10^{-24}\).

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Key Concepts

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

Photosynthesis
Photosynthesis is the remarkable process by which green plants, algae, and some bacteria convert light energy into chemical energy stored in glucose. This process occurs mainly in the chloroplasts of plant cells. During photosynthesis, plants absorb carbon dioxide from the air and water from the soil. Then, using sunlight, they convert these into glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) and oxygen (\(\text{O}_2\)).

Photosynthesis can be divided into two main stages:
  • Light-dependent reactions: These take place in the thylakoid membranes and require sunlight to split water molecules, releasing oxygen and transferring energy to ATP and NADPH.
  • Calvin cycle: This stage occurs in the stroma, where ATP and NADPH are used to synthesize glucose from carbon dioxide and water.
Understanding photosynthesis is crucial because it forms the base of the food chain, supporting life on Earth.
Chemical Equilibrium
Chemical equilibrium refers to the state where the concentrations of reactants and products remain constant over time in a reversible chemical reaction. At equilibrium, the forward and backward reactions occur at the same rate, leading to a balanced system.

For the photosynthesis reaction \(6 \text{H}_{2} \text{O}(g)+6 \text{CO}_{2}(g) \rightleftharpoons \text{C}_{6} \text{H}_{12} \text{O}_{6}(s)+6 \text{O}_{2}(g)\), the system reaches equilibrium when the rate of glucose and oxygen formation equals the rate at which they convert back into water and carbon dioxide.
  • Dynamic Equilibrium: Even though concentrations remain stable, the reactions continue to occur.
  • Le Chatelier's Principle: This principle explains how changes in concentration, temperature, or pressure can shift the equilibrium, affecting the amounts of reactants and products.
Chemical equilibrium is essential in understanding how reactions behave under different conditions and predicting the outcomes of changes in a system.
Reaction Stoichiometry
Reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical equation. It provides a way to calculate the amounts of substances involved in chemical reactions.

For the balanced photosynthesis equation, the stoichiometry tells us that:
  • 6 molecules of water react with 6 molecules of carbon dioxide.
  • This produces 1 molecule of glucose and 6 molecules of oxygen.
Stoichiometry is used to determine how changes in the quantity of one substance affect others in the reaction. Understanding these relationships allows scientists to predict amounts, optimize reactions, and calculate yields.
Glucose Formation
Glucose formation via photosynthesis is a critical process that sustains life. As a simple sugar, glucose serves as an energy source for plants and, indirectly, for animals that consume plants.

In photosynthesis, glucose is synthesized in the chloroplasts through a series of reactions using the Calvin cycle, which combines carbon dioxide and energy from ATP and NADPH. The process of forming glucose can be represented by the equation:\[6 \text{H}_2\text{O} + 6 \text{CO}_2 \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6 \text{O}_2\]
  • Energy Storage: The glucose formed is used to store energy for the plant, which can be converted into starch or cellulose.
  • Role in Metabolism: Glucose is a fundamental molecule in metabolic pathways like glycolysis and cellular respiration.
Through glucose formation, photosynthesis not only fuels plant growth but also impacts the entire ecosystem by providing food and oxygen.

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Most popular questions from this chapter

At \(125^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.25\) for the reaction $$ 2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ A 1.00-L flask containing \(10.0 \mathrm{~g} \mathrm{NaHCO}_{3}\) is evacuated and heated to \(125^{\circ} \mathrm{C}\). a. Calculate the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) after equilibrium is established. b. Calculate the masses of \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) present at equilibrium. c. Calculate the minimum container volume necessary for all of the \(\mathrm{NaHCO}_{3}\) to decompose.

An important reaction in the commercial production of hydrogen is $$ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) $$ How will this system at equilibrium shift in each of the five following cases? a. Gaseous carbon dioxide is removed. b. Water vapor is added. c. In a rigid reaction container, the pressure is increased by adding helium gas. d. The temperature is increased (the reaction is exothermic). e. The pressure is increased by decreasing the volume of the reaction container.

For a typical equilibrium problem, the value of \(K\) and the initial reaction conditions are given for a specific reaction, and you are asked to calculate the equilibrium concentrations. Many of these calculations involve solving a quadratic or cubic equation. What can you do to avoid solving a quadratic or cubic equation and still come up with reasonable equilibrium concentrations?

Le Ch芒telier's principle is stated (Section \(13.7\) ) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.

Nitric oxide and bromine at initial partial pressures of \(98.4\) and \(41.3\) torr, respectively, were allowed to react at \(300 . \mathrm{K}\). At equilibrium the total pressure was \(110.5\) torr. The reaction is $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ a. Calculate the value of \(K_{\mathrm{p}}\). b. What would be the partial pressures of all species if \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), both at an initial partial pressure of \(0.30 \mathrm{~atm}\), were allowed to come to equilibrium at this temperature?
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