Bioremediation is the use of microorganisms to degrade environmental pollutants. Many pollutants contain only carbon and hydrogen (oil being one example). The chemical reactions are complicated, but in general the microorganisms react the pollutant hydrocarbon with \(\mathrm{O}_{2}\) to produce \(\mathrm{CO}_{2}\) and other carbon-containing compounds that are incorporated into the organism's biomass. How would increasing levels of \(\mathrm{CO}_{2}\) in the environment affect the bioremediation reaction?

Short Answer

Expert verified
Increasing the levels of \(\mathrm{CO}_{2}\) in the environment would negatively affect the bioremediation reaction by slowing down the process and reducing the rate at which pollutant hydrocarbons are degraded. This is due to the application of Le Chatelier's Principle, which states that the reaction system will adjust to minimize the effects of increased \(\mathrm{CO}_{2}\) concentration, causing the equilibrium to shift to the left.

Step by step solution

01

Understand the Bioremediation Reaction

Bioremediation is the process in which microorganisms degrade environmental pollutants, such as hydrocarbons. The general reaction for bioremediation can be represented as follows: \[ {\rm Hydrocarbon \: (Pollutant) + \: O_2 \xrightarrow{Microorganism} CO_2 + Biomass} \] In other words, microorganisms use oxygen to convert hydrocarbon pollutants into carbon dioxide and their own biomass.
02

Apply Le Chatelier's Principle

By applying Le Chatelier's Principle, when the concentration of \(\mathrm{CO}_{2}\) in the environment increases, the reaction system will adjust to minimize the effects of this change.
03

Determine the Effect of Increased CO2 on the Reaction

Increasing the levels of \(\mathrm{CO}_{2}\) in the environment will cause a decrease in the rate of bioremediation, as the reaction will try to counteract the increase in \(\mathrm{CO}_{2}\) concentration. As a result, the equilibrium will shift to the left, reducing the rate at which the pollutant hydrocarbon is degraded, and slowing down the overall bioremediation process. This is an example of how rising \(\mathrm{CO}_{2}\) levels can have negative effects on natural processes that help maintain environmental balance. In conclusion, increasing the levels of \(\mathrm{CO}_{2}\) in the environment would negatively affect the bioremediation reaction by slowing down the process and reducing the rate at which pollutant hydrocarbons are degraded.

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Most popular questions from this chapter

The reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) has \(K_{p}=0.0870\) at \(300^{\circ} \mathrm{C}\). A flask is charged with \(0.50 \mathrm{~atm} \mathrm{PCl}_{3}, 0.50 \mathrm{~atm} \mathrm{Cl}_{2}\), and 0.20 atm \(\mathrm{PCl}_{5}\) at this temperature. (a) Use the reaction quotient to determine the direction the reaction must proceed to reach equilibrium. (b) Calculate the equilibrium partial pressures of the gases. (c) What effect will increasing the volume of the system have on the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture? (d) The reaction is exothermic. What effect will increasing the temperature of the system have on the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

A mixture of \(\mathrm{H}_{2}, \mathrm{~S},\) and \(\mathrm{H}_{2} \mathrm{~S}\) is held in a \(1.0-\mathrm{L}\) vessel at \(90{ }^{\circ} \mathrm{C}\) and reacts according to the equation: $$\mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$$ At equilibrium the mixture contains \(0.46 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(0.40 \mathrm{~g}\) \(\mathrm{H}_{2}\) (a) Write the equilibrium-constant expression for this reaction. (b) What is the value of \(K_{c}\) for the reaction at this temperature? (c) Why can we ignore the amount of \(\mathrm{S}\) when doing the calculation in part (b)?

When the following reactions come to equilibrium, does he equilibrium mixture contain mostly reactants or mostly broducts? a) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) ; K_{c}=1.5 \times 10^{-10}\) b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) ; K_{p}=2.5 \times 10^{9}\)

(a) How does a reaction quotient differ from an equilibrium constant? (b) If \(Q_{c}

Gaseous hydrogen iodide is placed in a closed container at \(425^{\circ} \mathrm{C},\) where it partially decomposes to hydrogen and iodine: \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) .\) At equilibrium it is found that \([\mathrm{HI}]=3.53 \times 10^{-3} \mathrm{M},\left[\mathrm{H}_{2}\right]=4.79 \times 10^{-4} \mathrm{M}\) and \(\left[\mathrm{I}_{2}\right]=4.79 \times 10^{-4} \mathrm{M} .\) What is the value of \(K_{c}\) at this temperature?

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