Question

Consider the following reaction:$$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$.Amounts of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}, \mathrm{H}_{2},\) and \(\mathrm{CO}_{2}\) are put into a flask so that the composition corresponds to an equilibrium position. If the CO placed in the flask is labeled with radioactive \(^{14} \mathrm{C},\) will \(^{14} \mathrm{C}\) be found only in CO molecules for an indefinite period of time? Explain.

Short answer

No, the radioactive \(^{14}\mathrm{C}\) will not be found only in CO molecules indefinitely. Due to the dynamic nature of chemical equilibrium, the forward and reverse reactions continue to occur at equal rates. As the \(^{14}\mathrm{C}\)-labeled CO molecules participate in the forward reaction, they will convert into \(^{14}\mathrm{C}\)-labeled CO₂ molecules. Thus, the \(^{14}\mathrm{C}\) is found in both CO and CO₂ molecules.

Step-by-step solution

Understand the dynamic nature of chemical equilibrium

Chemical equilibrium is reached when the rate of the forward reaction is equal to the rate of the reverse reaction. At this point, the concentrations of products and reactants remain constant. However, the reactions still continue to occur - reactants still convert into products and vice versa, maintaining the equilibrium state.

Consider the radioactive \(^{14}\mathrm{C}\) in the system

Initially, the radioactive \(^{14}\mathrm{C}\) is only present in CO molecules. However, since the forward and reverse reactions still occur at equilibrium, albeit at equal rates, there is a possibility that the \(^{14}\mathrm{C}\) will participate in the reaction and become distributed among different molecules.

Analyze the radioactive \(^{14}\mathrm{C}\) distribution in products

The given chemical reaction is: \(\mathrm{H}_{2}\mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) The radioactive \(^{14}\mathrm{C}\)-labeled CO molecules can participate in the forward reaction to form \(^{14}\mathrm{C}\) labeled CO₂ molecules: \(\mathrm{H}_{2}\mathrm{O}(g)+\mathrm{^{14}CO}(g) \rightarrow \mathrm{H}_{2}(g)+\mathrm{^{14}CO}_{2}(g)\)

Conclude whether \(^{14}\mathrm{C}\) remains only in CO molecules indefinitely

Since the equilibrium involves ongoing reactions, the radioactive \(^{14}\mathrm{C}\) will not remain only in CO molecules indefinitely. As CO molecules with \(^{14}\mathrm{C}\) participate in the forward reaction, they will convert into \(^{14}\mathrm{C}\)-labeled CO₂ molecules. Consequently, the \(^{14}\mathrm{C}\) is found not only in CO molecules but also in CO₂ molecules.

Key concepts

Chemical Reactions

Understanding chemical reactions is the foundation of studying chemistry. These are processes in which substances, known as reactants, transform into new substances called products. In the context of the given exercise,
\(\mathrm{H}_{2}\mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\)
is a chemical reaction where water vapor (\(\mathrm{H}_{2}\mathrm{O}(g)\)) and carbon monoxide (\(\mathrm{CO}(g)\)) are reactants that can combine to form hydrogen gas (\(\mathrm{H}_{2}(g)\)) and carbon dioxide (\(\mathrm{CO}_{2}(g)\)), the products.

This particular chemical reaction is reversible, meaning the reactants can form the products and the products can revert to the original reactants. This leads to a dynamic state when both forward and reverse reactions happen at the same rate, known as equilibrium. In this state, although the concentrations of reactants and products appear static to an observer, the substances are continuously interconverting. This dynamic essence of chemical reactions is crucial for both naturally occurring processes and synthetic applications.

Radioactive Isotopes

Radioactive isotopes are variants of elements that have unstable nuclei and can undergo radioactive decay, releasing energy in the process. The exercise involves a radioactive isotope,
\(^{14}\mathrm{C}\)
, which is a common tracer used in scientific studies due to its detectable radioactivity.

In the provided example,
\(^{14}\mathrm{C}\)
is initially part of the carbon monoxide molecule (\(\mathrm{CO}(g)\)) in a chemical reaction. Its radioactivity allows us to track its movement within the chemical system. As reactions proceed towards equilibrium, the radioactive isotope will not only illustrate the conversion between reactants and products but will also demonstrate the essential concept of dynamic equilibrium. The use of radioactive tracers, such as
\(^{14}\mathrm{C}\)
, has revolutionized our understanding of chemical processes by providing a way to visualize the invisible movements of atoms and molecules in real time.

Dynamic Equilibrium

Dynamic equilibrium is a balanced state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time. Despite this apparent constancy, the reaction is not static.

Using the provided exercise as an example, the presence of a radioactive isotope like
\(^{14}\mathrm{C}\)
offers a profound insight into equilibrium. It highlights that atoms and molecules are perpetually interchanging at an indivisible level. The movement of the radioactive isotope
\(^{14}\mathrm{C}\)
from CO to
\(^{14}\mathrm{CO}_{2}\)
and back, shows that the system is in constant motion, although macroscopically it seems unchanging. This principle can be extrapolated to understand natural systems at equilibrium, such as the balance of gases in the atmosphere.

Comprehending dynamic equilibrium enhances our ability to predict how systems will respond to changes in conditions, making it a powerful concept in fields like chemical engineering, environmental science, and pharmacology.