Problem 14 - Chemistry The Science in Context

Question

A mixture of \(^{13} \mathrm{CO},^{12} \mathrm{CO}_{2},\) and \(\mathrm{O}_{2}\) in a sealed reaction vessel was used to follow the reaction $$2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)$$ Analysis of the reaction mixture after 1 day revealed the presence of compounds with molar masses 28,29,32,44 and \(45 \mathrm{g} / \mathrm{mol} .\) Identify the compounds and account for their appearance.

Step-by-Step Solution

Answer

Question: Identify the compounds present in the sealed reaction vessel after 1 day and explain why they are present in the mixture. Answer: The compounds present in the sealed reaction vessel after 1 day are \(^{12} \mathrm{CO}\), \(^{13} \mathrm{CO}\), \(\mathrm{O}_{2}\), \(^{12} \mathrm{CO}_{2}\), and \(^{13} \mathrm{CO}_{2}\). They are present due to the reversibility of the given chemical reaction and the fact that the mixture is in a sealed vessel.

Step 1: List the initial compounds with their molecular weights

In the initial mixture, we have \(^{13} \mathrm{CO}\), \(^{12} \mathrm{CO}_{2}\), and \(\mathrm{O}_{2}\). The molecular weights of these compounds are as follows: - \(^{13} \mathrm{CO}\): 12 (C) + 1 (extra neutron) + 16 (O) = 29 g/mol - \(^{12} \mathrm{CO}_{2}\): 12 (C) + 2 * 16 (O) = 44 g/mol - \(\mathrm{O}_{2}\): 2 * 16 (O) = 32 g/mol

Step 2: Determine the products of the chemical reaction and their molecular weights

The balanced chemical equation provided is: \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)\). Since \(^{13} \mathrm{CO}\) and \(^{12} \mathrm{CO}_{2}\) are already in the initial mixture, we can assume that the products formed would have molecular weights of 28 and 45 g/mol. These can be: - \(^{12} \mathrm{CO}\): 12 (C) + 16 (O) = 28 g/mol - \(^{13} \mathrm{CO}_{2}\): 12 (C) + 1 (extra neutron) + 2 * 16 (O) = 45 g/mol

Step 3: Identify the compounds present after 1 day

We are given that the mixture after 1 day contains compounds with molar masses of 28, 29, 32, 44, and 45 g/mol. Comparing these values with the molecular weights calculated in Steps 1 and 2, we can identify the following compounds: - 28 g/mol: \(^{12} \mathrm{CO}\) - 29 g/mol: \(^{13} \mathrm{CO}\) - 32 g/mol: \(\mathrm{O}_{2}\) - 44 g/mol: \(^{12} \mathrm{CO}_{2}\) - 45 g/mol: \(^{13} \mathrm{CO}_{2}\)

Step 4: Explain the appearance of these compounds in the mixture

As the reaction takes place, some of the initial \(^{13} \mathrm{CO}\) and \(\mathrm{O}_{2}\) react to form \(^{13} \mathrm{CO}_{2}\), while some of the initial \(^{12} \mathrm{CO}_{2}\) reacts to form \(^{12} \mathrm{CO}\). This is due to the reversibility of the reaction, as represented by the double arrows in the balanced equation: \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)\). Since the reaction is happening in a sealed vessel, none of the substances can escape, and so they all remain in equilibrium. In conclusion, after 1 day, we have identified the following compounds in the sealed reaction vessel: \(^{12} \mathrm{CO}\), \(^{13} \mathrm{CO}\), \(\mathrm{O}_{2}\), \(^{12} \mathrm{CO}_{2}\), and \(^{13} \mathrm{CO}_{2}\). They are present due to the reversibility of the given chemical reaction and the fact that the mixture is in a sealed vessel.

Key Concepts

Isotopes in Chemistry

Isotopes are variants of a particular chemical element that differ in neutron number, and consequently, in nuclear mass. While all isotopes of an element have the same number of protons, contributing to their identical atomic number, the difference in the number of neutrons leads to distinct atomic masses. Understanding isotopes is crucial when analyzing chemical reactions involving elements that exist in multiple isotopic forms.

For example, carbon, the element central to the provided exercise, has several isotopes with carbon-12 (({}^{12})C) and carbon-13 (({}^{13})C) being the most stable and naturally occurring. The presence of ({}^{13})CO in the reaction vessel indicates that isotopic variants of compounds are participating in the chemical reaction, resulting in products with different molar masses. Identifying isotopes can impact the interpretation of a reaction's outcome as the mass shift due to isotopes becomes important in calculating the molar mass and identifying the compounds.

Molar Mass Calculation

Molar mass, often measured in grams per mole (g/mol), is a physical property defined as the mass of one mole of a substance. It corresponds to the atomic mass for an element expressed in atomic mass units (amu). In a chemical equation, calculating molar masses of reactants and products allows us to predict the mass of substances involved in the reactions and understand the stoichiometry - the quantitative relationship between reactants and products.

In the context of the given exercise, for instance, distinguishing between {}^{12}CO and {}^{13}CO is essential because the additional neutron in {}^{13}C adds mass, thus altering the overall molar mass of the molecule (({}^{13})CO has a molar mass of 29 g/mol compared to {}^{12}CO's 28 g/mol). Accurately computing these masses is key to not only identifying the substances present after a chemical reaction but also to scrutinize the changes in a system at equilibrium, as each isotope's mass will contribute differently to the total mass.

Chemical Reaction Analysis

Analyzing a chemical reaction involves looking at the reactants, products, and the balanced chemical equation to understand the nature of the reaction. When a reaction reaches equilibrium, it means that the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products.

In the exercise, the balanced chemical equation \(2 \mathrm{CO}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)\) indicates that carbon monoxide and oxygen combine to form carbon dioxide in a reversible reaction. Observing the molar masses of the resulting mixture helps in identifying the compounds present and their origins. For example, the appearance of a compound with a molar mass of 45 g/mol (({}^{13})CO₂) can be explained through the isotopic carbon dioxide formed from ({}^{13})CO and {}^{12}CO₂ reverts to {}^{12}CO due to the equilibrium process. By analyzing which compounds form and in what quantities, we gain insight into the reaction dynamics and kinetics at play.