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Chapter 13: Problem 33

Sulfur dioxide emissions in power-plant stack gases may react with carbon monoxide as follows: $$\mathrm{SO}_{2}(g)+3 \mathrm{CO}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\cos (g)$$ Write an equation relating each of the following pairs of rates: a. The rate of formation of \(\mathrm{CO}_{2}\) to the rate of consumption of CO b. The rate of formation of COS to the rate of consumption of \(\mathrm{SO}_{2}\) c. The rate of consumption of \(\mathrm{CO}\) to the rate of consumption of \(\mathrm{SO}_{2}\)

Short Answer

Expert verified
Question: Write equations for the relationship between the rates of formation and consumption of different compounds in the given balanced chemical reaction - \(\mathrm{SO}_{2}(g)+3 \mathrm{CO}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\cos (g)\) a) Rate of formation of CO鈧� and the rate of consumption of CO b) Rate of formation of COS and the rate of consumption of SO鈧� c) Rate of consumption of CO and the rate of consumption of SO鈧� Solution: a) \(\text{Rate}_{\mathrm{CO}_{2}} = \frac{2}{3} \times \text{Rate}_{\mathrm{CO}}\) b) \(\text{Rate}_{\mathrm{COS}} = \text{Rate}_{\mathrm{SO}_{2}}\) c) \(\text{Rate}_{\mathrm{CO}} = 3 \times \text{Rate}_{\mathrm{SO}_{2}}\)

Step by step solution

01

Identify the stoichiometry of the balanced chemical reaction

In the given balanced chemical reaction, we have the following stoichiometry: 1 mol of SO鈧� reacts with 3 moles of CO to form 2 moles of CO鈧� and 1 mole of COS. The balanced chemical reaction can be expressed as: \(\mathrm{SO}_{2}(g)+3 \mathrm{CO}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\cos (g)\)
02

Write an equation for the rate of formation of CO鈧� and the rate of consumption of CO (a)

Using stoichiometry, we can see that for every 2 moles of CO鈧� formed, 3 moles of CO are consumed. Therefore, the rate of formation of CO鈧� (\(\text{Rate}_{\mathrm{CO}_{2}}\)) is related to the rate of consumption of CO (\(\text{Rate}_{\mathrm{CO}}\)) as follows: \(\text{Rate}_{\mathrm{CO}_{2}} = \frac{2}{3} \times \text{Rate}_{\mathrm{CO}}\)
03

Write an equation for the rate of formation of COS and the rate of consumption of SO鈧� (b)

Using stoichiometry, we see that 1 mole of COS is formed for every 1 mole of SO鈧� consumed. Thus, the rate of formation of COS (\(\text{Rate}_{\mathrm{COS}}\)) is equal to the rate of consumption of SO鈧� (\(\text{Rate}_{\mathrm{SO}_{2}}\)): \(\text{Rate}_{\mathrm{COS}} = \text{Rate}_{\mathrm{SO}_{2}}\)
04

Write an equation for the rate of consumption of CO and the rate of consumption of SO鈧� (c)

From the stoichiometry of the balanced chemical reaction, 3 moles of CO are consumed for every 1 mole of SO鈧� consumed. Therefore, the rate of consumption of CO (\(\text{Rate}_{\mathrm{CO}}\)) is related to the rate of consumption of SO鈧� (\(\text{Rate}_{\mathrm{SO}_{2}}\)) as follows: \(\text{Rate}_{\mathrm{CO}} = 3 \times \text{Rate}_{\mathrm{SO}_{2}}\)

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Key Concepts

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

Stoichiometry
Understanding stoichiometry is key to solving chemical reaction problems. Stoichiometry involves the relationship between the quantities of reactants and products in a chemical reaction. In our example, the equation is \(\mathrm{SO}_{2}(g)+3 \mathrm{CO}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\cos (g)\). This tells us that:
  • 1 mole of \(\mathrm{SO}_2\) reacts with 3 moles of \(\mathrm{CO}\)
  • This produces 2 moles of \(\mathrm{CO}_2\) and 1 mole of \(\mathrm{COS}\)
The coefficients before the chemical formulas indicate these mole ratios. Stoichiometry helps predict how much of one substance is required to react with another and what amounts of products are formed.
Sulfur Dioxide
Sulfur dioxide (\(\mathrm{SO}_{2}\)) is a crucial reactant in this chemical equation. It is a gas commonly found as a pollutant, especially from burning fossil fuels. In the given reaction, \(\mathrm{SO}_2\) reacts with carbon monoxide to produce carbon dioxide (\(\mathrm{CO}_2\)) and carbonyl sulfide (\(\mathrm{COS}\)).
To understand its role:
  • It acts as a starting material or reactant.
  • One mole of \(\mathrm{SO}_2\) is consumed.
  • The reaction's stoichiometry tells us it aligns with the formation of one mole of \(\mathrm{COS}\).
This reveals \(\mathrm{SO}_{2}\)'s vital involvement in creating products through its interaction with \(\mathrm{CO}\).
Carbon Monoxide
Carbon monoxide (\(\mathrm{CO}\)) plays a dual role as a reactant and a substance usually regarded as a harmful gas. In our reaction, it combines with \(\mathrm{SO}_2\) to form products:
  • Three moles of \(\mathrm{CO}\) are needed for this process.
  • This results in the formation of two moles of \(\mathrm{CO}_2\) gas.
  • \(\mathrm{CO}\) is vital for driving the reaction forward by facilitating the transformation of \(\mathrm{SO}_2\).
Here, understanding how much \(\mathrm{CO}\) is consumed is important for calculating the rates of reaction.
Balanced Chemical Equation
A balanced chemical equation is crucial for understanding any chemical reaction. It accurately represents the reactants and products, ensuring that matter remains conserved. In our example:
\[\mathrm{SO}_{2}(g)+3 \mathrm{CO}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\cos (g)\]
This equation confirms that:
  • The number of each type of atom is equal on both sides (the law of conservation of mass).
  • 1 sulfur atom on both sides.
  • 4 oxygen atoms on both sides.
  • 3 carbon atoms on both sides.
Balancing a chemical equation allows us to understand and predict the behavior of the reaction. It helps in determining the rates at which substances are consumed and produced.

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Most popular questions from this chapter

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