Question

In the formation of sulphur trioxide by the contact process,\(2 \mathrm{SO}_{2}+\mathrm{O}_{2}=2 \mathrm{SO}_{3}\), the rate of reaction can be measured as \(-\mathrm{d}\left(\mathrm{SO}_{2}\right)=6.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \frac{\mathrm{s}^{-1}}{\mathrm{dt}}\). Here the incorrect statements are a. The rate of reaction expressed in terms of \(\mathrm{O}_{2}\) will be \(4.0 \times 10^{-4}\) mole \(\mathrm{L}^{-1} \mathrm{~s}^{-1}\) b. The rate of reaction expressed in terms of \(\mathrm{O}_{2}\) will be \(6.0 \times 10^{-6}\) mole \(\mathrm{L}^{-1} \mathrm{~s}^{-1}\) c. The rate of reaction expressed in terms of \(\mathrm{SO}_{3}\) will be \(6.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). d. The rate of reaction expressed in terms of \(\mathrm{O}_{2}\) will be \(3.0 \times 10^{-4}\) mole \(\mathrm{L}^{-1} \mathrm{~s}^{-1}\)

Short answer

Statements a and b are incorrect.

Step-by-step solution

Understand the Rate Expression

The given reaction is \(2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightarrow 2 \mathrm{SO}_{3}\). The rate of reaction is given in terms of \(-\mathrm{d} [\mathrm{SO}_{2}]/\mathrm{dt} = 6.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). Since there are two moles of \(\mathrm{SO}_{2}\) and two moles of \(\mathrm{SO}_{3}\) involved, this is the rate of disappearance of \(\mathrm{SO}_{2}\) and the rate of formation of \(\mathrm{SO}_{3}\).

Calculate Rate in terms of \(\mathrm{O}_{2}\)

For the rate of \(\mathrm{O}_{2}\), use the stoichiometric relationship from the balanced equation: \(\mathrm{O}_{2}\) is consumed at half the rate of \(\mathrm{SO}_{2}\). Thus, the rate for \(\mathrm{O}_{2}\) is \(-\frac{1}{2} \times 6.0 \times 10^{-4} = 3.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\).

Analyze Incorrect Statements

a. For statement (a), it claims \(4.0 \times 10^{-4}\) mole \(\mathrm{L}^{-1} \mathrm{~s}^{-1}\) for \(\mathrm{O}_{2}\), which is incorrect based on Step 2.b. For statement (b), the rate \(6.0 \times 10^{-6}\) mole \(\mathrm{L}^{-1} \mathrm{~s}^{-1}\) is incorrect for \(\mathrm{O}_{2}\).c. Statement (c) claims that \(\mathrm{SO}_{3}\) has a rate of \(6.0 \times 10^{-4}\) which is correct as it matches \(\mathrm{SO}_{2}\).d. Statement (d) correctly claims the rate of \(\mathrm{O}_{2}\) is \(3.0 \times 10^{-4}\), which is what we calculated.

Key concepts

Rate of Reaction

Understanding the rate of reaction is crucial in the study of chemical kinetics. The rate of reaction refers to how fast a reactant is consumed or a product is formed over time. In the given exercise, the rate refers to the speed at which \(\mathrm{SO}_2\) is transformed into \(\mathrm{SO}_3\). This is expressed as \(-\mathrm{d}[\mathrm{SO}_2]/\mathrm{dt} \). The negative sign indicates the concentration of \(\mathrm{SO}_2\) decreases as it is consumed in the reaction.
The units for the rate are moles per liter per second \(\left(\mathrm{mol}\,\mathrm{L}^{-1}\, \mathrm{s}^{-1}\right)\), reflecting how much concentration changes each second. It is important to understand that the rate for various reactants and products will differ due to their stoichiometric coefficients in the balanced reaction equation. The stoichiometry is what connects them, which we'll explore next.

Stoichiometry

Stoichiometry is the bridge that connects the amount of reactants to the products in a chemical reaction through their coefficients in the balanced equation. In the example provided, which is part of the contact process, the reaction is \(2 \mathrm{SO}_2 + \mathrm{O}_2 \rightarrow 2 \mathrm{SO}_3\). The coefficients indicate:

• Two moles of \(\mathrm{SO}_2\) are needed for every mole of \(\mathrm{O}_2\).
• Two moles of \(\mathrm{SO}_3\) are produced for every two moles of \(\mathrm{SO}_2\) that disappeared.

This balanced equation helps determine the rate at which \(\mathrm{O}_2\) is consumed. As determined, the consumption rate of \(\mathrm{O}_2\) is exactly half that of \(\mathrm{SO}_2\), since there is one mole of \(\mathrm{O}_2\) for every two moles of \(\mathrm{SO}_2\), thus explaining the calculation of \(3.0 \times 10^{-4} \, \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1}\) for \(\mathrm{O}_2\). Stoichiometry allows us to quantitatively connect these rates and understand the dynamics of the reaction.

Contact Process

The contact process is an industrial technique for the production of sulfuric acid, which is crucial in the manufacturing of fertilizers, chemicals, and many other industrial processes. It involves the catalytic oxidation of sulfur dioxide (\(\mathrm{SO}_2\)) to sulfur trioxide (\(\mathrm{SO}_3\)), followed by its conversion to sulfuric acid (\(\mathrm{H_2SO_4}\)) through absorption with water.
Given that sulfuric acid is one of the most highly produced chemicals worldwide, the efficiency of the contact process is of great importance.

• High conversion rates are essential to ensure economic viability.
• The reaction we've been discussing, where \(\mathrm{SO}_2\) and \(\mathrm{O}_2\) form \(\mathrm{SO}_3\), is crucial in this process.
• Understanding the kinetics, including the rate of reaction and stoichiometry, helps optimize the process to maximize yield.

Thus, the study of reaction kinetics in the contact process is pivotal for optimizing conditions such as temperature and pressure, ensuring the process is both economically and environmentally sustainable.