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2SO2+O2→2SO3 Rate of formation of SO3 according to the reaction  is 1.6×10−3 kg min−1. Hence rate at which SO2 reacts is:

Question

2SO₂ + O₂ → 2SO₃

Rate of formation of SO₃ according to the reaction is 1.6 × 10⁻³ kg min⁻¹. Hence rate at which SO₂ reacts is:

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Solution

1. Break Down the Problem

We start with the balanced chemical reaction: 2SO2+O22SO3 2 \text{SO}_2 + \text{O}_2 \rightarrow 2 \text{SO}_3 We need to find the rate at which SO2\text{SO}_2 reacts given the rate of formation of SO3\text{SO}_3.

2. Relevant Concepts

According to the stoichiometry of the reaction:

  • For every 2 moles of SO2\text{SO}_2 that react, 2 moles of SO3\text{SO}_3 are produced.
  • Thus, the rate of reaction for SO2\text{SO}_2 will be equal to the rate of formation of SO3\text{SO}_3 multiplied by the inverse of the stoichiometric coefficients.

3. Analysis and Detail

Let’s denote:

  • Rate of formation of SO3=1.6×103kg/min\text{SO}_3 = 1.6 \times 10^{-3} \, \text{kg/min}.
  • Let the rate at which SO2\text{SO}_2 reacts be RSO2R_{\text{SO}_2}.

From the stoichiometry of the reaction: Rate of SO2 reacts=22×Rate of formation of SO3 \text{Rate of } \text{SO}_2 \text{ reacts} = \frac{2}{2} \times \text{Rate of formation of } \text{SO}_3 This indicates that the rate at which SO2\text{SO}_2 reacts is the same as the rate at which SO3\text{SO}_3 forms.

Therefore, we can express it as: RSO2=1.6×103kg/min R_{\text{SO}_2} = 1.6 \times 10^{-3} \, \text{kg/min}

4. Verify and Summarize

Since the coefficient for SO2\text{SO}_2 and SO3\text{SO}_3 are equal in a one-to-one relationship based on their stoichiometric coefficients in the balanced equation, the rates are equal. Hence, no further adjustments are necessary.

Final Answer

The rate at which SO2\text{SO}_2 reacts is: RSO2=1.6×103kg/min R_{\text{SO}_2} = 1.6 \times 10^{-3} \, \text{kg/min}

This problem has been solved

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