In the following reaction, `2SO_2(g)+O_2(g)rarr2SO_3(g)` the rate of formation of SO3 is 100g/min. rate of disappearance of O2

Options

A. `50 " gmin"^(-1)`
B. `20 " gmin"^(-1)`
C. `100 " gmin"^(-1)`
D. `200 " gmin"^(-1)`

Answer

To solve the problem, we need to analyze the given chemical reaction and the rates of formation and disappearance of the substances involved.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The balanced equation for the reaction is:
\[ 2SO_2(g) + O_2(g) \rightarrow 2SO_3(g) \]

2. **Identify the Rates**:
We are given that the rate of formation of \( SO_3 \) is 100 grams per minute. We need to find the rate of disappearance of \( O_2 \).

3. **Express the Rates in Terms of Change**:
The rates of the reaction can be expressed as:
\[ -\frac{d[SO_2]}{dt} = -\frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt} \]
Here, the negative signs indicate that the concentrations of the reactants (\( SO_2 \) and \( O_2 \)) are decreasing, while the product (\( SO_3 \)) is increasing.

4. **Relate the Rate of Formation of \( SO_3 \) to the Rate of Disappearance of \( O_2 \)**:
From the stoichiometry of the reaction, we can relate the rates:
\[ -\frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt} \]

5. **Substitute the Given Rate**:
We know that:
\[ \frac{d[SO_3]}{dt} = 100 \text{ grams/min} \]
Therefore, substituting this into the equation gives:
\[ -\frac{d[O_2]}{dt} = \frac{1}{2} \times 100 = 50 \text{ grams/min} \]

6. **Conclusion**:
The rate of disappearance of \( O_2 \) is:
\[ -\frac{d[O_2]}{dt} = 50 \text{ grams/min} \]

### Final Answer:
The rate of disappearance of \( O_2 \) is **50 grams per minute**.

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