For the reaction: `aA + bB rarr cC+dD`
Rate `= (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt)`
The rate of formation of `SO_(3)` in the following reaction `2SO_(2) + O_(2) rarr 2SO_(3)` is `100 g min^(-1)`. Hence the rate of disappearance of `O_(2)` is

Options

A. `2 g min^(-1)`
B. `20 g min^(-1)`
C. `200 g min^(-1)`
D. `50 g min^(-1)`

Explanation

To solve the problem, we need to analyze the given reaction and the rate of formation of the product \( SO_3 \). The reaction is:

\[ 2SO_2 + O_2 \rightarrow 2SO_3 \]

### Step-by-Step Solution:

1. **Identify the Reaction Stoichiometry:**
The balanced chemical equation shows that 2 moles of \( SO_2 \) react with 1 mole of \( O_2 \) to produce 2 moles of \( SO_3 \).

2. **Write the Rate Expressions:**
From the stoichiometry of the reaction, we can write the rate expressions as follows:
\[ \text{Rate} = -\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{1}{1} \frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt} \]

3. **Given Rate of Formation of \( SO_3 \):**
We are given that the rate of formation of \( SO_3 \) is \( 100 \, \text{g/min} \). This can be expressed as:
\[ \frac{d[SO_3]}{dt} = 100 \, \text{g/min} \]

4. **Relate the Rate of Formation of \( SO_3 \) to the Rate of Disappearance of \( O_2 \):**
Using the rate expressions, we can relate the rate of disappearance of \( O_2 \) to the rate of formation of \( SO_3 \):
\[ -\frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt} \]

5. **Substitute the Given Rate:**
Substitute the known rate of formation of \( SO_3 \) into the equation:
\[ -\frac{d[O_2]}{dt} = \frac{1}{2} \times 100 \, \text{g/min} \]

6. **Calculate the Rate of Disappearance of \( O_2 \):**
\[ -\frac{d[O_2]}{dt} = 50 \, \text{g/min} \]
Therefore, the rate of disappearance of \( O_2 \) is:
\[ \frac{d[O_2]}{dt} = -50 \, \text{g/min} \]

### Final Answer:
The rate of disappearance of \( O_2 \) is \( 50 \, \text{g/min} \).

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