The rate of disappearance of `SO_(2)` in the reaction `2SO_(2) + O_(2) rarr 2SO_(3)` is `1.28 xx 10^(-3) g//sec` then the rate of formation of `SO_(3)` is

Options

A. `0.64 xx 10^(-5) g s^(-1)`
B. `0.32 xx 10^(-5) g s^(-1)`
C. `2.56 xx 10^(-5) g s^(-1)`
D. `1.28 xx 10^(-5) gs^(-1)`

Explanation

To solve the problem, we need to determine the rate of formation of \( SO_3 \) given the rate of disappearance of \( SO_2 \) in the reaction:

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

### Step-by-Step Solution:

1. **Write the Rate Expression**:
The rate of the reaction can be expressed in terms of the change in concentration of the reactants and products. For the given reaction, we can write:

\[ -\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt} \]

Here, the negative sign indicates the rate of disappearance of the reactants, while the positive sign indicates the rate of formation of the product.

2. **Relate the Rates**:
From the rate expression, we can relate the rate of disappearance of \( SO_2 \) to the rate of formation of \( SO_3 \):

\[ -\frac{1}{2} \frac{d[SO_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt} \]

This implies:

\[ -\frac{d[SO_2]}{dt} = \frac{d[SO_3]}{dt} \]

3. **Substitute the Given Rate**:
We are given that the rate of disappearance of \( SO_2 \) is:

\[ -\frac{d[SO_2]}{dt} = 1.28 \times 10^{-3} \, \text{g/s} \]

Therefore, we can substitute this value into our equation:

\[ \frac{d[SO_3]}{dt} = 1.28 \times 10^{-3} \, \text{g/s} \]

4. **Convert to Molarity**:
To express the rate of formation of \( SO_3 \) in molarity (M), we need to convert grams to moles. Assuming we know the molar mass of \( SO_3 \) (which is approximately 80 g/mol), we can convert the rate:

\[ \text{Rate of formation of } SO_3 = \frac{1.28 \times 10^{-3} \, \text{g/s}}{80 \, \text{g/mol}} = 1.6 \times 10^{-5} \, \text{mol/s} \]

5. **Final Answer**:
Thus, the rate of formation of \( SO_3 \) is:

\[ 1.6 \times 10^{-5} \, \text{mol/s} \]

### Summary:
The rate of formation of \( SO_3 \) is \( 1.6 \times 10^{-5} \, \text{mol/s} \).

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