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^(-3) g//sec`
B. `0.80 xx 10^(-3) g//sec`
C. `1.28 xx 10^(-3) g//sec`
D. `1.60 xx 10^(-3) g//sec`

Explanation

To solve the problem, we need to find 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. **Identify the Rate of Disappearance of \( SO_2 \)**:
The rate of disappearance of \( SO_2 \) is given as:
\[ \text{Rate of } SO_2 = 1.28 \times 10^{-3} \, \text{g/s} \]

2. **Write the Rate Expression**:
From the balanced chemical equation, we can express the rates as follows:
\[ -\frac{1}{2} \frac{d[SO_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt} \]
This implies:
\[ \frac{d[SO_3]}{dt} = -\frac{d[SO_2]}{dt} \]

3. **Convert the Rate of Disappearance to Moles**:
We need to convert the rate of disappearance of \( SO_2 \) from grams to moles. The molar mass of \( SO_2 \) is calculated as follows:
\[ \text{Molar mass of } SO_2 = 32 \, \text{(S)} + 16 \times 2 \, \text{(O)} = 64 \, \text{g/mol} \]
Therefore, the rate in moles per second is:
\[ \frac{d[SO_2]}{dt} = \frac{1.28 \times 10^{-3} \, \text{g/s}}{64 \, \text{g/mol}} = 2.0 \times 10^{-5} \, \text{mol/s} \]

4. **Calculate the Rate of Formation of \( SO_3 \)**:
Using the relationship from step 2:
\[ \frac{d[SO_3]}{dt} = -\frac{d[SO_2]}{dt} = 2.0 \times 10^{-5} \, \text{mol/s} \]

5. **Convert the Rate of Formation of \( SO_3 \) Back to Grams**:
The molar mass of \( SO_3 \) is:
\[ \text{Molar mass of } SO_3 = 32 \, \text{(S)} + 16 \times 3 \, \text{(O)} = 80 \, \text{g/mol} \]
Therefore, the rate of formation in grams per second is:
\[ \frac{d[SO_3]}{dt} = 2.0 \times 10^{-5} \, \text{mol/s} \times 80 \, \text{g/mol} = 1.60 \times 10^{-3} \, \text{g/s} \]

### Final Answer:
The rate of formation of \( SO_3 \) is:
\[ \boxed{1.60 \times 10^{-3} \, \text{g/s}} \]