For a reaction`2SO_(2) + O_(2) to 2SO_(3)` rate of consumption of `SO_(2)` is `6.4 xx 10^(-3)` kg/sec. the rate of formation of `SO_(3)` in same units will be

Options

A. `6.4 xx 10^(-3)`
B. `8 xx 10^(-3)`
C. `4 xx 10^(-3)`
D. `16 xx 10^(-3)`

Explanation

To find the rate of formation of \( SO_3 \) from the given reaction:

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

we start with the information provided in the question:

1. The rate of consumption of \( SO_2 \) is given as \( 6.4 \times 10^{-3} \) kg/sec.

### Step-by-step Solution:

1. **Write the Rate Expressions**:
For the reaction, we can express the rates of change of the reactants and products based on their stoichiometric coefficients. The rate of consumption of \( SO_2 \) and the rate of formation of \( SO_3 \) can be expressed as:
\[ -\frac{d[SO_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt} \]
Here, the negative sign indicates that \( SO_2 \) is being consumed, while \( SO_3 \) is being formed.

2. **Relate the Rates**:
From the stoichiometry of the reaction, we can relate the rate of consumption of \( SO_2 \) to the rate of formation of \( SO_3 \):
\[ -\frac{d[SO_2]}{dt} = \frac{2}{2} \frac{d[SO_3]}{dt} \]
This simplifies to:
\[ -\frac{d[SO_2]}{dt} = \frac{d[SO_3]}{dt} \]

3. **Substitute the Given Value**:
We know that the rate of consumption of \( SO_2 \) is \( 6.4 \times 10^{-3} \) kg/sec. Therefore, we can substitute this value into our equation:
\[ \frac{d[SO_3]}{dt} = -\left(-\frac{d[SO_2]}{dt}\right) = 6.4 \times 10^{-3} \text{ kg/sec} \]

4. **Conclusion**:
Thus, the rate of formation of \( SO_3 \) is:
\[ \frac{d[SO_3]}{dt} = 6.4 \times 10^{-3} \text{ kg/sec} \]

### Final Answer:
The rate of formation of \( SO_3 \) is \( 6.4 \times 10^{-3} \) kg/sec.

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