The gas-phase decomposition of NOBr is second order in [NOBr] with `k = 0.81 M^(-1)s^(-1)` at `10^(@)C` .Initial concentration of NOBr in the flask at `10^(@)C = 4.00 xx10^(-3)` M . In how many seconds does it take up `1.50 xx10^(-3)` M of this NOBr?
`2NOBr to 2NO +Br_(2)`

Explanation

To solve the problem, we will use the second-order reaction kinetics formula. The decomposition of NOBr is given as:

\[ 2 \text{NOBr} \rightarrow 2 \text{NO} + \text{Br}_2 \]

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Rate constant, \( k = 0.81 \, \text{M}^{-1}\text{s}^{-1} \)
- Initial concentration, \( [\text{NOBr}]_0 = 4.00 \times 10^{-3} \, \text{M} \)
- Final concentration, \( [\text{NOBr}]_t = 4.00 \times 10^{-3} \, \text{M} - 1.50 \times 10^{-3} \, \text{M} = 2.50 \times 10^{-3} \, \text{M} \)

2. **Use the Second-Order Kinetics Formula:**
The integrated rate law for a second-order reaction is given by:

\[ \frac{1}{[\text{NOBr}]_t} - \frac{1}{[\text{NOBr}]_0} = k \cdot t \]

3. **Substitute the Values into the Formula:**
\[ \frac{1}{2.50 \times 10^{-3}} - \frac{1}{4.00 \times 10^{-3}} = 0.81 \cdot t \]

4. **Calculate the Left Side:**
- Calculate \( \frac{1}{2.50 \times 10^{-3}} \):
\[ \frac{1}{2.50 \times 10^{-3}} = 400 \, \text{M}^{-1} \]
- Calculate \( \frac{1}{4.00 \times 10^{-3}} \):
\[ \frac{1}{4.00 \times 10^{-3}} = 250 \, \text{M}^{-1} \]
- Now, substitute these values:
\[ 400 - 250 = 0.81 \cdot t \]
- This simplifies to:
\[ 150 = 0.81 \cdot t \]

5. **Solve for \( t \):**
\[ t = \frac{150}{0.81} \approx 185.19 \, \text{s} \]

### Final Answer:
It takes approximately **185.19 seconds** for the concentration of NOBr to decrease from \( 4.00 \times 10^{-3} \, \text{M} \) to \( 2.50 \times 10^{-3} \, \text{M} \).

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