Consider the reaction ` 2NO_((g) ) +O_(2(g) ) to 2NO_(2(g)).`
if ` (d [NO_(2)])/(dt)=0.052 M//s ` then `-(d[O_(2)])/(dt)` will be

Options

A. `0.052 M//s`
B. `0.114M//s`
C. `0.026M//s`
D. `-0.026 M//s`

Explanation

To solve the problem, we need to analyze the given chemical reaction and apply the principles of chemical kinetics.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction is given as:
\[ 2 \text{NO}_{(g)} + \text{O}_{2(g)} \rightarrow 2 \text{NO}_{2(g)} \]

2. **Identify the Rate of Change:**
We are given that:
\[ \frac{d[\text{NO}_2]}{dt} = 0.052 \, \text{M/s} \]

3. **Use Stoichiometry to Relate Rates:**
From the balanced equation, we can see the stoichiometric coefficients:
- For \(\text{NO}_2\), the coefficient is 2.
- For \(\text{O}_2\), the coefficient is 1.

The relationship between the rates of change of the concentrations can be expressed as:
\[ -\frac{d[\text{O}_2]}{dt} = \frac{1}{2} \frac{d[\text{NO}_2]}{dt} \]

4. **Substitute the Given Rate:**
Now, we substitute the value of \(\frac{d[\text{NO}_2]}{dt}\) into the equation:
\[ -\frac{d[\text{O}_2]}{dt} = \frac{1}{2} \times 0.052 \, \text{M/s} \]

5. **Calculate the Rate of Change of \([\text{O}_2]\):**
Performing the calculation:
\[ -\frac{d[\text{O}_2]}{dt} = 0.026 \, \text{M/s} \]

6. **Final Answer:**
Therefore, the rate of change of \([\text{O}_2]\) is:
\[ \frac{d[\text{O}_2]}{dt} = -0.026 \, \text{M/s} \]

### Conclusion:
Thus, the value of \(-\frac{d[\text{O}_2]}{dt}\) is \(0.026 \, \text{M/s}\).

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