Consider the reaction
`2N_(2) O_(5(g) ) to 4NO_(2(g)) +O_(2(g))` in liquid bromine If `--(d[N_(2)O_(5) ])/(dt)=0.02 M s^(-1)` then `NO_(2)` is formed at _______.

Options

A. `0.01 M//s`
B. `0.02 M//s`
C. `0.04 M//s`
D. `0.08M//s`

Answer

To solve the problem, we need to determine the rate at which \(NO_2\) is formed from the given rate of disappearance of \(N_2O_5\). The balanced chemical equation for the reaction is:

\[ 2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g) \]

### Step-by-Step Solution:

1. **Identify the Rate of Disappearance of \(N_2O_5\)**:
We are given that the rate of disappearance of \(N_2O_5\) is:
\[ -\frac{d[N_2O_5]}{dt} = 0.02 \, M/s \]

2. **Write the Rate Expression**:
According to the stoichiometry of the reaction, the rate of the reaction can be expressed in terms of the change in concentration of the reactants and products. The general rate expression based on the balanced equation is:
\[ \text{Rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} \]

3. **Relate the Rates**:
From the stoichiometry of the reaction, we can relate the rate of formation of \(NO_2\) to the rate of disappearance of \(N_2O_5\):
\[ \frac{d[NO_2]}{dt} = 2 \left(-\frac{d[N_2O_5]}{dt}\right) \]

4. **Substitute the Given Value**:
Now we can substitute the value of \(-\frac{d[N_2O_5]}{dt}\) into the equation:
\[ \frac{d[NO_2]}{dt} = 2 \times 0.02 \, M/s \]

5. **Calculate the Rate of Formation of \(NO_2\)**:
\[ \frac{d[NO_2]}{dt} = 0.04 \, M/s \]

### Final Answer:
The rate at which \(NO_2\) is formed is \(0.04 \, M/s\).

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