If the instantaneous rate of appearance of `NO_2 (g)` is 0.0400 M/s at some moment in time, what is the rate of disappearance of `N_2O_5 (g)` in M/s? `[N_2O_5(g)rarr2NO_2(g)+1/2O_2(g)]`

Options

A. 0.02
B. 0.01
C. 0.04
D. 0.08

Answer

To solve the problem, we need to analyze the given chemical reaction and the relationship between the rates of disappearance and appearance of the reactants and products.

### Given Reaction:
\[ N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2}O_2(g) \]

### Step 1: Write the Rate Expressions
From the stoichiometry of the reaction, we can express the rates of disappearance of the reactants and the rates of appearance of the products as follows:

- The rate of disappearance of \( N_2O_5 \) is given by:
\[ -\frac{d[N_2O_5]}{dt} \]

- The rate of appearance of \( NO_2 \) is given by:
\[ \frac{d[NO_2]}{dt} \]

### Step 2: Relate the Rates Using Stoichiometry
From the balanced equation, we can see that for every 1 mole of \( N_2O_5 \) that disappears, 2 moles of \( NO_2 \) appear. Therefore, we can relate the rates as follows:
\[ -\frac{d[N_2O_5]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt} \]

### Step 3: Substitute the Given Rate of Appearance
We are given that the instantaneous rate of appearance of \( NO_2 \) is:
\[ \frac{d[NO_2]}{dt} = 0.0400 \, M/s \]

### Step 4: Calculate the Rate of Disappearance of \( N_2O_5 \)
Substituting the rate of appearance of \( NO_2 \) into the rate expression for \( N_2O_5 \):
\[ -\frac{d[N_2O_5]}{dt} = \frac{1}{2} \times 0.0400 \, M/s \]
\[ -\frac{d[N_2O_5]}{dt} = 0.0200 \, M/s \]

Thus, the rate of disappearance of \( N_2O_5 \) is:
\[ \frac{d[N_2O_5]}{dt} = -0.0200 \, M/s \]

### Final Answer
The rate of disappearance of \( N_2O_5 \) is:
\[ 0.0200 \, M/s \]

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