Express the rate of following reactions
` 2NO_2 to 2NO +O_2`

Explanation

To express the rate of the reaction \( 2NO_2 \rightarrow 2NO + O_2 \), we need to consider the changes in concentration of the reactants and products over time. Here's a step-by-step solution:

### Step 1: Write the general form of the rate expression
The rate of a chemical reaction can be expressed in terms of the change in concentration of the reactants and products. For the reaction \( 2NO_2 \rightarrow 2NO + O_2 \), we can express the rate as follows:

\[ \text{Rate} = -\frac{1}{2} \frac{d[NO_2]}{dt} = \frac{1}{2} \frac{d[NO]}{dt} = \frac{1}{1} \frac{d[O_2]}{dt} \]

### Step 2: Consider the stoichiometry of the reaction
The coefficients in the balanced equation indicate how the rates of change of the concentrations are related. For every 2 moles of \( NO_2 \) that react, 2 moles of \( NO \) and 1 mole of \( O_2 \) are produced. This gives us the following relationships:

- The rate of decrease of \( NO_2 \) is half the rate of formation of \( NO \).
- The rate of formation of \( O_2 \) is equal to the rate of formation of \( NO \) divided by 2.

### Step 3: Write the rate expressions for each species
Using the relationships from Step 2, we can express the rate of the reaction in terms of each species:

1. For \( NO_2 \):
\[ \text{Rate} = -\frac{1}{2} \frac{d[NO_2]}{dt} \]

2. For \( NO \):
\[ \text{Rate} = \frac{1}{2} \frac{d[NO]}{dt} \]

3. For \( O_2 \):
\[ \text{Rate} = \frac{d[O_2]}{dt} \]

### Step 4: Combine the expressions into a single rate expression
We can summarize the rate of the reaction as:

\[ \text{Rate} = -\frac{1}{2} \frac{d[NO_2]}{dt} = \frac{1}{2} \frac{d[NO]}{dt} = \frac{d[O_2]}{dt} \]

### Final Rate Expression
Thus, the final expression for the rate of the reaction is:

\[ \text{Rate} = -\frac{1}{2} \frac{d[NO_2]}{dt} = \frac{1}{2} \frac{d[NO]}{dt} = \frac{d[O_2]}{dt} \]