For the reaction `2N_2O_5(g) to 4NO_2(g)+O_2(g)` rate of reaction can be expressed as:

Options

A. `1/2 (d[N_2O_5])/(dt)`
B. `2 (d[N_2O_5])/(dt)`
C. `1/4 (d[NO_2])/(dt)`
D. `4 (d[NO_2])/(dt)`

Answer

To express the rate of the reaction \(2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)\), we need to follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ 2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g) \]

### Step 2: Define the rate of reaction
The rate of a reaction can be expressed in terms of the change in concentration of the reactants and products over time. For a general reaction:
\[ aA + bB \rightarrow cC + dD \]
the rate can be expressed as:
\[ \text{Rate} = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt} \]

### Step 3: Apply stoichiometry to the given reaction
For our specific reaction:
- \(A = N_2O_5\) with a stoichiometric coefficient of 2,
- \(B = NO_2\) with a stoichiometric coefficient of 4,
- \(C = O_2\) with a stoichiometric coefficient of 1.

Thus, we can express the rate of the reaction in terms of the concentrations of the reactants and products as follows:
\[ \text{Rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{1}{1} \frac{d[O_2]}{dt} \]

### Step 4: Rearranging the rate expressions
From the rate expressions derived:
1. For \(N_2O_5\):
\[ \text{Rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} \]
2. For \(NO_2\):
\[ \text{Rate} = \frac{1}{4} \frac{d[NO_2]}{dt} \]
3. For \(O_2\):
\[ \text{Rate} = \frac{d[O_2]}{dt} \]

### Step 5: Final expression for the rate of reaction
From the above expressions, we can conclude:
\[ \text{Rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt} \]

### Conclusion
The rate of the reaction can be expressed as:
\[ \text{Rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt} \]