For the reaction `4NH_(3) + 5O_(2) to 4NO + 6H_(2)O`, the rate of reaction with respect to `NH_(3)` is `2 xx 10^(-3) Ms^(-1)` . Then the rate of the reaction with respect to oxygen in `Ms^(-1)`

Options

A. `2 xx 10^(-3)`
B. `1.5 xx 10^(-3)`
C. `2.5 xx 10^(-3)`
D. `3 xx 10^(-3)`

Explanation

To determine the rate of the reaction with respect to oxygen (O₂) for the reaction:

\[ 4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O \]

given that the rate of reaction with respect to ammonia (NH₃) is \( 2 \times 10^{-3} \, \text{M s}^{-1} \), we can follow these steps:

### Step 1: Write the rate expression for the reaction

The rate of the reaction can be expressed in terms of the change in concentration of the reactants and products. For the given reaction, the rate expressions are:

\[ -\frac{1}{4} \frac{d[NH_3]}{dt} = -\frac{1}{5} \frac{d[O_2]}{dt} = \frac{1}{4} \frac{d[NO]}{dt} = \frac{1}{6} \frac{d[H_2O]}{dt} \]

### Step 2: Relate the rates of NH₃ and O₂

From the stoichiometry of the reaction, we can relate the rates of disappearance of NH₃ and O₂. The coefficients in front of NH₃ and O₂ are 4 and 5, respectively. Thus, we can write:

\[ -\frac{d[O_2]}{dt} = \frac{5}{4} \left(-\frac{d[NH_3]}{dt}\right) \]

### Step 3: Substitute the rate of NH₃ into the equation

We know that:

\[ -\frac{d[NH_3]}{dt} = 2 \times 10^{-3} \, \text{M s}^{-1} \]

Substituting this into the equation gives:

\[ -\frac{d[O_2]}{dt} = \frac{5}{4} \times 2 \times 10^{-3} \]

### Step 4: Calculate the rate of O₂

Now, calculate the rate:

\[ -\frac{d[O_2]}{dt} = \frac{5 \times 2 \times 10^{-3}}{4} = \frac{10 \times 10^{-3}}{4} = 2.5 \times 10^{-3} \, \text{M s}^{-1} \]

Thus, the rate of reaction with respect to oxygen is:

\[ \frac{d[O_2]}{dt} = -2.5 \times 10^{-3} \, \text{M s}^{-1} \]

### Final Answer

The rate of the reaction with respect to oxygen (O₂) is:

\[ 2.5 \times 10^{-3} \, \text{M s}^{-1} \]

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