For the reaction `4NH_(3) + 5O_(2) to 4NO + 6H_(2) O`, if the rate of disappearance of `NH_(3)` is `3.6 xx 10^(-3)` mol `L^(-1) s^(-1)`, what is the rate of formation of `H_(2) O`

Options

A. `5.4 xx 10^(-3) mol L^(-1) s^(-1)`
B. `3.6 xx 10^(-3) mol L^(-1) s^(-1)`
C. `4 xx 10^(-4) mol L^(-1) s^(-1)`
D. `0.6 xx 10^(-4) mol L^(-1) s^(-1)`

Explanation

To solve the problem, we need to determine the rate of formation of \( H_2O \) from the given rate of disappearance of \( NH_3 \) in the reaction:

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

### Step 1: Identify the rate of disappearance of \( NH_3 \)

The problem states that the rate of disappearance of \( NH_3 \) is given as:

\[ -\frac{d[NH_3]}{dt} = 3.6 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 2: 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 can be expressed as:

\[ \text{Rate} = -\frac{1}{4} \frac{d[NH_3]}{dt} = \frac{1}{6} \frac{d[H_2O]}{dt} \]

### Step 3: Relate the rates of disappearance and formation

From the rate expression, we can relate the rate of formation of \( H_2O \) to the rate of disappearance of \( NH_3 \):

\[ \frac{d[H_2O]}{dt} = -\frac{6}{4} \frac{d[NH_3]}{dt} \]

This simplifies to:

\[ \frac{d[H_2O]}{dt} = \frac{3}{2} \left(-\frac{d[NH_3]}{dt}\right) \]

### Step 4: Substitute the given value

Now, we can substitute the given value of \( -\frac{d[NH_3]}{dt} \):

\[ \frac{d[H_2O]}{dt} = \frac{3}{2} \times 3.6 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 5: Calculate the rate of formation of \( H_2O \)

Calculating the above expression:

\[ \frac{d[H_2O]}{dt} = \frac{3 \times 3.6}{2} \times 10^{-3} = \frac{10.8}{2} \times 10^{-3} = 5.4 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Conclusion

Thus, the rate of formation of \( H_2O \) is:

\[ \frac{d[H_2O]}{dt} = 5.4 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]