The differential rate law for the reaction, `4NH_3(g)+5O_2(g)rarr4NO(g)+6H_2O(g)`

Options

A. `-(d[NH_3])/(dt) =-(d[O_2])/(dt)=-(d[NO])/(dt)=-(d[H_2O])/(dt)`
B. `(d[NH_3])/(dt) =(d[O_2])/(dt)=-1/4(d[NO])/(dt)=-1/6(d[H_2O])/(dt)`
C. `1/4(d[NH_3])/(dt) =1/5(d[O_2])/(dt)=1/4(d[NO])/(dt)=1/6(d[H_2O])/(dt)`
D. `-1/4(d[NH_3])/(dt) =-1/5(d[O_2])/(dt)=1/4(d[NO])/(dt)=1/6(d[H_2O])/(dt)`

Answer

To derive the differential rate law for the reaction:

\[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2O(g) \]

we will follow these steps:

### Step 1: Write the general form of the rate law
The rate of a reaction can be expressed in terms of the change in concentration of the reactants and products. For the reaction, we can write:

\[ \text{Rate} = -\frac{1}{4} \frac{d[\text{NH}_3]}{dt} = -\frac{1}{5} \frac{d[\text{O}_2]}{dt} = \frac{1}{4} \frac{d[\text{NO}]}{dt} = \frac{1}{6} \frac{d[\text{H}_2O]}{dt} \]

### Step 2: Relate the rates to the stoichiometry
From the balanced equation, we can see the stoichiometric coefficients are 4 for NH₃, 5 for O₂, 4 for NO, and 6 for H₂O. This means that the rate of change of concentration for each species is related to its stoichiometric coefficient.

### Step 3: Write the rate expressions
Using the stoichiometric coefficients, we can express the rate of the reaction as follows:

\[ \text{Rate} = -\frac{1}{4} \frac{d[\text{NH}_3]}{dt} = -\frac{1}{5} \frac{d[\text{O}_2]}{dt} = \frac{1}{4} \frac{d[\text{NO}]}{dt} = \frac{1}{6} \frac{d[\text{H}_2O]}{dt} \]

### Step 4: Choose a standard form
We can choose one of the reactants (commonly the one with the lowest coefficient) to express the rate law. Here, we will use NH₃:

\[ \text{Rate} = -\frac{1}{4} \frac{d[\text{NH}_3]}{dt} \]

### Step 5: Finalize the differential rate law
Thus, the differential rate law for the reaction can be summarized as:

\[ -\frac{1}{4} \frac{d[\text{NH}_3]}{dt} = -\frac{1}{5} \frac{d[\text{O}_2]}{dt} = \frac{1}{4} \frac{d[\text{NO}]}{dt} = \frac{1}{6} \frac{d[\text{H}_2O]}{dt} \]

### Conclusion
The differential rate law for the given reaction is:

\[ -\frac{1}{4} \frac{d[\text{NH}_3]}{dt} = -\frac{1}{5} \frac{d[\text{O}_2]}{dt} = \frac{1}{4} \frac{d[\text{NO}]}{dt} = \frac{1}{6} \frac{d[\text{H}_2O]}{dt} \]