The differential rate law equation for the elementary reaction `A+2Boverset(k)to3C`, is

Options

A. `-(d[A])/(dt)=(d[B])/(dt)=(d[C])/(dt)=k[A][B]^(2)`
B. `-(d[A])/(dt)=(1)/(2)(d[b])/(dt)=(1)/(3)(d[C])/(dt)=k[A]^(2)[B]`
C. `-(d[A])/(dt)=(1)/(2)(d[b])/(dt)=(1)/(3)(d[C])/(dt)=k[A][B]^(2)`
D. None of these

Answer

To derive the differential rate law equation for the elementary reaction \( A + 2B \overset{k}{\rightarrow} 3C \), we can follow these steps:

### Step 1: Identify the Reaction
The reaction given is \( A + 2B \rightarrow 3C \). This is an elementary reaction, meaning that the rate law can be directly derived from the stoichiometry of the reaction.

### Step 2: Write the Rate Law Expression
For an elementary reaction, the rate law can be expressed as:
\[ \text{Rate} = k [A]^m [B]^n \]
where \( m \) and \( n \) are the stoichiometric coefficients of the reactants \( A \) and \( B \) respectively.

### Step 3: Assign Stoichiometric Coefficients
From the reaction:
- The stoichiometric coefficient of \( A \) is 1.
- The stoichiometric coefficient of \( B \) is 2.

Thus, we can write:
\[ \text{Rate} = k [A]^1 [B]^2 = k [A] [B]^2 \]

### Step 4: Express the Rate in Terms of Concentration Change
The rate of the reaction can also be expressed in terms of the change in concentration of the reactants and products. For reactants, we have:
\[ -\frac{d[A]}{dt} = \frac{1}{1} \text{Rate} = k [A] [B]^2 \]
For \( B \):
\[ -\frac{1}{2} \frac{d[B]}{dt} = \frac{1}{1} \text{Rate} = k [A] [B]^2 \]
For products, we have:
\[ \frac{1}{3} \frac{d[C]}{dt} = \frac{1}{1} \text{Rate} = k [A] [B]^2 \]

### Step 5: Combine the Expressions
From the above equations, we can express the rate of change of concentration for each reactant and product:
\[ -\frac{d[A]}{dt} = k [A] [B]^2 \]
\[ -\frac{1}{2} \frac{d[B]}{dt} = k [A] [B]^2 \]
\[ \frac{1}{3} \frac{d[C]}{dt} = k [A] [B]^2 \]

### Final Rate Law Equation
Thus, the differential rate law equation for the reaction \( A + 2B \rightarrow 3C \) is:
\[ \text{Rate} = k [A] [B]^2 \]