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-by-Step Solution:

1. **Identify the Reaction**: The given reaction 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.

2. **Write the Rate Law**: For an elementary reaction, the rate of the reaction can be expressed in terms of the concentration of the reactants raised to the power of their coefficients in the balanced equation. Thus, the rate law can be written as:
\[ \text{Rate} = k[A]^1[B]^2 \]
where \( k \) is the rate constant.

3. **Express the Rate in Terms of Change in Concentration**: The rate of the reaction can also be expressed in terms of the change in concentration of the reactants and products over time:
\[ \text{Rate} = -\frac{d[A]}{dt} = -\frac{1}{2}\frac{d[B]}{dt} = \frac{1}{3}\frac{d[C]}{dt} \]

4. **Combine the Expressions**: From the rate law and the change in concentration, we can equate the expressions:
\[ k[A][B]^2 = -\frac{d[A]}{dt} = -\frac{1}{2}\frac{d[B]}{dt} = \frac{1}{3}\frac{d[C]}{dt} \]

5. **Final Differential Rate Law Equation**: Therefore, we can express the differential rate law equation as:
\[ -\frac{d[A]}{dt} = k[A][B]^2 \]
\[ -\frac{1}{2}\frac{d[B]}{dt} = k[A][B]^2 \]
\[ \frac{1}{3}\frac{d[C]}{dt} = k[A][B]^2 \]

### Conclusion:
The differential rate law equation for the reaction \( A + 2B \rightarrow 3C \) is:
\[ -\frac{d[A]}{dt} = k[A][B]^2 \]