If for the reaction `A to B" rate" =-(d[A])/(dt)=2(d[B])/(dt)` then, rate law is

Options

A. `-(d[A])/(dt)=k[A]^(2)`
B. `(d[B])/(dt)=k[A] `
C. `(d[A])/(dt)=k[A]^(2)`
D. `-(d[B])/(dt)=k[B]^(2)`

Answer

To derive the rate law for the reaction \( A \rightarrow B \) given the relationship between the rates of change of concentrations, we can follow these steps:

### Step 1: Understand the given relationship
The problem states that:
\[ -\frac{d[A]}{dt} = 2\frac{d[B]}{dt} \]
This indicates how the concentration of A decreases relative to the concentration of B increases.

### Step 2: Express the rate in terms of concentration
From the relationship, we can express the rate of the reaction in terms of the concentration of A:
\[ -\frac{d[A]}{dt} = 2\frac{d[B]}{dt} \]
This implies that for every 1 mole of A that reacts, 2 moles of B are produced.

### Step 3: Define the rate law
The general form of the rate law for a reaction is given by:
\[ \text{Rate} = k[A]^n \]
where \( k \) is the rate constant, \( [A] \) is the concentration of the reactant, and \( n \) is the order of the reaction with respect to A.

### Step 4: Determine the order of the reaction
From the stoichiometry of the reaction, we see that the rate of disappearance of A is twice the rate of formation of B. This suggests that the reaction is first order with respect to A, but since the stoichiometry shows that 2 moles of B are produced for every mole of A consumed, we can infer that the order of the reaction is 2 with respect to A.

### Step 5: Write the final rate law expression
Thus, the rate law can be expressed as:
\[ \text{Rate} = k[A]^2 \]

### Summary
The rate law for the reaction \( A \rightarrow B \) is:
\[ \text{Rate} = k[A]^2 \]