For the reaction `2AtoB+3C`, if `-(d[A])/(dt)=k_(1)[A]^(2),-(d[B])/(dt)=k_(2)[A]^(2),-(d[C])/(dt)=k_(3)[A]^(2)` the correct reaction between `k_(1),k_(2)` and `k_(3)` is :

Options

A. `k_(1)=k_(2)=k_(3)`
B. `2k_(1)=k_(2)=3k_(2)`
C. `4k_(1)=k_(2)=3k_(2)`
D. `(k_(1))/(2)=k_(2)=(k_(3))/(3)`

Answer

To solve the problem, we need to analyze the given reaction and the rate equations provided. The reaction is:

\[ 2A \rightarrow B + 3C \]

From the stoichiometry of the reaction, we can derive the relationships between the rates of change of the concentrations of A, B, and C.

### Step 1: Write the rate expressions based on stoichiometry

For the reaction, the rate of change of concentration can be expressed as:

\[ -\frac{d[A]}{dt} = k_1[A]^2 \]

\[ -\frac{d[B]}{dt} = k_2[A]^2 \]

\[ -\frac{d[C]}{dt} = k_3[A]^2 \]

### Step 2: Relate the rates using stoichiometric coefficients

From the stoichiometry of the reaction, we know that:

- For every 2 moles of A that react, 1 mole of B is produced.
- For every 2 moles of A that react, 3 moles of C are produced.

Thus, we can write the relationships as follows:

\[ -\frac{1}{2} \frac{d[A]}{dt} = \frac{1}{1} \frac{d[B]}{dt} = \frac{1}{3} \frac{d[C]}{dt} \]

### Step 3: Substitute the rate expressions into the stoichiometric relationships

Substituting the rate expressions into the stoichiometric relationships, we get:

1. From A to B:
\[ -\frac{1}{2} \left(-\frac{d[A]}{dt}\right) = \frac{d[B]}{dt} \]
This gives:
\[ \frac{1}{2} k_1[A]^2 = k_2[A]^2 \]

2. From A to C:
\[ -\frac{1}{2} \left(-\frac{d[A]}{dt}\right) = \frac{1}{3} \frac{d[C]}{dt} \]
This gives:
\[ \frac{1}{2} k_1[A]^2 = \frac{1}{3} k_3[A]^2 \]

### Step 4: Cancel out \([A]^2\) and derive the relationships

Since \([A]^2\) is common in all expressions, we can cancel it out:

1. From \( \frac{1}{2} k_1 = k_2 \)
2. From \( \frac{1}{2} k_1 = \frac{1}{3} k_3 \)

### Step 5: Express the relationships in terms of \(k_1\), \(k_2\), and \(k_3\)

From the first equation:
\[ k_2 = \frac{1}{2} k_1 \]

From the second equation:
\[ k_3 = \frac{3}{2} k_1 \]

### Final Relationship

Thus, we can summarize the relationships as:
\[ k_1 = 2k_2 \quad \text{and} \quad k_1 = \frac{2}{3}k_3 \]

This leads us to the conclusion that:
\[ \frac{k_1}{2} = k_2 = \frac{k_3}{3} \]

### Conclusion

The correct relationship among \(k_1\), \(k_2\), and \(k_3\) is:
\[ \frac{k_1}{2} = k_2 = \frac{k_3}{3} \]