For a general chemical change 2A 3B `to` products, the rate of reaction with respect to A is `r_1` and that with respect to B is `r_2`. The rates `r_1 and r_2` are related as :

Options

A. `3r_1=2r_2`
B. `r_1=r_2`
C. `2r_1=3r_2`
D. `r_1^2=2r_2^2`

Answer

To solve the problem of relating the rates of reaction with respect to A and B in the chemical change \( 2A + 3B \rightarrow \text{products} \), we can follow these steps:

### Step 1: Write the Rate Expressions
For the reaction, we can express the rates of disappearance of reactants A and B as follows:

- The rate of disappearance of A can be expressed as:
\[ r_1 = -\frac{1}{2} \frac{d[A]}{dt} \]
- The rate of disappearance of B can be expressed as:
\[ r_2 = -\frac{1}{3} \frac{d[B]}{dt} \]

### Step 2: Relate the Rates
From the rate expressions, we can set up the relationship between \( r_1 \) and \( r_2 \):
\[ -\frac{1}{2} \frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt} \]

### Step 3: Eliminate the Negative Signs
Since both rates are negative (as they represent the disappearance of reactants), we can eliminate the negative signs from both sides:
\[ \frac{1}{2} \frac{d[A]}{dt} = \frac{1}{3} \frac{d[B]}{dt} \]

### Step 4: Cross-Multiply
To find a direct relationship between \( r_1 \) and \( r_2 \), we can cross-multiply:
\[ 3 \frac{d[A]}{dt} = 2 \frac{d[B]}{dt} \]

### Step 5: Substitute \( r_1 \) and \( r_2 \)
Now, substituting \( r_1 \) and \( r_2 \) into the equation:
\[ 3 \left(-2r_1\right) = 2 \left(-3r_2\right) \]
This simplifies to:
\[ 3r_1 = 2r_2 \]

### Final Relation
Thus, the final relationship between the rates of reaction with respect to A and B is:
\[ 3r_1 = 2r_2 \]