For the chemical reaction of the type
`2A to 1/2B `
the correct relationship amongst the rate expressions is

Options

A. `-2(d[A])/(dt)=+I/2(d[B])/(dt)`
B. `+2 (d[A])/(dt) =-1/2 (d[R])/(dt)=+2(d[B])/(dt)`
C. `+2 (d[A])/(dt) =-1/2 (d[R])/(dt)=+2(d[B])/(dt)`
D. `+1/2 (d[A])/(dt)=-2(d[B])/(dt)`

Answer

To solve the problem regarding the rate expressions for the chemical reaction \(2A \rightarrow \frac{1}{2}B\), we need to derive the correct relationships among the rate expressions of the reactants and products.

### Step-by-Step Solution:

1. **Write the Reaction**:
The given reaction is:
\[ 2A \rightarrow \frac{1}{2}B \]

2. **Define Rate of Reaction**:
The rate of a reaction is generally defined in terms of the change in concentration of reactants and products over time. For the disappearance of reactants and the formation of products, we can express this as:
\[ \text{Rate} = -\frac{d[A]}{dt} = \frac{d[B]}{dt} \]

3. **Incorporate Stoichiometry**:
For the reaction \(2A \rightarrow \frac{1}{2}B\), we need to account for the stoichiometry:
- The stoichiometric coefficient for \(A\) is 2.
- The stoichiometric coefficient for \(B\) is \(\frac{1}{2}\).

Therefore, the rate expressions can be written as:
\[ -\frac{1}{2} \frac{d[A]}{dt} = \frac{1}{\frac{1}{2}} \frac{d[B]}{dt} \]

4. **Rearranging the Rate Expressions**:
From the stoichiometric relationships, we can express the rates as:
\[ -\frac{1}{2} \frac{d[A]}{dt} = 2 \frac{d[B]}{dt} \]

This indicates that:
\[ \frac{d[B]}{dt} = -\frac{1}{4} \frac{d[A]}{dt} \]

5. **Final Rate Expression**:
Thus, we can summarize the relationships as:
\[ -\frac{1}{2} \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt} \]

This leads us to conclude that:
\[ \frac{d[A]}{dt} = -2 \frac{d[B]}{dt} \]

6. **Identifying the Correct Option**:
Based on the derived relationships, we can now match these expressions with the given options in the question. The correct option will show that the rate of disappearance of \(A\) is twice the rate of formation of \(B\), with appropriate signs.