`2A+ 2C rarr2B`, rate of reaction `(+d(B))/(dt)` is equal to

Options

A. `-3/2(d(A))/(dt)`
B. `-(d(A))/(dt)`
C. `-1/3(d(A))/(dt)`
D. `+2(d(A))/(dt)`

Answer

To determine the rate of reaction for the given chemical equation \(2A + 2C \rightarrow 2B\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction and Stoichiometry**:
The balanced chemical equation is:
\[ 2A + 2C \rightarrow 2B \]
Here, the stoichiometric coefficients are 2 for A, 2 for C, and 2 for B.

2. **Write the Rate of Reaction**:
The rate of reaction can be expressed in terms of the change in concentration of the reactants and products. The general form is:
\[ R = -\frac{1}{\text{stoichiometric coefficient}} \frac{d[\text{reactant}]}{dt} = \frac{1}{\text{stoichiometric coefficient}} \frac{d[\text{product}]}{dt} \]

3. **Express the Rate for Each Component**:
For reactant A:
\[ R = -\frac{1}{2} \frac{d[A]}{dt} \]
For reactant C:
\[ R = -\frac{1}{2} \frac{d[C]}{dt} \]
For product B:
\[ R = \frac{1}{2} \frac{d[B]}{dt} \]

4. **Set the Rates Equal**:
Since all expressions represent the same rate of reaction \(R\), we can set them equal to each other:
\[ -\frac{1}{2} \frac{d[A]}{dt} = -\frac{1}{2} \frac{d[C]}{dt} = \frac{1}{2} \frac{d[B]}{dt} \]

5. **Relate the Rate of Change of B to A**:
We want to express the rate of formation of B in terms of the rate of change of A. From the equation:
\[ \frac{1}{2} \frac{d[B]}{dt} = -\frac{1}{2} \frac{d[A]}{dt} \]
We can multiply through by 2 to simplify:
\[ \frac{d[B]}{dt} = -\frac{d[A]}{dt} \]

6. **Final Expression**:
Therefore, the rate of formation of B in terms of the rate of change of A is:
\[ \frac{d[B]}{dt} = -\frac{d[A]}{dt} \]

### Conclusion:
The rate of reaction \(\frac{d[B]}{dt}\) is equal to \(-\frac{d[A]}{dt}\).