If decompositon reaction `A(g)toB(g) follows first order linetics then the graph of rate of formation (R ) of B against time t will be :

Options

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Answer

To solve the problem, we need to analyze the decomposition reaction \( A(g) \to B(g) \) that follows first-order kinetics. We want to determine the graph of the rate of formation of \( B \) against time \( t \).

### Step-by-Step Solution:

1. **Understanding First-Order Kinetics**:
- For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. This can be expressed mathematically as:
\[ -\frac{d[A]}{dt} = k[A] \]
- Here, \( k \) is the rate constant, and \( [A] \) is the concentration of \( A \).

2. **Relating the Formation of B**:
- The rate of formation of \( B \) is equal to the rate of consumption of \( A \):
\[ \frac{d[B]}{dt} = -\frac{d[A]}{dt} \]
- Therefore, we can write:
\[ \frac{d[B]}{dt} = k[A] \]

3. **Expressing Concentration of A**:
- For a first-order reaction, the concentration of \( A \) at any time \( t \) can be expressed as:
\[ [A] = [A_0] e^{-kt} \]
- Here, \( [A_0] \) is the initial concentration of \( A \).

4. **Substituting into the Rate Equation**:
- Substituting the expression for \( [A] \) into the rate of formation of \( B \):
\[ \frac{d[B]}{dt} = k[A_0] e^{-kt} \]
- This shows that the rate of formation of \( B \) is also an exponential function of time.

5. **Graphing the Rate of Formation**:
- The equation \( \frac{d[B]}{dt} = k[A_0] e^{-kt} \) indicates that the rate of formation of \( B \) decreases exponentially over time.
- Therefore, if we plot the rate of formation of \( B \) (on the y-axis) against time \( t \) (on the x-axis), we will get a graph that starts at a maximum value (when \( t = 0 \)) and decreases towards zero as time progresses.

### Conclusion:
The graph of the rate of formation of \( B \) against time \( t \) will be an exponential decay curve.