Which is correct graph :

Options

A.
B.
C.
D.

Explanation

To determine which graph is correct for the given reactions, we need to analyze each option based on the principles of chemical kinetics.

### Step-by-Step Solution:

1. **Understanding Second-Order Reactions**:
- For a second-order reaction, the half-life (\(t_{1/2}\)) is inversely proportional to the initial concentration (\(C_0\)):
\[ t_{1/2} \propto \frac{1}{C_0} \]
- This implies that if we plot \(t_{1/2}\) against \(C_0\), we will get a rectangular hyperbola. Therefore, **Option A** is correct.

**Hint**: Recall that for second-order reactions, the half-life is dependent on the concentration of the reactant.

2. **Analyzing First-Order Reactions**:
- For a first-order reaction, the rate of change of concentration is given by:
\[ -\frac{dC}{dt} = kC \]
- Integrating this gives:
\[ C = C_0 e^{-kt} \]
- Taking the natural logarithm, we find:
\[ \ln(-\frac{dC}{dt}) = \ln(kC_0) - kt \]
- This indicates a linear relationship where the slope is negative. Thus, **Option B** is also correct.

**Hint**: Remember that for first-order reactions, the natural logarithm of the concentration change over time leads to a straight line.

3. **Graph of ln(-dC/dt) vs. 1/T**:
- For any reaction, we can express the rate constant \(k\) using the Arrhenius equation:
\[ k = A e^{-\frac{E_a}{RT}} \]
- Substituting this into the expression for \(-\frac{dC}{dt}\) leads to:
\[ \ln(-\frac{dC}{dt}) = n \ln(C) + \ln(A) - \frac{E_a}{RT} \]
- At constant concentration, the graph of \(\ln(-\frac{dC}{dt})\) versus \(\frac{1}{T}\) will also be linear with a negative slope. Therefore, **Option C** is correct.

**Hint**: Consider how temperature affects the rate constant and subsequently the rate of reaction.

4. **Analyzing the Ratio of Times for First-Order Reactions**:
- For a first-order reaction, the relationship between time and concentration is given by:
\[ t = \frac{1}{k} \ln\left(\frac{C_0}{C}\right) \]
- If we consider the concentrations at different times, we can express:
\[ t_{0.75} = \frac{1}{k} \ln(4) \quad \text{and} \quad t_{0.5} = \frac{1}{k} \ln(2) \]
- The ratio \( \frac{t_{0.75}}{t_{0.5}} \) simplifies to:
\[ \frac{t_{0.75}}{t_{0.5}} = \frac{\ln(4)}{\ln(2)} = 2 \]
- This indicates that the ratio is constant and does not depend on the initial concentration, thus **Option D** is incorrect.

**Hint**: Remember that for first-order reactions, the time taken to reach a certain concentration is logarithmic and can be compared directly.

### Conclusion:
The correct graphs are **Option A, B, and C**. Option D is incorrect as it does not reflect the expected behavior of first-order reactions.