A graph plotted between `log t_((1)/(2))` vs log concentration is a straight line. What is your conclusion?
[Given : n = order of reaction]

Options

A. `n=1, t_((1)/(2))=(1)/(k.a)`
B. `n=2, t_((1)/(2))=(1)/(a)`
C. `n=1, t_((1)/(2))=(0.693)/(k)`
D. `n=0, t_((1)/(2))=(a)/(2k)`

Answer

To solve the problem, we need to analyze the relationship between the half-life of a reaction (t₁/₂) and the concentration of the reactants for different orders of reactions. The question states that a graph of log t₁/₂ versus log concentration is a straight line. Let's derive the conclusion step by step.

### Step 1: Understanding the relationship for different orders of reactions
For different orders of reactions, the half-life (t₁/₂) is expressed in terms of the initial concentration ([A]) and the rate constant (k) as follows:

1. **Zero-order reaction**:
\[ t_{1/2} = \frac{[A]_0}{2k} \]

2. **First-order reaction**:
\[ t_{1/2} = \frac{0.693}{k} \]

3. **Second-order reaction**:
\[ t_{1/2} = \frac{1}{k[A]_0} \]

### Step 2: Analyzing the zero-order reaction
For a zero-order reaction, we can express the half-life in terms of the initial concentration:
\[ t_{1/2} = \frac{[A]_0}{2k} \]
Taking logarithms on both sides:
\[ \log t_{1/2} = \log \left(\frac{[A]_0}{2k}\right) = \log [A]_0 - \log(2k) \]
This shows that:
\[ \log t_{1/2} = \log [A]_0 - \log(2k) \]
This equation indicates that if we plot log t₁/₂ against log [A], we will get a straight line with a slope of 1 and a y-intercept of \(-\log(2k)\).

### Step 3: Analyzing the first-order reaction
For a first-order reaction, the half-life is constant and does not depend on the concentration:
\[ t_{1/2} = \frac{0.693}{k} \]
Taking logarithms:
\[ \log t_{1/2} = \log(0.693) - \log(k) \]
This indicates that the plot of log t₁/₂ versus log [A] will not yield a straight line, as the half-life is independent of concentration.

### Step 4: Analyzing the second-order reaction
For a second-order reaction:
\[ t_{1/2} = \frac{1}{k[A]_0} \]
Taking logarithms:
\[ \log t_{1/2} = \log\left(\frac{1}{k}\right) - \log[A]_0 \]
This also indicates that the plot will not yield a straight line, as the relationship is not linear.

### Conclusion
Since the graph of log t₁/₂ versus log concentration is a straight line, we can conclude that the reaction is of **zero order**. The relationship derived confirms that the half-life is directly proportional to the initial concentration, which is characteristic of zero-order kinetics.

### Final Answer
The conclusion is that the order of the reaction is **zero** (n = 0), and the relationship is given by:
\[ t_{1/2} = \frac{[A]_0}{2k} \]