The plot of In `(c_(0))/(c_(0)-x)` against `t` is a straight line, showing the reaction to be a

Options

A. Zero order reaction
B. First order reaction
C. Second order reaction
D. Half order reaction

Explanation

To determine the order of the reaction based on the plot of \(\ln\left(\frac{c_0}{c_0 - x}\right)\) against \(t\), we can follow these steps:

### Step 1: Understand the Integrated Rate Law for First-Order Reactions
For a first-order reaction, the integrated rate law can be expressed as:
\[ \ln\left(\frac{[A]_0}{[A]}\right) = kt \]
where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is the rate constant.

### Step 2: Define Variables
Let:
- \(c_0\) = initial concentration of the reactant
- \(x\) = amount of reactant that has reacted at time \(t\)
- The concentration of the reactant at time \(t\) can be expressed as:
\[ c = c_0 - x \]

### Step 3: Substitute into the Integrated Rate Law
Substituting \(c_0\) and \(c\) into the integrated rate law gives:
\[ \ln\left(\frac{c_0}{c_0 - x}\right) = kt \]

### Step 4: Analyze the Plot
The equation \(\ln\left(\frac{c_0}{c_0 - x}\right) = kt\) indicates that if we plot \(\ln\left(\frac{c_0}{c_0 - x}\right)\) against time \(t\), we will obtain a straight line. The slope of this line will be equal to the rate constant \(k\).

### Step 5: Conclusion
Since the plot of \(\ln\left(\frac{c_0}{c_0 - x}\right)\) against \(t\) is a straight line, this indicates that the reaction follows first-order kinetics.

### Final Answer
The reaction is a **first-order reaction**.

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