For a lst order reaction , a straight line is obtained if you plot

Options

A. log conc. vs time
B. conc. vs time
C. 1/conc. vs time
D. log conc. vs 1/time

Answer

To solve the question regarding the plot for a first-order reaction, we can break it down into the following steps:

### Step-by-Step Solution:

1. **Understanding First-Order Reactions**:
A first-order reaction is characterized by the rate of reaction being directly proportional to the concentration of one reactant. The general form of a first-order reaction can be represented as:
\[ A \rightarrow B \]

2. **Rate Law for First-Order Reaction**:
The rate law for a first-order reaction can be expressed as:
\[ \text{Rate} = k[A] \]
where \( k \) is the rate constant and \([A]\) is the concentration of reactant A.

3. **Integrated Rate Equation**:
The integrated rate equation for a first-order reaction is given by:
\[ \ln[C_t] = \ln[C_0] - kt \]
where:
- \( [C_t] \) is the concentration at time \( t \),
- \( [C_0] \) is the initial concentration,
- \( k \) is the rate constant,
- \( t \) is the time.

4. **Rearranging the Equation**:
Rearranging the integrated rate equation gives us:
\[ \ln[C_t] = -kt + \ln[C_0] \]
This equation is in the form of \( y = mx + b \), where:
- \( y = \ln[C_t] \)
- \( m = -k \) (the slope)
- \( x = t \)
- \( b = \ln[C_0] \) (the y-intercept)

5. **Plotting the Graph**:
When we plot \( \ln[C_t] \) (y-axis) against time \( t \) (x-axis), we will obtain a straight line with a slope of \(-k\). This indicates that for a first-order reaction, the plot of the natural logarithm of concentration versus time yields a straight line.

### Conclusion:
For a first-order reaction, a straight line is obtained if you plot \(\ln[C_t]\) versus time \(t\).

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