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

Options

A. (a) log conc. vs time
B. (b) conc. vs time
C. (c) 1/conc. vs time
D. (d) log conc. vs 1/time

Answer

To determine what kind of graph yields a straight line for a first-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Rate Law for First-Order Reactions**:
A first-order reaction can be expressed in terms of the concentration of the reactant. The rate law is given by:
\[ \text{Rate} = k[A] \]
where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.

2. **Integrated Rate Equation**:
For a first-order reaction, the integrated rate equation is:
\[ \ln[A] = -kt + \ln[A_0] \]
where \( [A_0] \) is the initial concentration, \( [A] \) is the concentration at time \( t \), and \( k \) is the rate constant.

3. **Rearranging the Equation**:
We can rearrange the equation to express it in a linear form:
\[ \ln[A] = -kt + \ln[A_0] \]
This can be rewritten as:
\[ y = mx + c \]
where \( y = \ln[A] \), \( m = -k \), \( x = t \), and \( c = \ln[A_0] \).

4. **Plotting the Graph**:
- On the **y-axis**, we plot \( \ln[A] \) (the natural logarithm of the concentration).
- On the **x-axis**, we plot time \( t \).
- The slope of the line will be equal to \(-k\), and the y-intercept will be \(\ln[A_0]\).

5. **Conclusion**:
Therefore, for a first-order reaction, a straight line is obtained when we plot \( \ln[A] \) versus time \( t \).

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

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