Question

Identify the correct graph for a first-order reaction (A → P) Given:
- \( x \)-axis = time (\( t \))
- \( a \) = Initial concentration of A
- \( (a-x) \) = Concentration of A at time \( t \)

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

- First-order reactions follow logarithmic decay behavior.
- A plot of \(\log(a-x)\) vs. \( t \) gives a straight line with a negative slope.
- This property is useful in determining the rate constant from experimental data.

Answer 1

The Correct Option is D

To understand the graphical representation of a first-order reaction, we need to consider the kinetics of such reactions. In a first-order reaction, the rate of reaction depends linearly on the concentration of one reactant.

The rate equation for a first-order reaction can be expressed as:

\[-\frac{d[A]}{dt} = k[A]\]

Where:

• \( [A] \) is the concentration of reactant \( A \)
• \( k \) is the rate constant

Integrating this equation, we get the expression:

\[\ln [A] = \ln [A_0] - kt\]

Where:

• \( [A_0] \) is the initial concentration of \( A \)
• \( t \) is the time

This equation can be rearranged into the linear form of a straight line:

\[\ln [A] = -kt + \ln [A_0]\]

Comparing it to the standard line equation \( y = mx + c \), where \( y = \ln [A] \), \( m = -k \), \( x = t \), and \( c = \ln [A_0] \), we see that a plot of \( \ln [A] \) versus \( t \) should give a straight line with a slope equal to \( -k \).

Therefore, the graphical representation of a first-order reaction should depict a straight line when plotting the natural logarithm of concentration against time.

Now, examining the given options, the graph that correctly represents a first-order reaction is:

This graph shows a linear decrease in \( \ln [A] \) versus time (\( t \)), which is characteristic of a first-order reaction.