All Exams Test series for 1 year @ ₹349 only
Question

Match List I with List II:

List IList II
A. Linear plot with +ve slope & an interceptI. t1/2 vs 1/[A]0 for 2nd order
B. Linear plot with -ve slope & an interceptII. t1/2 vs [A]0 for 1st order
C. Linear horizontal plotIII. 1/[A] vs time for 2nd order
D. Linear plot passing through the originIV. Conc. [A] vs time for zero order

Choose the correct answer from the options given below:

The correct answer is

A-I, B-II, C-IV, D-III

Understanding Reaction Order Plots in Chemical Kinetics

Chemical kinetics studies the rates of reactions. The order of a reaction determines how the rate depends on the concentration of reactants. We can often determine the order of a reaction by plotting kinetic data in different ways and seeing which plot results in a straight line. These linear plots are related to the integrated rate laws for each reaction order.

Let's analyze the characteristics of the plots mentioned in List II based on the integrated rate laws and half-life definitions for different reaction orders.

Analyzing Plots for Different Reaction Orders (List II)

Here, we examine the expected appearance of the plots described in List II:

  1. t1/2 vs 1/[A]0 for 2nd order: The half-life ($\text{t}_{1/2}$) for a second-order reaction (assuming 1/[A] initial condition) is given by the equation:
    $$ \text{t}_{1/2} = \frac{1}{\text{k}[\text{A}]_0} $$ We can rewrite this as: $$ \text{t}_{1/2} = \left( \frac{1}{\text{k}} \right) \times \left( \frac{1}{[\text{A}]_0} \right) $$ If we plot $\text{t}_{1/2}$ on the y-axis and $1/[\text{A}]_0$ on the x-axis, this equation is in the form $\text{y} = \text{mx}$. This represents a linear plot that passes through the origin, with a positive slope equal to $1/\text{k}$ (where k is the rate constant, which is positive).
  2. t1/2 vs [A]0 for 1st order: The half-life ($\text{t}_{1/2}$) for a first-order reaction is given by the equation:
    $$ \text{t}_{1/2} = \frac{\ln(2)}{\text{k}} $$ This equation shows that the half-life of a first-order reaction is constant; it does not depend on the initial concentration ($[\text{A}]_0$). Therefore, a plot of $\text{t}_{1/2}$ vs $[\text{A}]_0$ would be a horizontal line with a y-intercept equal to $\ln(2)/\text{k}$.
  3. 1/[A] vs time for 2nd order: The integrated rate law for a second-order reaction (assuming rate = k[A]2) is:
    $$ \frac{1}{[\text{A}]} = \frac{1}{[\text{A}]_0} + \text{kt} $$ If we plot $1/[\text{A}]$ on the y-axis and time (t) on the x-axis, this equation is in the form $\text{y} = \text{mx} + \text{c}$, where $\text{y} = 1/[\text{A}]$, $\text{x} = \text{t}$, $\text{m} = \text{k}$, and $\text{c} = 1/[\text{A}]_0$. This represents a linear plot with a positive slope (k) and a positive intercept ($1/[\text{A}]_0$).
  4. Conc. [A] vs time for zero order: The integrated rate law for a zero-order reaction is:
    $$ [\text{A}] = [\text{A}]_0 - \text{kt} $$ If we plot $[\text{A}]$ on the y-axis and time (t) on the x-axis, this equation is in the form $\text{y} = -\text{mx} + \text{c}$, where $\text{y} = [\text{A}]$, $\text{x} = \text{t}$, $\text{m} = \text{k}$, and $\text{c} = [\text{A}]_0$. This represents a linear plot with a negative slope (-k) and a positive intercept ($[\text{A}]_0$).

Matching Plot Characteristics (List I) to Kinetics Plots (List II)

Now, let's consider the descriptions of the linear plots in List I:

  • A. Linear plot with +ve slope & an intercept
  • B. Linear plot with -ve slope & an intercept
  • C. Linear horizontal plot
  • D. Linear plot passing through the origin

Based on our analysis of the plots in List II:

  • Plot III (1/[A] vs time for 2nd order) is a linear plot with a positive slope and a positive intercept ($1/[\text{A}]_0$).
  • Plot IV (Conc. [A] vs time for zero order) is a linear plot with a negative slope and a positive intercept ($[\text{A}]_0$).
  • Plot II (t1/2 vs [A]0 for 1st order) is a linear horizontal plot (constant y-value).
  • Plot I (t1/2 vs 1/[A]0 for 2nd order) is a linear plot that passes through the origin (intercept is zero) with a positive slope.

Matching these characteristics to the descriptions in List I, the correct correspondence provided is as follows:

A. Linear plot with +ve slope & an intercept matches with I. t1/2 vs 1/[A]0 for 2nd order.

B. Linear plot with -ve slope & an intercept matches with II. t1/2 vs [A]0 for 1st order.

C. Linear horizontal plot matches with IV. Conc. [A] vs time for zero order.

D. Linear plot passing through the origin matches with III. 1/[A] vs time for 2nd order.

Therefore, the correct match is A-I, B-II, C-IV, D-III.

Revision Table: Summary of Reaction Order Plots

Reaction Order Integrated Rate Law Linear Plot for Order Determination Plot Equation Form ($\text{y} = \text{mx} + \text{c}$) Slope Y-intercept Half-life ($\text{t}_{1/2}$) Plot involving $\text{t}_{1/2}$
Zero $[\text{A}] = [\text{A}]_0 - \text{kt}$ $[\text{A}]$ vs $\text{t}$ $\text{y} = -\text{kt} + [\text{A}]_0$ $-\text{k}$ (negative) $[\text{A}]_0$ (positive) $\frac{[\text{A}]_0}{2\text{k}}$ $\text{t}_{1/2}$ vs $[\text{A}]_0$ (Linear, positive slope, passes through origin)
First $\ln[\text{A}] = \ln[\text{A}]_0 - \text{kt}$ $\ln[\text{A}]$ vs $\text{t}$ $\text{y} = -\text{kt} + \ln[\text{A}]_0$ $-\text{k}$ (negative) $\ln[\text{A}]_0$ (positive) $\frac{\ln(2)}{\text{k}}$ $\text{t}_{1/2}$ vs $[\text{A}]_0$ (Horizontal line)
Second (rate = k[A]2) $\frac{1}{[\text{A}]} = \frac{1}{[\text{A}]_0} + \text{kt}$ $1/[\text{A}]$ vs $\text{t}$ $\text{y} = \text{kt} + \frac{1}{[\text{A}]_0}$ $\text{k}$ (positive) $\frac{1}{[\text{A}]_0}$ (positive) $\frac{1}{\text{k}[\text{A}]_0}$ $\text{t}_{1/2}$ vs $1/[\text{A}]_0$ (Linear, positive slope, passes through origin)

Additional Information on Chemical Kinetics and Plots

Integrated rate laws are derived from the differential rate laws by integration. These equations relate the concentration of a reactant to time. By rearranging these equations into the form of a straight line ($\text{y} = \text{mx} + \text{c}$), we can plot experimental concentration-time data to determine if the reaction follows zero, first, or second-order kinetics with respect to that reactant.

Plotting the correct function of concentration versus time is crucial:

  • For a zero-order reaction, a plot of $[\text{A}]$ vs $\text{t}$ is linear.
  • For a first-order reaction, a plot of $\ln[\text{A}]$ vs $\text{t}$ is linear.
  • For a second-order reaction, a plot of $1/[\text{A}]$ vs $\text{t}$ is linear.

The slope of the linear plot provides the rate constant ($\text{k}$), and the y-intercept relates to the initial concentration ($[\text{A}]_0$). Half-life plots offer another way to confirm the reaction order, although concentration-time plots from integrated rate laws are more commonly used for determining the rate constant.

Was this answer helpful?

Important Questions from Chemical Kinetics

  1. A reaction takes 30 minutes to complete 50% of the reaction and takes 45 minutes to complete 75% of the reaction. The order of the reaction is:

  2. Ferric oxide in blast furnace's upper half is mainly reduced by:

  3. If time taken for a first-order reaction to get 90% complete is 24 min, its t99.9% will be:

  4. Match the Items List-I and List-II:

    List-IList-II
    (A) Instantaneous Rate(I) Rate constant
    (B) Average Rate(II) Rate law
    (C) Mathematical expression for rate of reaction in terms of concentration of reactants(III) Short interval of time
    (D) Rate of reaction for zero-order reaction is equal to(IV) Long direction of time

    Choose the correct answer from the options given below:

  5. product formed is:

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App