JEE Challenger
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ChemHardAdv 2018 • Paper 2

First Order Gas Phase Reaction Pressure Time Relationships

Consider a first-order gaseous reaction A(g)→2B(g)+C(g)\text{A(g)} \rightarrow 2\text{B(g)} + \text{C(g)} occurring at a constant volume and a temperature of 300 K300\text{ K}. The total pressures initially (t=0t = 0) and at time tt are denoted as P0P_0 and PtP_t, respectively. Initially, only component A\text{A} is present with an initial concentration [A]0[\text{A}]_0, and t1/3t_{1/3} denotes the time taken for the partial pressure of A\text{A} to drop to 1/3rd1/3^{\text{rd}} of its initial value. Identify the correct option(s). (Assume all gases behave as ideal gases)

Options

A
Option A Diagram 1
Correct
B
Option B Diagram 1
C
Option C Diagram 1
D
Option D Diagram 1
Correct

Topics & Concepts

Chemical KineticsOrder of ReactionRate of Reaction

Step-by-Step Solution

To determine the correct options, we analyze the reaction kinetics and pressure relationships for the given first-order reaction:

A(g)→2B(g)+C(g)\text{A(g)} \rightarrow 2\text{B(g)} + \text{C(g)}

1. Pressure-Time Relationship (Analysis of Graphs A and C)

Let the initial pressure of gas A\text{A} at time t=0t = 0 be P0P_0. At time tt, let xx be the decrease in the partial pressure of A\text{A}. The partial pressures of each species are:

A(g)→2B(g)+C(g)At t=0:P000At time t:P0−x2xx\begin{array}{rccc} & \text{A(g)} & \rightarrow & 2\text{B(g)} & + & \text{C(g)} \\ \text{At } t = 0: & P_0 & & 0 & & 0 \\ \text{At time } t: & P_0 - x & & 2x & & x \end{array}

The total pressure PtP_t at time tt is the sum of the partial pressures: Pt=(P0−x)+2x+x=P0+2xP_t = (P_0 - x) + 2x + x = P_0 + 2x

Solving for xx: x=Pt−P02x = \frac{P_t - P_0}{2}

Now, the partial pressure of A\text{A} at time tt is: PA=P0−x=P0−(Pt−P02)=3P0−Pt2P_\text{A} = P_0 - x = P_0 - \left(\frac{P_t - P_0}{2}\right) = \frac{3P_0 - P_t}{2}

For a first-order reaction, the integrated rate law in terms of partial pressure is: ln⁡(P0PA)=kt\ln\left(\frac{P_0}{P_\text{A}}\right) = kt

Substituting PA=3P0−Pt2P_\text{A} = \frac{3P_0 - P_t}{2}: ln⁡(2P03P0−Pt)=kt\ln\left(\frac{2P_0}{3P_0 - P_t}\right) = kt

ln⁡(2P0)−ln⁡(3P0−Pt)=kt\ln(2P_0) - \ln(3P_0 - P_t) = kt

ln⁡(3P0−Pt)=−kt+ln⁡(2P0)\ln(3P_0 - P_t) = -kt + \ln(2P_0)

This equation is of the form y=mx+cy = mx + c, where:

  • y=ln⁡(3P0−Pt)y = \ln(3P_0 - P_t)
  • x=Time (t)x = \text{Time } (t)
  • Slope m=−k\text{Slope } m = -k (negative slope)
  • Intercept c=ln⁡(2P0)\text{Intercept } c = \ln(2P_0)

Thus, a plot of ln⁡(3P0−Pt)\ln(3P_0 - P_t) vs Time\text{Time} is a straight line with a negative slope.

  • Option (A) is correct.
  • Option (C) is incorrect.

2. Dependence of t1/3t_{1/3} on [A]0[\text{A}]_0 (Analysis of Graph B)

t1/3t_{1/3} is the time required for the partial pressure of A\text{A} to reach 13\frac{1}{3} of its initial value (PA=13P0P_\text{A} = \frac{1}{3}P_0).

Using the first-order rate equation: k⋅t1/3=ln⁡(P013P0)=ln⁡3k \cdot t_{1/3} = \ln\left(\frac{P_0}{\frac{1}{3}P_0}\right) = \ln 3

t1/3=ln⁡3kt_{1/3} = \frac{\ln 3}{k}

For a first-order reaction, t1/3t_{1/3} is independent of the initial concentration [A]0[\text{A}]_0 or initial pressure P0P_0. The plot of t1/3t_{1/3} vs [A]0[\text{A}]_0 should be a horizontal line parallel to the [A]0[\text{A}]_0-axis.

  • Option (B) is incorrect.

3. Dependence of Rate Constant kk on [A]0[\text{A}]_0 (Analysis of Graph D)

The rate constant kk depends only on temperature (Arrhenius equation) and is independent of the concentration of reactants. Since the temperature is kept constant at 300 K300\text{ K}, kk remains constant regardless of [A]0[\text{A}]_0.

Therefore, the plot of Rate constant vs [A]0[\text{A}]_0 is a horizontal straight line.

  • Option (D) is correct.

Conclusion:

The correct options are A and D.

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