For a first order reaction A(g) - 2B(g) + C(g) at constant volume and 300 K, the total pressure at

Q: Why do we use partial pressures instead of concentration?

For ideal gases at constant volume and temperature, partial pressure is directly proportional to the number of moles (or concentration) according to $P = (n/V)RT$. It makes the math much cleaner when dealing with gas-phase reactions!

Q: How did we get $P_t = P_0 + 2P$?

By Dalton's Law of Partial Pressures. The total pressure is the sum of the partial pressures of all gases present at time $t$. So, $P_t = P_A + P_B + P_C = (P_0 - P) + 2P + P = P_0 + 2P$.

Q: Why is graph C physically impossible?

Because the reaction produces more moles of gas than it consumes ($1 \rightarrow 3$), the total pressure $P_t$ will always be greater than the initial pressure $P_0$. This makes $(P_0 - P_t)$ a negative number, and the natural logarithm of a negative number is undefined!

Q: What exactly is $t_{1/3}$? Is it the time when 1/3 of the reactant has reacted?

Read carefully! The question says it's the time required for the partial pressure to *reach* $1/3^{\text{rd}}$ of its initial value. So, $P_A = P_0/3$, meaning $2/3^{\text{rd}}$ has actually reacted.

Q: Why is the rate constant $k$ independent of the initial concentration?

The rate constant $k$ is a fundamental property of the reaction that depends only on the temperature and activation energy (Arrhenius equation). It does not care how much reactant you start with!

Question

For a first order reaction A(g) $ightarrow$ 2B(g) + C(g) at constant volume and 300 K, the total pressure at the beginning (t = 0) and at time t are $P_0$ and $P_t$, respectively. Initially, only A is present with concentration $[A]_0$, and $t_{1/3}$ is the time required for the partial pressure of A to reach $1/3^{	ext{rd}}$ of its initial value. The correct option(s) is (are) :-
(Assume that all these gases behave as ideal gases)

Sigma Percentile · Accepted Answer

### The Beauty of Gas-Phase Kinetics

Chemical kinetics isn't just about abstract concentrations; it's about measurable, physical realities. In this classic JEE Advanced problem, we are tasked with connecting the microscopic decomposition of a gas to the macroscopic pressure we can read on a gauge.

We are given a first-order reaction occurring at a constant volume and temperature:

$$A(g) ightarrow 2B(g) + C(g)$$

Because the gases behave ideally, we can use their partial pressures as a direct proxy for their molar concentrations. This is a massive simplification that allows us to track the progress of the reaction purely through pressure changes.

### Analyzing the Setup and Stoichiometry

Let's break down what happens inside our closed vessel. At the very beginning ($t = 0$), only gas A is present. Let's call its initial pressure $P_0$. At this moment, the products B and C haven't formed yet, so their pressures are zero.

As time ticks forward to some time $t$, a certain amount of gas A reacts. Let's denote the drop in its pressure as $P$. Because of the stoichiometry of the reaction, for every mole of A that disappears, two moles of B and one mole of C are created.

Therefore, at time $t$, the partial pressures are:
*   $P_A = P_0 - P$
*   $P_B = 2P$
*   $P_C = P$

According to Dalton's Law of Partial Pressures, the total pressure $P_t$ in the vessel is simply the sum of the individual partial pressures:

$$P_t = (P_0 - P) + 2P + P = P_0 + 2P$$

### The Master Equation: Finding $P_A$

To use our rate laws, we need to know the partial pressure of our reactant A at time $t$. We know $P_A = P_0 - P$, but we don't know what $P$ is. However, we *can* measure the total pressure $P_t$. Let's express $P$ in terms of $P_t$:

$$2P = P_t - P_0 \implies P = \frac{P_t - P_0}{2}$$

Now, we substitute this back into our expression for $P_A$:

$$P_A = P_0 - \left(\frac{P_t - P_0}{2}ight) = \frac{2P_0 - P_t + P_0}{2} = \frac{3P_0 - P_t}{2}$$

This is the algebraic heart of the problem. We have successfully linked the unmeasurable partial pressure of A to the easily measurable total pressure $P_t$.

### Applying the First-Order Rate Law

For a first-order reaction, the integrated rate law is given by:

$$k = \frac{1}{t} \ln\left(\frac{P_0}{P_A}ight)$$

Let's substitute our hard-earned expression for $P_A$ into this equation:

$$kt = \ln\left(\frac{P_0}{\frac{3P_0 - P_t}{2}}ight) = \ln\left(\frac{2P_0}{3P_0 - P_t}ight)$$

Using the properties of logarithms ($\ln(a/b) = \ln(a) - \ln(b)$), we can expand this:

$$kt = \ln(2P_0) - \ln(3P_0 - P_t)$$

### Graphical Analysis: Decoding the Options

Now comes the fun part—testing the given graphs against our derived mathematics.

**Evaluating Option (A):**
Let's rearrange our rate law equation to isolate the term plotted on the y-axis of graph A:

$$\ln(3P_0 - P_t) = -kt + \ln(2P_0)$$

Notice the structure? It perfectly mirrors the equation of a straight line, **$y = mx + c$**. Here, our $y$-variable is $\ln(3P_0 - P_t)$, our $x$-variable is time $t$, the y-intercept $c$ is $\ln(2P_0)$, and crucially, the slope $m$ is **$-k$**. A negative slope means the line goes downwards as time progresses. This perfectly matches the visual in Option (A). Thus, **(A) is correct**.

**Evaluating Option (B):**
Graph B plots $t_{1/3}$ against the initial concentration $[A]_0$. The problem defines $t_{1/3}$ as the time required for the partial pressure of A to reach $1/3^{	ext{rd}}$ of its initial value. So, at $t = t_{1/3}$, $P_A = P_0 / 3$.

Let's plug this into our rate law:

$$k = \frac{1}{t_{1/3}} \ln\left(\frac{P_0}{P_0/3}ight) = \frac{\ln 3}{t_{1/3}}$$

Rearranging for $t_{1/3}$ gives:

$$t_{1/3} = \frac{\ln 3}{k}$$

Since the rate constant $k$ is a constant at a given temperature, $t_{1/3}$ is also a **constant**. It does not depend on the initial concentration $[A]_0$ at all! The graph should be a perfectly horizontal line. However, Option (B) shows a decreasing line, making it incorrect.

**Evaluating Option (C):**
Graph C plots $\ln(P_0 - P_t)$ against time. Let's think about the physics here. The reaction converts 1 mole of gas into 3 moles of gas. Therefore, as the reaction proceeds, the total pressure $P_t$ must continuously increase. This means $P_t$ is always greater than $P_0$.

Consequently, the term $(P_0 - P_t)$ will always be a **negative number**. In mathematics, the natural logarithm of a negative number is undefined in the real plane. Therefore, plotting $\ln(P_0 - P_t)$ is physically and mathematically meaningless. Option (C) is incorrect.

**Evaluating Option (D):**
Graph D plots the rate constant $k$ against the initial concentration $[A]_0$. This is a fundamental concept check. The rate constant $k$ is an intrinsic property of the reaction that depends strictly on the **temperature** and the activation energy (as described by the Arrhenius equation). It is completely independent of how much reactant you start with. Therefore, the graph must be a horizontal straight line. Option (D) shows exactly this, making **(D) correct**.

### Final Verdict

By systematically combining stoichiometry, Dalton's Law, and first-order kinetics, we have rigorously proven that the correct graphical representations are **(A) and (D)**.

Select Answer:

Visualized Solution

$A (g) \to 2 B (g) + C (g)$

$P_{t} = P_{0} + 2 P$

$P_{A} = \frac{3 P _{0} - P _{t}}{2}$

$k = \frac{1}{t} ln (\frac{P _{0}}{P _{A}})$

$ln (3 P_{0} - P_{t}) = - k t + ln (2 P_{0})$

$t_{1/3} = \frac{l n 3}{k}$

$Evaluating (C) and (D)$

$Final Conclusion$

The Sigma Insight: Order and Molecularity

The Beauty of Gas-Phase Kinetics

Analyzing the Setup and Stoichiometry

The Master Equation

Applying the First-Order Rate Law

Graphical Analysis

Final Verdict

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A(g)→2B(g)+C(g)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Pt​=P0​+2P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PA​=23P0​−Pt​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=t1​ln(PA​P0​​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ln(3P0​−Pt​)=−kt+ln(2P0​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1/3​=kln3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating (C) and (D)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion