First Order Reaction via Pressure Data

 

❓ Question

Reaction:

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

First order reaction

Given:

t (min)	Pressure (mm Hg)
10	160
∞	240

Find the incorrect statement.

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🖼 Question Image

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✍️ Short Solution

In gas reactions:

$$P \propto n$$

So pressure can be used as a substitute for concentration.

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🔷 Step 1 — Initial Pressure Idea 💯

At $t = 0$:

Only A present → let pressure = $P_0$

Reaction:

$$A \rightarrow 3 \text{ moles (2B + C)}$$

So pressure increases during reaction.

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🔷 Step 2 — Use Final Pressure

At $t = \infty$:

All A converted

Final pressure:

$$P_\infty = 3P_0$$

Given:

$$240 = 3P_0$$
$$P_0 = 80$$

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🔷 Step 3 — Pressure of A at Time t

At time t:

$$P_t = 160$$

Let pressure of A remaining = $P_A$

Total pressure:

$$P_t = P_A + \text{products}$$

Using relation:

$$P_A = 3P_0 - 2P_t$$
$$P_A = 240 - 2(160)$$
$$P_A = -80 \quad ❌ (wrong approach)$$

👉 Use correct relation below.

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🔷 Step 4 — Correct Relation

For reaction:

$$A \rightarrow 2B + C$$

Let initial = $P_0$, decomposed = x

Then:

Total pressure:

$$P_t = P_0 + 2x$$

At completion:

$$P_\infty = P_0 + 2P_0 = 3P_0$$

So:

$$x = \frac{P_t - P_0}{2}$$

Remaining A:

$$P_A = P_0 - x$$
$$= P_0 - \frac{P_t - P_0}{2}$$
$$= \frac{3P_0 - P_t}{2}$$

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🔷 Step 5 — Calculate $P_A$

$$P_A = \frac{240 - 160}{2}$$ $$P_A = 40$$

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🔷 Step 6 — First Order Formula

$$k = \frac{1}{t} \ln \frac{P_0}{P_A}$$
$$k = \frac{1}{10} \ln \frac{80}{40}$$
$$k = \frac{1}{10} \ln 2$$

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✅ Final Takeaway

$$\boxed{k = \frac{\ln 2}{10}}$$

Use this to check options → find incorrect one.

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⭐ Golden JEE Insight

For gas reactions:

👉 Always relate pressure to moles

👉 Use:

$$P_\infty = n_{final} \times P_0$$

👉 Then derive $P_A$ carefully.