First Order Reaction Rate Constant Using Pressure

 

❓ Question

A(g) → B(g) + C(g) is a first-order reaction.
The reaction is followed using pressure method.

Let:

• $P_t$ = total pressure at time $t$

• $P_\infty$ = total pressure at infinite time

• Reaction started with only A

Which of the following expressions gives the correct rate constant $k$?

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 First order reaction
👉 Pressure relation in gaseous reactions
👉 Integrated rate law.

Main idea:

For first-order reaction:

$$\boxed{ k=\frac1t\ln\frac{a}{a-x} }$$

Here pressures are used instead of concentrations.

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🔷 Step 1 — Initial and Final Pressure 💯

Reaction:

$$A \rightarrow B+C$$

Suppose initial pressure of $A$:

$$P_0$$

Initially only $A$ is present.

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At completion:

1 mole $A$ gives:

2 moles products.

Hence total pressure doubles:

$$\boxed{ P_\infty=2P_0 }$$

Thus:

$$P_0=\frac{P_\infty}{2}$$

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🔷 Step 2 — Pressure at Time $t$

Let pressure decrease of $A$ be:

$$x$$

Then:

$$A=P_0-x$$
$$B=x,\quad C=x$$

Total pressure:

$$P_t=(P_0-x)+x+x$$
$$P_t=P_0+x$$

Thus:

$$x=P_t-P_0$$

Remaining pressure of $A$:

$$P_A=P_0-x$$
$$=P_0-(P_t-P_0)$$
$$=2P_0-P_t$$

Since:

$$P_\infty=2P_0$$

we get:

$$\boxed{ P_A=P_\infty-P_t }$$

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🔷 Step 3 — Apply First Order Formula

For first order:

$$k=\frac1t\ln\frac{P_0}{P_A}$$

Substitute:

$$P_0=\frac{P_\infty}{2}$$

and

$$P_A=P_\infty-P_t$$

Thus:

$$k= \frac1t \ln \left( \frac{P_\infty/2}{P_\infty-P_t} \right)$$
$$\boxed{ k= \frac1t \ln \left( \frac{P_\infty}{2(P_\infty-P_t)} \right) }$$

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🔷 Step 4 — JEE Trap Alert 🚨

❌ Final pressure directly initial pressure maan lena

❌ Remaining pressure of $A$ incorrectly calculate karna

Remember:

For:

$$A\rightarrow B+C$$

total moles increase.

So total pressure increases with time.

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✅ Final Answer

$$\boxed{ k= \frac1t \ln \left( \frac{P_\infty}{2(P_\infty-P_t)} \right) }$$

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