Question

Given a first order reaction which takes place as follows:

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

Total pressure observed is as follows:

Which is/are correct statement(s)? [Given $ln2 = 0.7$]

• A. Rate constant of appearance of $B$ is equal to $7 \times 10^{-3} min^{-1}$
• B. Half life period of reaction is equal to 100 min.
• C. 75% of reaction is completed in 200 min.
• D. 87.5% of reaction is completed in 100 min.

Answer 1

Answer: (A), (B), (C)

Rate constant of appearance of B is equal to 7×10−3min−1, Half life period of reaction is equal to 100 min., 75% of reaction is completed in 200 min.

Answer 2

We start with the reaction

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

Let the initial number of moles of A be 1. If a fraction $x$ has reacted at time $t$, then:

• Moles of unreacted $A$: $1-x$
• Moles of products: $3x$ (since 1 mole of $A$ gives 2 moles of $B$ and 1 mole of $C$)
• Total moles at time $t$: $(1-x)+3x = 1+2x$

Since the reaction occurs in a closed container at constant temperature and volume, pressure is proportional to the number of moles.

Determining the Pressures

1. At $t=0$ (Apparatus I):

   The manometer uses a liquid with density $d = 6.8\;\text{g/cm}^3$. A column of $152\;\text{cm}$ of that liquid is equivalent to a column of mercury (density $13.6\;\text{g/cm}^3$) of height:

   $$h_{\text{Hg}} = 152 \times \frac{6.8}{13.6} = 152 \times 0.5 = 76\; \text{cm}$$

   Since $76\;\text{cm Hg}$ corresponds to $1\;\text{atm}$, the initial pressure $P_0$ is $1\;\text{atm}$.

2. At $t=100$ min (Apparatus II):

   The manometer now uses mercury directly. The $76\;\text{cm}$ difference when one arm is open to the atmosphere (1 atm) indicates that the gas in the bulb exerts a pressure of:

   $$P = 1\;\text{atm} + 76\;\text{cm Hg equivalent} = 1\;\text{atm} + 1\;\text{atm} = 2\;\text{atm}.$$

Relating Pressure to Reaction Progress

Since pressure is proportional to the moles:

$$P(t) = P_0 (1+2x)$$

At $t=100$ min, $P=2\;\text{atm}$:

$$2 = 1(1+2x) \quad \Longrightarrow \quad 1+2x=2 \quad \Longrightarrow \quad x=0.5.$$

So, 50% of the reactant has reacted at 100 min.

For a first order reaction, the fraction reacted is given by:

$$1-x = e^{-kt}.$$

At $x = 0.5$ (i.e. 50% reacted):

$$e^{-k(100)} = 0.5 \quad \Longrightarrow \quad k = \frac{\ln 2}{100} \approx \frac{0.693}{100} \approx 7 \times 10^{-3}\;\text{min}^{-1}.$$

Thus, the half-life $t_{1/2}$ is 100 min.

Checking the Statements

1. Rate constant of appearance of $B$ is equal to $7 \times 10^{-3} \; \text{min}^{-1}$.

   Since the reaction is first order, the rate constant calculated is $k \approx 7 \times 10^{-3}\;\text{min}^{-1}$.

   → Correct.

2. Half life period of reaction is equal to 100 min.

   We found $t_{1/2} = 100$ min.

   → Correct.

3. 75% of reaction is completed in 200 min.

   For $t=200$ min:

   $$e^{-k(200)} = e^{-0.693 \times 2} = e^{-1.386} \approx 0.25.$$

   So, fraction reacted is $1-0.25 = 0.75$ or 75%.

   → Correct.

4. 87.5% of reaction is completed in 100 min.

   At $t=100$ min we already have 50% conversion, not 87.5%.

   → Incorrect.