At 518°C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr s–1 when 5% has reacted and 0.5 Torr s–1 when 33% had reacted. The order of the reaction is :

Engineering

Chemistry

Order of Reaction and Law of Mass Action

Methods of Determination of Order of Reaction

Question

At 518°C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr s^–1 when 5% has reacted and 0.5 Torr s^–1 when 33% had reacted. The order of the reaction is :

Rate = k(P)ⁿ

1 = k (0.95 P0)n

0.5 = k (0.67 P0)ⁿ

$\Rightarrow \frac{1}{0 .5} = {(\frac{0 .95 P_{0}}{0 .67 P_{0}})}^{n}$

⇒ 2 = (1.4)ⁿ

⇒ 2' = (2^1/2)ⁿ ⇒ n = 2

This question involves determining the order of a reaction using rate data at different extents of conversion. The decomposition of acetaldehyde is given with initial pressure and rates at two different reaction percentages.

Key Concept: For a reaction of order $n$ , the rate law is $r a t e = k \times {[A]}^{n}$ , where $[A]$ is the concentration (or partial pressure for gases) of the reactant. Since the reaction involves gaseous acetaldehyde, we can use partial pressures directly, as rate is given in Torr/s.

Step 1: Express the concentration in terms of pressure. For a gas, partial pressure is proportional to concentration. So, we can write the rate law as: $r a t e = k \times {(P_{A})}^{n}$ , where $P_{A}$ is the partial pressure of acetaldehyde at any time.

Step 2: Find the partial pressure at each given conversion. The initial pressure $P_{0} = 363 Torr$ . When $x %$ has reacted, the partial pressure of acetaldehyde remaining is $P_{A} = P_{0} \times (1 - \frac{x}{100})$ .

At 5% reacted: $P_{A 1} = 363 \times (1 - 0.05) = 363 \times 0.95 = 344.85 Torr$ . Rate $r_{1} = 1.00 Torr \times s^{- 1}$ .
At 33% reacted: $P_{A 2} = 363 \times (1 - 0.33) = 363 \times 0.67 = 243.21 Torr$ . Rate $r_{2} = 0.50 Torr \times s^{- 1}$ .

Step 3: Write the rate law for both points and take the ratio to eliminate k. $\frac{r_{1}}{r_{2}} = \frac{k \times {(P_{A 1})}^{n}}{k \times {(P_{A 2})}^{n}} = {(\frac{P_{A 1}}{P_{A 2}})}^{n}$

Substitute the values: $\frac{1.00}{0.50} = {(\frac{344.85}{243.21})}^{n}$ , which simplifies to $2 = {(1.418)}^{n}$ .

Step 4: Solve for n. Take logarithm on both sides: $\log 2 = n \times \log 1.418$ . $n = \frac{\log 2}{\log 1.418} = \frac{0.3010}{0.1517} \approx 1.984 \approx 2$ .

Thus, the order of the reaction is 2.

[A] (mol L^–1)	[B] (mol L^–1)	Initial Rate (mol L^–1S^–1)
0.05	0.05	0.045
0.10	0.05	0.090
0.20	0.10	0.72

Detailed Solution

Related Topics and Formulae

Detailed Solution

Related Topics and Formulae

Student who searched for similar questions

Questions 1

Questions 2

Questions 3

Questions 4

Questions 5