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% had reacted and 0.5 Torr s⁻¹ when 33% had reacted. The order of the reaction is:

विकल्प

• 2

• 3

• 1

• 0

Answer 1

2

Explanation:

Given: Initial pressure of acetaldehyde = P₀ = 363 Torr

At 5% decomposition:

Pressure remaining = 0.95 × 363 = 344.85 Torr

Rate = 1.00 Torr s⁻¹

At 33% decomposition:

Pressure remaining = 0.67 × 363 = 243.21 Torr

Rate = 0.5 Torr s⁻¹

By using the general rate law expression:

Rate ∝ [P]ⁿ

So for two sets of data:

`"Rate"_1/"Rate"_2 = ([P]_1/[P]_2)^n`

⇒ `1.00/0.5 = (344.85/243.21)^n`

⇒ 2 = (1.417)ⁿ

log(2) = n log(1.417)

⇒ 0.3010 = n × 0.1515

⇒ n = `0.3010/0.1515`

n = 1.99 ≈ 2