Question

At $518^{\circ} C$ , the rate of decomposition of 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 -

• A. 2
• B. 3
• C. 1
• D. 0

Answer 1

The Correct Option is A

The problem requires determining the order of the reaction of gaseous acetaldehyde decomposition. We know that the reaction rate and the corresponding percentage decomposition values are provided. Let's solve this step-by-step:

1. 5\% \text{ decomposition corresponds to a pressure drop of } 0.05 \times 363 = 18.15 \text{ Torr}.
2. 33\% \text{ decomposition corresponds to a pressure drop of } 0.33 \times 363 = 119.79 \text{ Torr}.
3. Given rates,
   • \text{Rate}_{5\%} = 1 \text{ Torr s}^{-1} \text{ at } 18.15 \text{ Torr}
   • \text{Rate}_{33\%} = 0.5 \text{ Torr s}^{-1} \text{ at } 119.79 \text{ Torr}
4. Rate of reaction (r)\propto [\text{Acetaldehyde}]^n where n is the order of the reaction.
5. Using the equation for reaction rates and assuming the rates are proportional to the decomposition pressures, compare the rates:
   • 1 \, \text{Torr s}^{-1} \propto (363 - 18.15)^n
   • 0.5 \, \text{Torr s}^{-1} \propto (363 - 119.79)^n
6. Dividing the first equation by the second gives: \frac{1}{0.5} = \left(\frac{344.85}{243.21}\right)^n
7. Simplifying: 2 = \left(\frac{344.85}{243.21}\right)^n
8. Taking logarithms on both sides: n = \frac{\log(2)}{\log(344.85/243.21)}
9. Calculating the values:
   • \log(2) \approx 0.3010
   • \log(344.85/243.21) \approx 0.1505
   Thus, n = \frac{0.3010}{0.1505} \approx 2.

Thus, the order of the reaction is 2.