Home
Class 12
CHEMISTRY
The half-life of decomposition of gaseou...

The half-life of decomposition of gaseous `CH_(3)CHO` at initial pressure of `365 mm` and `170 mm` of `Hg` were `420 sec` and `880 sec` respectively. The order of the reaction is:

A

2

B

4

C

6

D

8

Text Solution

AI Generated Solution

The correct Answer is:
To determine the order of the reaction for the decomposition of gaseous CH₃CHO based on the given half-lives at different initial pressures, we can follow these steps: ### Step 1: Understand the relationship between half-life and initial concentration The half-life (t₁/₂) of a reaction is related to the initial concentration (or pressure for gases) and the order of the reaction (n) by the following relationship: \[ t_{1/2} \propto P_0^{(1-n)} \] where \( P_0 \) is the initial pressure. ### Step 2: Set up the equation for two different pressures Given: - \( t_{1/2,1} = 420 \, \text{sec} \) at \( P_{0,1} = 365 \, \text{mm Hg} \) - \( t_{1/2,2} = 880 \, \text{sec} \) at \( P_{0,2} = 170 \, \text{mm Hg} \) We can write the relationship for the two half-lives: \[ \frac{t_{1/2,1}}{t_{1/2,2}} = \left( \frac{P_{0,1}}{P_{0,2}} \right)^{(1-n)} \] ### Step 3: Substitute the values into the equation Substituting the known values: \[ \frac{420}{880} = \left( \frac{365}{170} \right)^{(1-n)} \] ### Step 4: Simplify the left side Calculating the left side: \[ \frac{420}{880} = \frac{21}{44} \approx 0.4773 \] ### Step 5: Simplify the right side Calculating the right side: \[ \frac{365}{170} = 2.147 \] ### Step 6: Set up the equation Now we have: \[ 0.4773 = (2.147)^{(1-n)} \] ### Step 7: Take logarithms to solve for n Taking logarithms on both sides: \[ \log(0.4773) = (1-n) \cdot \log(2.147) \] ### Step 8: Calculate the logarithms Calculating the logarithms: - \( \log(0.4773) \approx -0.320 \) - \( \log(2.147) \approx 0.332 \) ### Step 9: Substitute and solve for n Substituting the values: \[ -0.320 = (1-n)(0.332) \] Now, rearranging to solve for n: \[ 1 - n = \frac{-0.320}{0.332} \] \[ 1 - n \approx -0.963 \] \[ n \approx 1 + 0.963 \] \[ n \approx 1.963 \] ### Step 10: Round to the nearest whole number Since n must be a whole number, we round n to 2. ### Conclusion Thus, the order of the reaction is: **n = 2**
To determine the order of the reaction for the decomposition of gaseous CH₃CHO based on the given half-lives at different initial pressures, we can follow these steps: ### Step 1: Understand the relationship between half-life and initial concentration The half-life (t₁/₂) of a reaction is related to the initial concentration (or pressure for gases) and the order of the reaction (n) by the following relationship: \[ t_{1/2} \propto P_0^{(1-n)} \] where \( P_0 \) is the initial pressure. ### Step 2: Set up the equation for two different pressures ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • CARBONYL COMPOUNDS (ALDEHYDES & KETONES ) & CARBOXYLIC ACID

    RESONANCE ENGLISH|Exercise Part -IV|23 Videos

  • CHEMISTRY IN EVERYDAY LIFE

    RESONANCE ENGLISH|Exercise ORGANIC CHEMISTRY(Chemistry in every day life)|31 Videos

Similar Questions

Explore conceptually related problems

The half life period for a reaction at initial concentration of 0.2 and 0.4 mol L ^(-1) are 100 s and 200 s respectively. The order of the reaction is

The half-life periofs of a reaction at initial concentration 0.1 mol L^(-1) and 0.5 mol L^(-1) are 200s and 40 s , respectively. The order of the reaction is

At 500 K, the half-life period of a gaseous reaction at the initial pressure of 80 kPa is 350 sec. When the pressure is 40 kPa, the half life period is 175 sec. The order of reaction is

The half -lie period for the decomposition of AB_(2)(g) at 100mm pressure is 10min and at 200mm pressure is 5 min, the order of reaction is

In the thermal decomposition of a gaseous substance, the time taken for the decomposition of half of the reactants was 105 minutes when the initial pressure was 750 mm and 950 minutes when the initial pressure was 250 mm. Find the order of reaction.

In the thermal decomposition of a gaseous substance, the time taken for the decomposition of half of the reactants was 105 minutes when the initial pressure was 750 mm and 950 minutes when the initial pressure was 250 mm. Find the order of reaction.

In the thermal decomposition of a gaseous substance, the time taken for the decomposition of half of the reactants was 105 minutes when the initial pressure was 750 mm and 950 minutes when the initial pressure was 250 mm. Find the order of reaction.

At 300 K, half life of a gaseous reactant initially at 58 KPa is 320 min. When the pressure is 29 KPa, the half life is 160 mm. The order of the reaction is

The half-life periof for catalystic decompoistion of AB_(3) at 50 mm is found to be 4 hr and at 100 mm it is 2.0 hr . The order of reaction is

The half life period for catalytic decomposition of XY_3 at 100 mm is found to be 8 hrs and at 200 mm it is 4 hrs. The order of reaction is

Home
Profile