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Example-1

Example-1
The half life of first-order decomposition of NH4NO3 is 2.10 h at 288 K temperature: NH4NO3(aq)  N2O(g) + 2H2O(l). If 6.2 g of NH4NO3 is allowed to decompose, the time required for NH4NO3 to decompose 90% and the volume of dry N2O produced at this point measured at STP are, respectively:
  1. 6.978 h, 2.016 L
  2. 0.319 h, 2.12 L
  3. 0.319 h, 2.016 L
  4. None of these
Solution (A)
k = Description: 34985.png
 
Let t be the time for 90% decomposition. So
a = 100% x = 90 or (a – x) = 10
k = Description: 34995.png
⇒ t = Description: 35004.png
Description: 35012.png taken = Description: 35021.png
 
As per question,
Number of moles of N2O produced = 0.1 × Description: 35031.png
Volume of N2O produced at STP = 0.09 × 22.4 = 2.016 L
 

Example-2

Example-2
A fresh radioactive mixture contains short lived species A and B. Both emitting α-particles initially at 8000 α-particles per minute. After 20 min, they emit at the rate of 3500 α-particles per minute. If the half lives of the species A and B are 10 min and 500 h, respectively, then the ratio of activities of A : B in the initial mixture was:
  1. 4 : 6
  2. 6 : 4
  3. 3 : 4
  4. 3 : 1
Solution (D)
Let initial activities of A and B are A0 and B0
(∵ after 2 half lives of activity of A will remain Description: 35041.png)
A0 + B0 = 8000 and also Description: 35048.png = 3500
 
We can assume that activity of B remains constant due to larger half life. So
Description: 35062.png Description: 35070.png
 

Example-3

Example-3
The decomposition of a gaseous substance (A) to yield gaseous products (B) and (C) follows first-order kinetics. If initially only (A) is present and 10 min after the start of the reaction the pressure of (A) is 200 mm Hg and that of overall mixture is 300 mm Hg, then the rate constant for 2A → B + 3C is:
  1. (1/600) ln 1.25 s–1
  2. (2.303/10) log 1.5 min–1
  3. (1/10) ln 1.25 s–1
  4. None of these
Solution (A)
2A → B + 3C
Description: 35078.png
 
Given P0– 2P′ = 200
P0 + 2P′ = 300
 
So, 2P0 = 500 and P′ = 25
Description: 35089.png
Description: 35098.png
Description: 35108.png
 

Example-4

Example-4
The reaction A(s) → 2B(g) + C(g) is first order. The pressures after 20 min and after very long time are 150 mm Hg and 225 mm Hg, respectively. The value of rate constant and pressure after 40 min are:
  1. 0.05 ln 1.5 min–1, 200 mm Hg
  2. 0.5 ln 2 min–1, 300 mm Hg
  3. 0.05 ln 3 min–1, 300 mm Hg
  4. 0.05 ln 3 min–1, 200 mm Hg
Solution (D)
A(s) →2B(g) + C
Description: 35119.png P0 = 0
Pt = 150 P = 225
k = Description: 35128.png
Description: 35137.png
Description: 35144.png
Description: 35151.png
Description: 35159.png
 

Example-5

Example-5
The time elapsed for a certain reaction between 33% and 67% completion of a first-order reaction is 30 min. What is the time needed for 25% completion?
  1. 150.5 min
  2. 12.5 min
  3. 180.5 min
  4. 165.5 min
Solution (B)
Description: 35170.png
Description: 35179.png
Description: 35188.png
Description: 35197.png
Description: 35206.png
Description: 35217.png
= 100 [log 4 – log 3]
= 100[.6020 × 0.4771]
= 12.49 = 12.5 min
 




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