# Example-1

Example-1

The half life of first-order decomposition of NH

_{4}NO_{3}is 2.10 h at 288 K temperature: NH_{4}NO_{3(aq)}N_{2}O_{(g)}+ 2H_{2}O_{(l)}. If 6.2 g of NH_{4}NO_{3}is allowed to decompose, the time required for NH_{4}NO_{3}to decompose 90% and the volume of dry N_{2}O produced at this point measured at STP are, respectively:- 6.978 h, 2.016 L
- 0.319 h, 2.12 L
- 0.319 h, 2.016 L
- None of these

Solution (A)

*k*=

Let

*t*be the time for 90% decomposition. So*a*= 100%

*x*= 90 or (

*a*â€“

*x*) = 10

*k*=

â‡’

*t*= taken =

As per question,

Number of moles of N

_{2}O produced = 0.1 Ã—Volume of N

_{2}O produced at STP = 0.09 Ã— 22.4 = 2.016 L# Example-2

Example-2

A fresh radioactive mixture contains short lived species

*A*and*B*. Both emitting*Î±*-particles initially at 8000*Î±*-particles per minute. After 20 min, they emit at the rate of 3500*Î±*-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:- 4 : 6
- 6 : 4
- 3 : 4
- 3 : 1

Solution (D)

Let initial activities of

*A*and*B*are*A*_{0}and*B*_{0}(âˆµ after 2 half lives of activity of

*A*will remain )*A*

_{0}+

*B*

_{0}= 8000 and also = 3500

We can assume that activity of

*B*remains constant due to larger half life. So# 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 2*A*â†’*B*+ 3*C*is:- (1/600) ln 1.25 s
^{â€“1} - (2.303/10) log 1.5 min
^{â€“1} - (1/10) ln 1.25 s
^{â€“1} - None of these

Solution (A)

2

*A*â†’*B*+ 3*C*Given

*P*_{0}â€“ 2*Pâ€²*= 200*P*

_{0}+ 2

*Pâ€²*= 300

So, 2

*P*_{0}= 500 and*Pâ€²*= 25# Example-4

Example-4

The reaction

*A*_{(s)}â†’ 2*B*_{(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:- 0.05 ln 1.5 min
^{â€“1}, 200 mm Hg - 0.5 ln 2 min
^{â€“1}, 300 mm Hg - 0.05 ln 3 min
^{â€“1}, 300 mm Hg - 0.05 ln 3 min
^{â€“1}, 200 mm Hg

Solution (D)

*A*

_{(s)}â†’2

*B*

_{(g)}+

*C*

*P*

_{0}= 0

*P*= 150

_{t}*P*

_{âˆž}= 225

*k*=

# 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?

- 150.5 min
- 12.5 min
- 180.5 min
- 165.5 min

Solution (B)

= 100 [log 4 â€“ log 3]

= 100[.6020 Ã— 0.4771]

= 12.49 = 12.5 min