10) In a rural bus stand, the arrival of buses is Poisson distributed with an average arrival rate of 2.10 buses per hour. a) Compute the average arrival time between buses. b) What is the probability that at least 5 hours will elapse between bus arrivals? c) What is the probability that at least 2 hours will elapse between bus arrivals?
Answers
Step-by-step explanation:
Time = 33.33 s
Explanation:
Given:
The reaction is a first order reaction
Time taken for 50% completion = 10 s
To Find:
Time required for 90% completion
Solution:
Let initial concentration be R₀ and final concentration be R
By given,
R = 50/100 × R₀ = 0.5 R₀
First finding the rate constant of the reaction,
For a first order reaction rate constant is given as,
\sf k=\dfrac{2.303}{t} \: log\: \dfrac{[R_0]}{R}k=
t
2.303
log
R
[R
0
]
Substitute the data,
\sf k=\dfrac{2.303}{10} \: log\: \dfrac{R_0}{0.5\times R_0}k=
10
2.303
log
0.5×R
0
R
0
\sf k=\dfrac{2.303}{10} \: log\: \dfrac{1}{0.5}k=
10
2.303
log
0.5
1
\sf k=0.2303\times (1-0.7)k=0.2303×(1−0.7)
\sf k=0.2303\times 0.3k=0.2303×0.3
\sf k =0.06909\: s^{-1}k=0.06909s
−1
Now finding the time required to complete 90% of the reaction,
Final concentration R = R₀ - 0.9 R₀ = 0.1 R₀
Substitute in the above formula,
\sf 0.06909=\dfrac{2.303}{t} \: log\dfrac{R_0}{0.1\:R_0}0.06909=
t
2.303
log
0.1R
0
R
0
\sf 0.06909=\dfrac{2.303}{t} \: log\dfrac{1}{0.1}0.06909=
t
2.303
log
0.1
1
\sf 0.06909=\dfrac{2.303}{t} \times 10.06909=
t
2.303
×1
\sf t=\dfrac{2.303}{0.06909}t=
0.06909
2.303
\sf t=33.33\: st=33.33s
Hence to complete 90% of the reaction the time required is 33.33 s