Questions of chemistry 12th eg. (3) for 1st order reaction the initial concentration of reactant is 16 Min & t1/2 of the reaction is 13.86 × 10^(-3) min then calculate?
(i) initial rate of reaction?
(ii) rate after 27.72 ×10^(-3) min ?
Answers
Answered by
0
Answer:
rate after 27.72×10^(-3)min
Answered by
0
Explanation:
ANSWER
For Ist order reaction, the half-life (t
1/2
) is:
t
1/2
=
k
0.693
....(i)
∵k=
t
2.303
log
N
[N
0
]
where N
0
= Initial concentration
N= Final concentration
∴k=
45
2.303
log
0.05−0.015M
0.05M
k=
45
2.303
log
0.035
0.05
=
45
2.303
log
7
10
=
45
2.303
[log10−log7]
=
45
2.303
[1−0.8450]
=
45
2.303
×0.154
k=0.00792
Put the value of k in equation (i), we get
t
1/2
=
0.00792
0.693
t
1/2
=87.5min
Hence, the correct option is B
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