A flask contains a mixture of A and B. Both the compounds decomposes by first order kinetics. The half lives are 60 min and 15 min for A and B respectively , if the initial concentration of A and B are equal , how long it will take for the concentration of A to be three times that of B.
Answers
Answer:
For first order reaction,
ln C = ln C₀ - Kt
⇒ ln (C/C₀) = - Kt
where C is concentration of the compounds.
half life of A = 54 min
ln(1/2) = -Ka×54
⇒ln 2 = Ka×54
⇒Ka = (ln 2)/54
half life of B = 18 min
ln(1/2) = -Kb×18
⇒ln 2 = Kb×18
⇒Kb = (ln 2)/18
Initial concentration was same
A₀ = B₀
Let at time t, concentration of A becomes 4 times of B
At = 4 Bt
A_t =A_0\ e^{-k_at} =A_0\ e^{- \frac{ln2}{54} t}A
t
=A
0
e
−k
a
t
=A
0
e
−
54
ln2
t
B_t =B_0\ e^{-k_bt} =B_0\ e^{- \frac{ln2}{18} t}B
t
=B
0
e
−k
b
t
=B
0
e
−
18
ln2
t
Dividing both, we get
\begin{gathered}\frac{A_t}{B_t} = \frac{A_0\ e^{- \frac{ln2}{54} t} }{B_0\ e^{- \frac{ln2}{18} t} }= e^{-ln2*t( \frac{1}{54} - \frac{1}{18} )} \\ \\ 4=e^{0.693t( \frac{1}{18} - \frac{1}{54} )}=e^{0.693t( \frac{2}{27} )}= e^{0.0513t} \\ \\ ln\ 4=0.0513t \\ \\ t = \frac{1.386}{0.0513} =27\ minutes\end{gathered}
B
t
A
t
=
B
0
e
−
18
ln2
t
A
0
e
−
54
ln2
t
=e
−ln2∗t(
54
1
−
18
1
)
4=e
0.693t(
18
1
−
54
1
)
=e
0.693t(
27
2
)
=e
0.0513t
ln 4=0.0513t
t=
0.0513
1.386
=27 minutes
Explanation:
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