Correct options are (A) and (C)
A + B → C
\(\frac{-d[A]}{dt} = \frac{-d[B]}{dt}=\frac{[C]}{dt}\) = K [A] [B]
\(-\frac{[A]}{dt}=k[A][B]\)
\(-\frac{[A]}{dt}=\)k'dt (k' = k[B])
\(\int\limits_{[A]o}^{[A]_t}\frac{d[A]}{[A]}\) = \(-\int\limits_0^t\)k'dt
k't = \(ln\frac{[A]_0}{[A]_t}\)
k' = \(\frac{2.303}tlog\frac{[A]_0}{[A]t}\)----(1)
(A) Putting the values of k', t and [A]0 in equation (1) we got---
2.0 x 10-3 x 6.93 = \(\frac{2.303}{100}log{0.1}{[A]_t}\)
⇒ [A]t = 0.025 M
= 2.5 x 10-2 M
it means after 100 sec. the concentration of A become approximately 2.5 x10-2 M.
(B) Similarly for B
k" t = 2.303 log\(\frac{[B]_o}{[B]_t}\) (k"= k[A])
2.0 x 10-3 x 0.1 = \(\frac{2.303}{100}log\frac{6.93}{[B]_t}\)
log\(\frac{6.93}{[B]_t} = \frac{0.2\times0.1}{2.303}\)
log[B]100 = 10-(0.832)
[B]100 = 6.79 M
After 100 sec concentration of [B] become 6.79 M.
(C) half - life for A - using equation (1)
k't = 2.303 log\(\cfrac{[A]_0}{\frac{[A]_0}{2}}\)
t = \(\frac{2.303\times logz}{k'}\)
= \(\frac{0.693}{k[B]}\)
= \(\frac{0.693}{2.0\times10^{-3}\times6.93}\)
t = 50 sec
it means, the half life of A is 50 sec.
(D) half-life for B--
k" = \(\frac{0.693}{t^{1/2}}\)
t1/2 = \(\frac{0.693}{k"}\) (k" = k[A])
= \(\frac{0.693}{2.0\times10^{-3}\times0.1}\)
= 3465 sec
the half life of B is 3465 sec.
Hence, option A and C is right statment.