JEE Chemistry- Chemical Kinetics

Question

A reaction A + B → C, follows first-order kinetics with respect to each reactant, with an overall rate constant, k = 2.0 × 10^-3 M^-1 s^-1. The initial concentrations of A and B are 0.10 M and 6.93 M, respectively. Then choose the correct statement(s)-

(A) The concentration of A remaining after 100 seconds is approximately 2.5 × 10^-2 M. 

(B) The concentration of B remaining after 100 seconds is approximately 1.73 M. 

(C)The half-life for A is about 50 seconds. 

(D) The half-life for B is about 50 seconds

Answer 1

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" = \(\frac{0.693}{t^{1/2}}\) 

t_1/2 = \(\frac{0.693}{k"}\)  (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.