For a first order reaction (A) ----- Products the concentration of A changes from 0 1M to 0 025M - Chemistry - Chemical Kinetics

Question

For a first order reaction (A) -----> Products the
concentration of A changes from 0 1M to 0 025M in 40minutes The
rate of reaction when the concentration of A is 0 01A is :
(A) 1 73 x 10-5M/min
(B) 3 47 x 10-4M/min
(C) 3 47 x 10-5M/min
(D) 1 73 x 10-4M/min

Kanika Dalal · Accepted Answer

To find rate of the reaction we first need to find the rate constant, which can be done as follows:

$k = \frac{2.303}{t} \log \frac{{[A]}_{1}}{{[A]}_{2}}$

where, t = 40min

[A]₁ = 0.1 M

[A]₂ = 0.025 M

$k = \frac{2.303}{40} \log \frac{0.1}{0.025} k = \frac{2.303}{40} \log 4 k = \frac{2.303}{40} \log {(2)}^{2} k = \frac{2.303}{40} 2 \log 2 k = \frac{2.303 \times 2 \times 0.3010}{40} k = 0.0346 k = 3.4 \times 10^{- 2} \min^{- 1}$

For a first order reaction Rate law can be written as:

Rate = k [A]

When [A] = 0.01 M

Rate = 3.4 x 10^-2 x 0.01

Rate = 3.4 x 10^-4 M/min

For a first order reaction (A) -----> Products the concentration of A changes from 0.1M to 0.025M in 40minutes. The rate of reaction when the concentration of A is 0.01A is : (A) 1.73 x 10-5M/min (B) 3.47 x 10-4M/min (C) 3.47 x 10-5M/min (D) 1.73 x 10-4M/min

For a first order reaction (A) -----> Products the concentration of A changes from 0.1M to 0.025M in 40minutes. The rate of reaction when the concentration of A is 0.01A is :

(A) 1.73 x 10^-5M/min

(B) 3.47 x 10^-4M/min

(C) 3.47 x 10^-5M/min

(D) 1.73 x 10^-4M/min