For a first order reaction, (A)→ product. Concentration of A, changes from 0.1 M to 0.025 M in 40 minutes the rate of reaction of when concentration of A is 0.01 M is :

Question

For a first order reaction, (A)→ product. Concentration of A, changes from 0.1 M to 0.025 M in 40 minutes the rate of reaction of when concentration of A is 0.01 M is : (A) 3.47×10-4 M /min-1 (B) 3.47×10-5 M /min-1 (C) 1.73×10-4 M /min-1 (D) 1.73×10-5 M /min

Tardigrade Admin · Accepted Answer

For first order reaction, k=(2.303/t) log(a/a-x)=(2.303/40)log(0.1/0.025)=3.46×10-2 Rate= [k]A=3.46×10-2×0.01=3.46×10-4

Q. For a first order reaction, $(A) \to$ product. Concentration of $A$ , changes from $0.1 M$ to $0.025 M$ in $40$ minutes the rate of reaction of when concentration of $A$ is $0.01 M$ is :

$3.47 \times 1 0^{- 4} M / mi n^{- 1}$

$3.47 \times 1 0^{- 5} M / mi n^{- 1}$

$1.73 \times 1 0^{- 4} M / mi n^{- 1}$

$1.73 \times 1 0^{- 5} M / min$

For first order reaction,
$k = \frac{2.303}{t} l o g \frac{a}{a - x} = \frac{2.303}{40} l o g \frac{0.1}{0.025} = 3.46 \times 1 0^{- 2}$
Rate= $[k] A = 3.46 \times 1 0^{- 2} \times 0.01 = 3.46 \times 1 0^{- 4}$

Q. For a first order reaction, (A)→ product. Concentration of A, changes from 0.1M to 0.025M in 40 minutes the rate of reaction of when concentration of A is 0.01M is :

3.47×10−4M/min−1

3.47×10−5M/min−1

1.73×10−4M/min−1

1.73×10−5M/min