A first-order reaction was started with a decimolar solution of the reactant, 8 minutes and 20 seconds later its concentration was found to be M/100. So the rate of the reaction is

Question

A first-order reaction was started with a decimolar solution of the reactant, 8 minutes and 20 seconds later its concentration was found to be M/100. So the rate of the reaction is (A) 2.303 × 10-5 sec-1 (B) 2.303 × 10-4 sec-1 (C) 4.606 × 10-3 sec-1 (D) 2.606 × 10-5 sec-1

Tardigrade Admin · Accepted Answer

For first order reactionK=(2.303/t)log(a/a-x) Given a=(1/10)=1m a-x=(1/100)=0.1m;t=500sec ∴ K=(2.303/500)log (10/01)=(2.303/500)log 10 =(2.303/500)=0.004606=4.6×10-3sec-1

Q. A first-order reaction was started with a decimolar solution of the reactant, $8$ minutes and $20$ seconds later its concentration was found to be $M /100.$ So the rate of the reaction is

$2.303 \times 1 0^{- 5} se c^{- 1}$

$2.303 \times 1 0^{- 4} se c^{- 1}$

$4.606 \times 1 0^{- 3} se c^{- 1}$

$2.606 \times 1 0^{- 5} se c^{- 1}$

For first order reaction $K = \frac{2.303}{t} l o g \frac{a}{a - x}$
Given $a = \frac{1}{10} = 1 m$
$a - x = \frac{1}{100} = 0.1 m; t = 500 sec$
$∴ K = \frac{2.303}{500} l o g \frac{10}{01} = \frac{2.303}{500} l o g 10$
$= \frac{2.303}{500} = 0.004606 = 4.6 \times 1 0^{- 3} se c^{- 1}$

Q. A first-order reaction was started with a decimolar solution of the reactant, 8 minutes and 20 seconds later its concentration was found to be M/100. So the rate of the reaction is

2.303×10−5sec−1

2.303×10−4sec−1

4.606×10−3sec−1

2.606×10−5sec−1