The rate of a reaction increases four-fold when the concentration of reactant is increased 16 times. If the rate of reaction is 4 × 10-6 mol L -1 s -1 when the concentration of the reactant is 4 × 10-4 mol L -1. The rate constant of the reaction will be

Question

The rate of a reaction increases four-fold when the concentration of reactant is increased 16 times. If the rate of reaction is 4 × 10-6 mol L -1 s -1 when the concentration of the reactant is 4 × 10-4 mol L -1. The rate constant of the reaction will be (A) 2 × 10-4 mol 1 / 2 L -1 / 2 s -1 (B) 1 × 10-2 s -1 (C) 2 × 10-4 mol -1 / 2 L 1 / 2 s -1 (D) 25 mol -1 L min -1

Tardigrade Admin · Accepted Answer

Rate propto √ text Concentration =k √ text Concentration k=( text Rate /( text Concentration )1 / 2) =(4 × 10-6/(4 × 10-4)(1)2)=(4 × 10-6/2 × 10-2) =2 × 10-4 mol 1 / 2 L -1 / 2 s -1

Q. The rate of a reaction increases four-fold when the concentration of reactant is increased $16$ times. If the rate of reaction is $4 \times 1 0^{- 6} m o l L^{- 1} s^{- 1}$ when the concentration of the reactant is $4 \times 1 0^{- 4} m o l L^{- 1}$ . The rate constant of the reaction will be

$2 \times 1 0^{- 4} m o l^{1/2} L^{- 1/2} s^{- 1}$

$1 \times 1 0^{- 2} s^{- 1}$

$2 \times 1 0^{- 4} m o l^{- 1/2} L^{1/2} s^{- 1}$

$25 m o l^{- 1} L min^{- 1}$

Rate $\propto Concentration = k Concentration$
$k = \frac{Rate}{( Concentration ) ^{1/2}}$
$= \frac{4 \times 1 0 ^{- 6}}{( 4 \times 1 0 ^{- 4} ) ^{\frac{1}{2}}} = \frac{4 \times 1 0 ^{- 6}}{2 \times 1 0 ^{- 2}}$
$= 2 \times 1 0^{- 4} m o l^{1/2} L^{- 1/2} s^{- 1}$

Q. The rate of a reaction increases four-fold when the concentration of reactant is increased 16 times. If the rate of reaction is 4×10−6molL−1s−1 when the concentration of the reactant is 4×10−4molL−1. The rate constant of the reaction will be

2×10−4mol1/2L−1/2s−1

1×10−2s−1

2×10−4mol−1/2L1/2s−1

25mol−1Lmin−1