The rate of formation of a dimer in a second order dimerisation reaction is 9.1 × 10-6 mol L-1s-1 at 0.01 mol L-1 monomer concentration. What will be the rate constant for the reaction?

Question

The rate of formation of a dimer in a second order dimerisation reaction is 9.1 × 10-6 mol L-1s-1 at 0.01 mol L-1 monomer concentration. What will be the rate constant for the reaction? (A) 9.1 × 10-2 L mol-1 s-1 (B) 9.1 × 10-6 L mol-1 s-1 (C) 3× 10-4 L mol-1 s-1 (D) 27.3 × 10-2 L mol-1 s-1

Tardigrade Admin · Accepted Answer

2A arrow A2 Rate of formation of dimer = k[A]2 k=(Rate of formation of dimer/[A]2) k=(9.1×10-6 mol L-1s-1/(0.01 mol L-1)2) =9.1×10-2 L mol-1 s-1

Q. The rate of formation of a dimer in a second order dimerisation reaction is $9.1 \times 1 0^{- 6} m o l L^{- 1} s^{- 1}$ at $0.01 m o l L^{- 1}$ monomer concentration. What will be the rate constant for the reaction?

$9.1 \times 1 0^{- 2} L m o l^{- 1} s^{- 1}$

$9.1 \times 1 0^{- 6} L m o l^{- 1} s^{- 1}$

$3 \times 1 0^{- 4} L m o l^{- 1} s^{- 1}$

$27.3 \times 1 0^{- 2} L m o l^{- 1} s^{- 1}$

$2 A \to A_{2}$
Rate of formation of dimer $= k [A]^{2}$
$k = \frac{R a t e o f f or ma t i o n o f d im er}{[ A ] ^{2}}$
$k = \frac{9.1 \times 1 0 ^{- 6} m o l L ^{- 1} s ^{- 1}}{( 0.01 m o l L ^{- 1} ) ^{2}}$
$= 9.1 \times 1 0^{- 2} L m o l^{- 1} s^{- 1}$

Q. The rate of formation of a dimer in a second order dimerisation reaction is 9.1×10−6molL−1s−1 at 0.01molL−1 monomer concentration. What will be the rate constant for the reaction?

9.1×10−2Lmol−1s−1

9.1×10−6Lmol−1s−1

3×10−4Lmol−1s−1

27.3×10−2Lmol−1s−1