The unit, mol L-1 s-1 is meant for the rate constant of the reaction having the order

Question

The unit, mol L-1 s-1 is meant for the rate constant of the reaction having the order (A) 0 (B) 2 (C) 1 (D) 3

Tardigrade Admin · Accepted Answer

General unit for rate constant = big((mol/L) big)1-n s-1 (where, n = order of reaction) Given, unit for rate constant = big((mol/L) big) s-1 On comparing, big((mol/L) big)1-n s-1 = big((mol/L) big) s-1 1 - n = 1 ∴ n = 0 Hence, the order of reaction is zero.

Q. The unit, $m o l L^{- 1} s^{- 1}$ is meant for the rate constant of the reaction having the order

0

2

1

3

4

General unit for rate constant = $(\frac{m o l}{L})^{1 - n} s^{- 1}$
(where, n = order of reaction)
Given, unit for rate constant = $(\frac{m o l}{L}) s^{- 1}$
On comparing,
$(\frac{m o l}{L})^{1 - n} s^{- 1} = (\frac{m o l}{L}) s^{- 1}$
$1 - n = 1$
$∴ n = 0$
Hence, the order of reaction is zero.

Q. The unit, molL−1s−1 is meant for the rate constant of the reaction having the order

0

2

1

3

4

General unit for rate constant = (Lmol​)1−ns−1 (where, n = order of reaction) Given, unit for rate constant = (Lmol​)s−1 On comparing, (Lmol​)1−ns−1=(Lmol​)s−1 1−n=1 ∴n=0 Hence, the order of reaction is zero.

Q. The unit, $m o l L^{- 1} s^{- 1}$ is meant for the rate constant of the reaction having the order

General unit for rate constant = $(\frac{m o l}{L})^{1 - n} s^{- 1}$
(where, n = order of reaction)
Given, unit for rate constant = $(\frac{m o l}{L}) s^{- 1}$
On comparing,
$(\frac{m o l}{L})^{1 - n} s^{- 1} = (\frac{m o l}{L}) s^{- 1}$
$1 - n = 1$
$∴ n = 0$
Hence, the order of reaction is zero.