Consider following two reactions A longrightarrow Product -(d[A]/d t)=k1[A]0 B longrightarrow Product -(d[B]/d t)=k2[B] k1 and k2 are expressed in terms of molarity ( mol L -1) and time ( s -1) as :

Question

Consider following two reactions A longrightarrow Product -(d[A]/d t)=k1[A]0 B longrightarrow Product -(d[B]/d t)=k2[B] k1 and k2 are expressed in terms of molarity ( mol L -1) and time ( s -1) as : (A) s -1, M s -1 L -1 (B) M s -1, M s -1 (C) s -1, M -1 s -1 (D) M s -1, s -1

Tardigrade Admin · Accepted Answer

Unit of rate constant k=(1/ text Time ) ⋅ (1/( text conc. )n-1) For k1 n=0 k=(1/t) ⋅ (1/( operatornamecon c)0-1) hence conc ⋅ t-1 ≡ M ⋅ sec -1 k2 n=1 k=(1/t) ⋅ ( 1/( operatornamecon c)1-1)=(1/t)=s-1

Q. Consider following two reactions
$A ⟶$ Product $- \frac{d [ A ]}{d t} = k_{1} [A]^{0}$
$B ⟶$ Product $- \frac{d [ B ]}{d t} = k_{2} [B]$
$k_{1}$ and $k_{2}$ are expressed in terms of molarity $(m o l L^{- 1})$ and time $(s^{- 1}) a s :$

$s^{- 1}, M s^{- 1} L^{- 1}$

$M s^{- 1}, M s^{- 1}$

$s^{- 1}, M^{- 1} s^{- 1}$

$M s^{- 1}, s^{- 1}$

Unit of rate constant $k = \frac{1}{Time} \cdot \frac{1}{( conc. ) ^{n - 1}}$

For $k_{1} n = 0 k = \frac{1}{t} \cdot \frac{1}{( con c ) ^{0 - 1}}$ hence

conc $\cdot t^{- 1} \equiv M \cdot sec^{- 1}$

$k_{2} n = 1 k = \frac{1}{t} \cdot \frac{1}{( con c ) ^{1 - 1}} = \frac{1}{t} = s^{- 1}$

Q. Consider following two reactions A⟶ Product −dtd[A]​=k1​[A]0 B⟶ Product −dtd[B]​=k2​[B] k1​ and k2​ are expressed in terms of molarity (molL−1) and time (s−1)as:

s−1,Ms−1L−1

Ms−1,Ms−1

s−1,M−1s−1

Ms−1,s−1