In the following reaction; xA → yB log10 [ - (d[A]/dt) ] = log10 [ (d[B]/dt) ] + 0.3010 'A' and 'B' respectively can be :

Question

In the following reaction; xA → yB log10 [ - (d[A]/dt) ] = log10 [ (d[B]/dt) ] + 0.3010 'A' and 'B' respectively can be : (A) n-Butane and Iso-butane (B) 2C2H4 and C4H8 (C) N2O4 and NO2 (D) C2H2 and C6H6

Tardigrade Admin · Accepted Answer

log (-d[A]/dt) = log (d[B]/dt) +0.3010 (-d[A]/dt) = 2×(d[B]/dt) (1/2) ×(-d[A]/dt) = (d[B]/dt) 2A → B 2C2H4 → C4H8

Q. In the following reaction; $x A \to y B$
$lo g_{10} [- \frac{d [ A ]}{d t}] = lo g_{10} [\frac{d [ B ]}{d t}] + 0.3010$
'A' and 'B' respectively can be :

n-Butane and Iso-butane

$2 C_{2} H_{4}$ and $C_{4} H_{8}$

$N_{2} O_{4}$ and $N O_{2}$

$C_{2} H_{2}$ and $C_{6} H_{6}$

$lo g \frac{- d [ A ]}{d t} = lo g \frac{d [ B ]}{d t} + 0.3010$
$\frac{- d [ A ]}{d t} = 2 \times \frac{d [ B ]}{d t}$
$\frac{1}{2} \times \frac{- d [ A ]}{d t} = \frac{d [ B ]}{d t}$
$2 A \to B$
$2 C_{2} H_{4} \to C_{4} H_{8}$

Q. In the following reaction; xA→yB log10​[−dtd[A]​]=log10​[dtd[B]​]+0.3010 'A' and 'B' respectively can be :

n-Butane and Iso-butane

2C2​H4​ and C4​H8​

N2​O4​ and NO2​

C2​H2​ and C6​H6​

logdt−d[A]​=logdtd[B]​+0.3010 dt−d[A]​=2×dtd[B]​ 21​×dt−d[A]​=dtd[B]​ 2A→B 2C2​H4​→C4​H8​

Q. In the following reaction; $x A \to y B$
$lo g_{10} [- \frac{d [ A ]}{d t}] = lo g_{10} [\frac{d [ B ]}{d t}] + 0.3010$
'A' and 'B' respectively can be :

$2 C_{2} H_{4}$ and $C_{4} H_{8}$

$N_{2} O_{4}$ and $N O_{2}$

$C_{2} H_{2}$ and $C_{6} H_{6}$

$lo g \frac{- d [ A ]}{d t} = lo g \frac{d [ B ]}{d t} + 0.3010$
$\frac{- d [ A ]}{d t} = 2 \times \frac{d [ B ]}{d t}$
$\frac{1}{2} \times \frac{- d [ A ]}{d t} = \frac{d [ B ]}{d t}$
$2 A \to B$
$2 C_{2} H_{4} \to C_{4} H_{8}$