The Arrhenius equation for two first order equation A longrightarrow B and C longrightarrow D is given by k 1=1012 e -81.28( kJ ) / RT k 2=1011 e -43.10( kJ ) / RT At what temperature (in Kelvin) k 1 becomes equal to k 2. The unit of activation energy is kJ / mol Use: In 10=2.3 and R=8.3 J / K / mol, where k1 and k2 are rate constant.

Question

The Arrhenius equation for two first order equation A longrightarrow B and C longrightarrow D is given by k 1=1012 e -81.28( kJ ) / RT k 2=1011 e -43.10( kJ ) / RT At what temperature (in Kelvin) k 1 becomes equal to k 2. The unit of activation energy is kJ / mol Use: In 10=2.3 and R=8.3 J / K / mol, where k1 and k2 are rate constant. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

log 1012-(81.28/2.303 RT )= log 10 1011-(43.10/2.303 RT ) 1=-(43.10/ RT )+(81.28/ RT ) × (1/2.303) T =(38.18/2.30 × 8.31 × 10-3)=2000 K

log1012−2.303RT81.28​=log10​1011−2.303RT43.10​ 1=−RT43.10​+RT81.28​×2.3031​ T=2.30×8.31×10−338.18​=2000K

$lo g 1 0^{12} - \frac{81.28}{2.303 RT} = lo g_{10} 1 0^{11} - \frac{43.10}{2.303 RT}$
$1 = - \frac{43.10}{RT} + \frac{81.28}{RT} \times \frac{1}{2.303}$
$T = \frac{38.18}{2.30 \times 8.31 \times 1 0 ^{- 3}} = 2000 K$