For reaction A arrow B, the rate constant k1=A1(e- Ea1 / RT) and for the reaction X arrow Y, the rate constant k2=A2(e- Ea2 / RT)⋅ If A1=109,A2=1010,Ea1=1200cal/mol and Ea2=1800cal/mol, then the temperature at which k1=k2 is (x/2 . 303)K. Find Y=(x/100) (Given; R=2cal/K-mol )

Question

For reaction A arrow B, the rate constant k1=A1(e- Ea1 / RT) and for the reaction X arrow Y, the rate constant k2=A2(e- Ea2 / RT)⋅ If A1=109,A2=1010,Ea1=1200cal/mol and Ea2=1800cal/mol, then the temperature at which k1=k2 is (x/2 . 303)K. Find Y=(x/100) (Given; R=2cal/K-mol ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

A arrow B K1=A1× e(- Ea1/RT) x arrow y K2=A2× e(- Ea2/RT) if k1=k2 109e(- 1200/RT)=1010e(- 1800/RT) (- 1200/RT)=(- 1800/RT)+ln10 T=(300/2 . 303)K

A→B K1​=A1​×eRT−Ea1​​​ x→y K2​=A2​×eRT−Ea2​​ if k1​=k2​ 109eRT−1200​=1010eRT−1800​ RT−1200​=RT−1800​+ln10 T=2.303300​K

$A \to B$
$K_{1} = A_{1} \times e^{\frac{- E _{a_{1}}}{RT}}$
$x \to y$
$K_{2} = A_{2} \times e^{\frac{- E a _{2}}{RT}}$
if $k_{1} = k_{2}$
$1 0^{9} e^{\frac{- 1200}{RT}} = 1 0^{10} e^{\frac{- 1800}{RT}}$
$\frac{- 1200}{RT} = \frac{- 1800}{RT} + l n 10$
$T = \frac{300}{2.303} K$