Catalyst A reduces the activation energy for a reaction by 10 kJ mol -1 at 300 K. The ratio of rate constants, ( k text T, Catalysed / k T text , Uncatalysed ) is e x . The value of x is [nearest integer] [Assume theat the pre-exponential factor is same in both the cases. Given R =8.31 J K -1 mol -1 ]

Question

Catalyst A reduces the activation energy for a reaction by 10 kJ mol -1 at 300 K. The ratio of rate constants, ( k text T, Catalysed / k T text , Uncatalysed ) is e x . The value of x is [nearest integer] [Assume theat the pre-exponential factor is same in both the cases. Given R =8.31 J K -1 mol -1 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

K =A e (- Ea / RT ) K textcat=A e-(Ea 1/R T), K text uncat. =A e-(E a/R T) ( K text cat / K text uncat. )= e ( E a - E a / RT ) = e (10 × 1000/8.31 × 300)= e 4.009= e x ∴ x =4

Q. Catalyst A reduces the activation energy for a reaction by 10kJmol−1 at 300K. The ratio of rate constants, kT, Uncatalysed k T, Catalysed ​ is ex. The value of x is ____[nearest integer] [Assume theat the pre-exponential factor is same in both the cases. Given R=8.31JK−1mol−1 ]

K=AeRT−Ea​ Kcat​=Ae−RTEa​1​, Kuncat. ​=Ae−RTEa​ Kuncat. ​Kcat ​​=eRTEa​−Ea​​ =e8.31×30010×1000​=e4.009=ex ∴x=4

$K = A e^{\frac{- E a}{RT}}$
$K_{cat} = A e^{- \frac{E _{a} ^{1}}{RT}},$
$K_{uncat.} = A e^{- \frac{E a}{RT}}$
$\frac{K _{cat}}{K _{uncat.}} = e^{\frac{E _{a} - E _{a}}{RT}}$
$= e^{\frac{10 \times 1000}{8.31 \times 300}} = e^{4.009} = e^{x}$
$∴ x = 4$