CBSE Class 12 · 2005 · All India · Chemical KineticsShort Answer

The activation energy of a reaction is 75.2 kJ mol

^{- 1}

in the absence of a catalyst and 50.14 kJ mol

^{- 1}

with a catalyst. How many times will the rate of the reaction grow in the presence of the catalyst if the reaction proceeds at 25°C? (R = 8.314 J K

^{- 1}

mol

^{- 1}

)

Answer: ≈ 2.5 × 10⁴ times

Solution

With the same frequency factor,

\frac{k _{c a t}}{k} = e^{(E_{a} - E_{a}^{'}) / R T}

\frac{E _{a} - E _{a}^{'}}{R T} = \frac{( 75.2 - 50.14 ) \times 1 0 ^{3}}{8.314 \times 298} = \frac{25060}{2477.6} = 10.11

\frac{k _{c a t}}{k} = e^{10.11}

\approx 2.5 \times 1 0^{4} times

CBSE Class 12 Board · 2005 · All India · Subjective