JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Chemical Kinetics

If the activation energy of a reaction is

80.9 kJ mol^{- 1}

, the fraction of molecules at 700 K, having enough energy to react to form products is

e^{- x}

. The value of

x

is ____ (Rounded-off to the nearest integer).
[use

R = 8.31 JK^{- 1} mol^{- 1}

]

Answer: 14

Solution

The fraction of molecules with energy ≥ activation energy is given by the Arrhenius factor:

f = e^{- E_{a} / R T}

. Here

E_{a} = 80.9 \times 1 0^{3}

J/mol,

T = 700

R = 8.31

J K

^{- 1}

mol

^{- 1}

. Calculate

x = \frac{E _{a}}{R T} = \frac{80900}{8.31 \times 700} = \frac{80900}{5817} \approx 13.9 \approx 14

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Integer answer