JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Chemical Kinetics

For a reaction, given below is the graph of

ln k

\frac{1}{T}

. The activation energy for the reaction is equal to ____ cal mol

^{- 1}

. (Nearest integer).
(Given :

R = 2 cal K^{- 1} mol^{- 1}

)

Answer: 8

Solution

From the Arrhenius equation:

k = A e^{- E_{a} / R T}

Taking natural logarithm:

ln k = - \frac{E _{a}}{R} \cdot \frac{1}{T} + ln A

This is a linear equation where the slope =

- \frac{E _{a}}{R}

From the graph, the slope =

\frac{Δ ( l n k )}{Δ ( 1/ T )} = \frac{- 20}{5} = - 4

Therefore:

- \frac{E _{a}}{R} = - 4

E_{a} = 4 R = 4 \times 2 = 8

cal mol

^{- 1}

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Integer answer