JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Chemical Kinetics

The activation energy of one of the reactions in a biochemical process is

532611 J mol^{- 1}

. When the temperature falls from 310 K to 300 K, the change in rate constant observed is

k_{300} = x \times 1 0^{- 3} k_{310}

. The value of x is
[Given:

ln 10 = 2.3

R = 8.3 J K^{- 1} mol^{- 1}

]

Answer: 1

Solution

Using the Arrhenius equation:

ln (\frac{k _{2}}{k _{1}}) = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

Let

k_{1} = k_{300}

and

k_{2} = k_{310}

T_{1} = 300 K

T_{2} = 310 K

ln (\frac{k _{310}}{k _{300}}) = \frac{532611}{8.3} (\frac{1}{300} - \frac{1}{310}) = \frac{532611}{8.3} \times \frac{10}{300 \times 310}

= \frac{532611 \times 10}{8.3 \times 93000} = 6.9 = 3 \times 2.3 = 3 ln 10 = ln 1 0^{3}

Therefore:

\frac{k _{310}}{k _{300}} = 1 0^{3}

, so

k_{300} = k_{310} \times 1 0^{- 3}

Comparing with

k_{300} = x \times 1 0^{- 3} k_{310}

, we get

x = 1

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Integer answer