The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is \(1.0 \times 10^{-3} \mathrm{~s}^{-1}\) and the activation energy \(\mathrm{E}_a=11.488 \mathrm{~kJ} \mathrm{~mol}^{-1}\), the rate constant at 200 K is \_\_\_\_ \(\times 10^{-5} \mathrm{~s}^{-1}\). (Round off to the nearest Integer). [Given R \(=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) ]

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Chemical Kinetics

The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is

1.0 \times 1 0^{- 3} s^{- 1}

and the activation energy

E_{a} = 11.488 kJ mol^{- 1}

, the rate constant at 200 K is ____

\times 1 0^{- 5} s^{- 1}

. (Round off to the nearest Integer).
[Given R

= 8.314 J mol^{- 1} K^{- 1}

]

Answer: 10

Two rate constants of the same reaction are tied by the Arrhenius equation:

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

With

k_{1} = 1.0 \times 1 0^{- 3} s^{- 1}

T_{1} = 300

K and

T_{2} = 200

ln \frac{k _{2}}{1.0 \times 1 0 ^{- 3}} = \frac{11488}{8.314} (\frac{1}{300} - \frac{1}{200}) = 1381.8 \times (- \frac{1}{600}) = - 2.303

Since

ln 10 = 2.303

, the ratio

k_{2} / k_{1}

1 0^{- 1}

, giving

k_{2} = 1.0 \times 1 0^{- 4} s^{- 1}

.
The rate constant at 200 K is

10 \times 1 0^{- 5} s^{- 1}

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Integer answer