The first order rate constant for the decomposition of $\mathrm{CaCO}_{3}$ at 700 K is $6.36 \times 10^{-3} \mathrm{~s}^{-1}$ and activation energy is $209 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Its rate constant (in $\mathrm{s}^{-1}$) at 600 K is $x \times 10^{-6}$. The value of x is ____. (nearest integer). [Given $\mathrm{R}=8.31 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} ; \log 6.36 \times 10^{-3}=-2.19,10^{-4.79}=1.62 \times 10^{-5}$]

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Chemical Kinetics

The first order rate constant for the decomposition of

CaCO_{3}

at 700 K is

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

and activation energy is

209 kJ mol^{- 1}

. Its rate constant (in

s^{- 1}

) at 600 K is

x \times 1 0^{- 6}

. The value of x is ____. (nearest integer).
[Given

R = 8.31 J K^{- 1} mol^{- 1}; lo g 6.36 \times 1 0^{- 3} = - 2.19, 1 0^{- 4.79} = 1.62 \times 1 0^{- 5}

]

Answer: 16

The Arrhenius equation in two-temperature form gives

lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}}) = \frac{209000}{2.303 \times 8.31} (\frac{1}{700} - \frac{1}{600}) = - 2.60.

With

lo g k_{1} = lo g (6.36 \times 1 0^{- 3}) = - 2.19

lo g k_{2} = - 2.19 - 2.60 = - 4.79, k_{2} = 1 0^{- 4.79} = 1.62 \times 1 0^{- 5} s^{- 1} .

This equals

16.2 \times 1 0^{- 6} s^{- 1}

, so

x = 16

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Integer answer