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JEE Main 202116 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
16 March
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2021 · 16 March · Morning Shift · Chemistry · Chemical Kinetics

The decomposition of formic acid on gold surface follows first...

The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300K300K is 1.0×10−3s−11 . 0 \times {10}^{- 3}{s}^{- 1} and the activation energy Ea=11.488kJmol−1{E}_{a} = 11 . 488kJ{mol}^{- 1}, the rate constant at 200K200K is _________ ×10−5s−1\times {10}^{- 5}{s}^{- 1}. (Round of to the Nearest Integer).

(Given R=8.314Jmol−1K−1R = 8 . 314J{mol}^{- 1}{K}^{- 1})

Correct answer

0

Step-by-step solution

K300=10−3K200=?{K}_{300} = {10}^{- 3}{K}_{200} = ?

Ea=11.488KJ/moleR=8.314J/mole−K{E}_{a} = 11 . 488KJ / moleR = 8 . 314J / mole - K

so ℓn(K300K200)=EaR(1200−1300)ℓn\left(\dfrac{{K}_{300}}{{K}_{200}}\right) = \dfrac{{E}_{a}}{R}\left(\dfrac{1}{200} - \dfrac{1}{300}\right)

ℓn(K300K200)=11.488×1000×1008.314×200×300ℓn\left(\dfrac{{K}_{300}}{{K}_{200}}\right) = \dfrac{11 . 488 \times 1000 \times 100}{8 . 314 \times 200 \times 300}

=2.303= 2 . 303

=ℓn10= ℓn10

so K300K200=10\dfrac{{K}_{300}}{{K}_{200}} = 10

K200=110×K300=10−4{K}_{200} = \dfrac{1}{10} \times {K}_{300} = {10}^{- 4}

=10×10−5sec⁡−1= 10 \times {10}^{- 5}\sec^{- 1}

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