The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0×10−3 s−1 and the activation energy Ea=11.488 kJ mol−1, the rate constant at 200 K is _____×10−1 . (Round off to the nearest Integer).(Given : R=8.314 J mol−1 K−1 )

Name: Video solution 1: The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0×10−3 s−1 and the activation energy Ea=11.488 kJ mol−1, the rate constant at 200 K is _____×10−1 . (Round off to the nearest Integer).(Given : R=8.314 J mol−1 K−1 )
Uploaded: 2022-06-12T10:04:45.000Z
Duration: 9 min
Description: Video solution 1: The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0×10−3 s−1 and the activation energy Ea=11.488 kJ mol−1, the rate constant at 200 K is _____×10−1 . (Round off to the nearest Integer).(Given : R=8.314 J mol−1 K−1 )

T1=300 K K1−1×10−3 s−1 T2=200 K K2=?log K1K2=2.303REa[ T11− T21]log10−3K2=2.303×8.31411.488×103×(600−1)log K210−3=1log K210−3=log10 K210−3=10 K2=1010−3=10−4=10×10−5 s−1

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Theories of chemical kinetics

The decomposition of formic acid on gold surface follows first

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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 _____ $\times 1 0^{- 1}$ . (Round off to the nearest Integer).
(Given : $R = 8.314 J mol^{- 1} K^{- 1}$ )

Views: 5,793 students

Updated on: Jul 19, 2022

Exam:

JEE Mains 2021 Mar 16 Shift 1

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Michael Discussed

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

300 K

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^{- 1}

. (Round off to the nearest Integer).
(Given :

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

)

9 mins ago

Discuss this question LIVE

9 mins ago

Text solutionVerified

T_{1} = 300 K K_{1} - 1 \times 1 0^{- 3} s^{- 1} T_{2} = 200 K K_{2} = ? lo g \frac{K _{2}}{K _{1}} = \frac{E _{a}}{2.303 R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}] lo g \frac{K _{2}}{1 0 ^{- 3}} = \frac{11.488 \times 1 0 ^{3}}{2.303 \times 8.314} \times (\frac{- 1}{600}) lo g \frac{1 0 ^{- 3}}{K _{2}} = 1 lo g \frac{1 0 ^{- 3}}{K _{2}} = lo g 10 \frac{1 0 ^{- 3}}{K _{2}} = 10 K_{2} = \frac{1 0 ^{- 3}}{10} = 1 0^{- 4} = 10 \times 1 0^{- 5} s^{- 1}

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Question Text	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 _____ $\times 1 0^{- 1}$ . (Round off to the nearest Integer). (Given : $R = 8.314 J mol^{- 1} K^{- 1}$ )
Updated On	Jul 19, 2022
Topic	Chemical Kinetics
Subject	Chemistry
Class	Class 12
Answer Type	Text solution:1 Video solution: 2
Upvotes	237
Avg. Video Duration	6 min

Text solutionVerified

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The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0×10−3 s−1 and the activation energy Ea​=11.488 kJ mol−1, the rate constant at 200 K is _____×10−1 . (Round off to the nearest Integer).(Given : R=8.314 J mol−1 K−1 )

Text solutionVerified

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Practice questions from Chemical Kinetics in the same exam

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