In a first-order decomposition reaction, 50.0% of a compound decomposes in 14.5

Question

In a first-order decomposition reaction, 50.0% of a compound decomposes in 14.5 mins. What is the rate constant of the reaction? (Round to 4th decimal place)
K = _________min^-1

How long does it take for 82.0% of the compound to decompose? (Round to 1st decimal place)
t = ___________min In a first-order decomposition reaction, 50.0% of a compound decomposes in 14.5 mins. What is the rate constant of the reaction? (Round to 4th decimal place)
K = _________min^-1

How long does it take for 82.0% of the compound to decompose? (Round to 1st decimal place)
t = ___________min In a first-order decomposition reaction, 50.0% of a compound decomposes in 14.5 mins. What is the rate constant of the reaction? (Round to 4th decimal place)
K = _________min^-1

How long does it take for 82.0% of the compound to decompose? (Round to 1st decimal place)
t = ___________min

Explanation / Answer

Part A

he has given t1/2 = 14.5 min

there is a direct relation

k = 0.693 / t1/2

k = 0.693 / 14.5

k = 0.0478 min-1

Part B

formula is

k = (2.303 / t) log(a0/a)

where a0 = intial concentration = 100 %

a = final concentration = 100 - 82 = 16%

0.0478 min-1= (2.303 / t) log (100 / 16)

t = 1.833 / 0.0478

t = 38.3 min

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