Question

a.) Technetium-99m (99mTc) is commonly used as a radioisotope for medical purposes. The “m” denotes it as metastable since its nucleus is in an excited state, which relaxes to its ground nuclear state by emitting gamma radiation with a half-life of 6.01 hours. For a 1.00 μg sample of 99mTc, how many hours will it take the sample to decompose to 0.275 μg? Enter your answer to the tenths place.

b.)The reaction of stratospheric ozone with chlorine radicals has an activation energy of 39.1 kJ/mol. If the rate constant for the reaction is 196.7 1/M∙s at -65.0 °C, at what temperature, in Kelvin, will the rate constant be doubled? Enter your answer to the ones place.

Answer 1

Given data: Half life of 99 mTc = 6.01 hours

Initial amount of 99 mTc = 1.00g

Amount of 99 mTc at time t = 0.275 g

We have , decay constant ( ) = 0.693 / t _1/2

Where, t _1/2 is a half life radioactive element.

Decay constant of 99 mTc = = 0.693 / t _1/2

= 0.693 / 6.01 hours

=0.1153 hr ^-1

We have equation : t = - 1/ ln ( N _t / N ₀)

Where N _t is a amount of radioactive element at time t and N ₀ is a amount of radioactive element at time t = 0

Therefore, t = - 1 / 0.1153 hr ^-1 ln ( 0.275 g / 1.00g )

= - 1 / 0.1153 hr ^-1 ln 0.275

= - 1 / 0.1153 hr ^-1 x ( -1.291 )

= 1.291 / 0.1153 hr ^-1

t = 11.20 hours

ANSWER : Sample will take 11.20 hours to decompose from 1.00 g to 0.275 g

2) Given Activation Energy of reaction Ea = 39.1 kJ = 39100 J

Rate constant of reaction at temperature T ₁₍ k ₁)= 196.7 1 / M s

Rate constant of reaction at temperature (T ₂ ) =K ₂=2 k ₁

Temperature of reaction = - 65.0 ⁰ C = 208.15 K  

Temperature of reaction ( T ₂ ) = ?

We have Arrhenius equation at two different temperatures.

log k 2 / k 1 = Ea / 2.303 R [ 1/ T ₁ - 1 / T ₂ )

Here K ₂=2 k ₁ .

log 2 k ₁ ./ k ₁ .= 39100 / 2.303 x ( 8.314 ) [ 1/ 208.15 K - 1 / T ₂ ]

log 2 = 39100 / 19.15 [ 0.00480 - 1 / T ₂ ]

0.3010 = 2041.8 [ 0.00480 - 1 / T ₂ ]

0.3010 = 9.8 - 2041.8 / T ₂

- 2041.8 / T ₂ = 0.301 - 9.8 = - 9.499

2041.8 /T ₂ ​​​​​​​ = 9.499

Therefore, T ₂ ​​​​​​​ = 2041.8 / 9.499 = 214.9 K

ANSWER : Temperature at which rate will double = 214.9 K