更新1:
I understand how to get to .0123 min^-1 but it has to be in s^-1 and .74^-1 (.0123*60) isnt the correct answer
Help! rate constant with half life at specific temp?
2018-07-09 8:41 pm
At 415 ∘C, (CH2)2O decomposes in the gas phase, (CH2)2O(g)→CH4(g)+CO(g). If the reaction is first order with a half-life of 56.3 min at this temperature, calculate the rate constant
回答 (3)
2018-07-09 10:35 pm
For a first-order reaction, half-life, t = ln(2) / k
Rate constant, k = ln(2) / t = ln(2) / (56.3 × 60 s) = 2.05 × 10⁻⁴ s⁻¹ = 0.000205 s⁻¹
Rate constant, k = ln(2) / t = ln(2) / (56.3 × 60 s) = 2.05 × 10⁻⁴ s⁻¹ = 0.000205 s⁻¹
2018-07-10 3:52 am
Rate constant.....
For first order kinetics k = ln2 / t½
To get k in reciprocal seconds you don't multiply the rate constant in reciprocal minutes by 60. You divide by 60. If you include the units and use the unit-factor method, it will make a great deal more sense.
k = 0.0123 1/min x (1 min / 60 seconds) = 2.05x10^-4 1/s
When t½ = 56.3 minutes
k = 0.693 / 56.3 minutes = 0.0123 min⁻¹
k = 0.693 / (56.3 min x (60 s / 1 min) = 2.05x10^-4 s⁻¹
k = 0.693 / (56.3 min x 1 hr / 60 min) = 0.739 hr⁻¹
k = 0.693 / (56.3 min x 1 fortnight / 20160 min) = 248 fortnight⁻¹
Each of these values of k is equivalent to each of the others. Only the time units are different. If anyone told you that any of these are incorrect, then they are working a different problem.
For first order kinetics k = ln2 / t½
To get k in reciprocal seconds you don't multiply the rate constant in reciprocal minutes by 60. You divide by 60. If you include the units and use the unit-factor method, it will make a great deal more sense.
k = 0.0123 1/min x (1 min / 60 seconds) = 2.05x10^-4 1/s
When t½ = 56.3 minutes
k = 0.693 / 56.3 minutes = 0.0123 min⁻¹
k = 0.693 / (56.3 min x (60 s / 1 min) = 2.05x10^-4 s⁻¹
k = 0.693 / (56.3 min x 1 hr / 60 min) = 0.739 hr⁻¹
k = 0.693 / (56.3 min x 1 fortnight / 20160 min) = 248 fortnight⁻¹
Each of these values of k is equivalent to each of the others. Only the time units are different. If anyone told you that any of these are incorrect, then they are working a different problem.
2018-07-09 8:55 pm
Half life formula for a first order reaction is:
t = (ln 2) / k
Plugging in:
56.3 min = (ln 2) / k
k = (ln 2) / (56.3 min)
k = 0.0123 min⁻¹
t = (ln 2) / k
Plugging in:
56.3 min = (ln 2) / k
k = (ln 2) / (56.3 min)
k = 0.0123 min⁻¹
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