Asked by bobjahng2
follows first-order kinetics and has an activation energy of 244 kJ mol-1. At 684 K, k = 0.00000508 s-1. What is the half-life (min) for this reaction at 427 °C ? Round your answer to 3 significant figures.
I really don't know what to do
I really don't know what to do
Answers
Answered by
DrBob222
You have k at one T (684) and you can calculate k at another T (427) with the Arrhenius equation. Then knowing it is first order, you can use
ln(No/N) = kt.
You can use any convenient number (such as 100) for No, use half that for N (50 if you use 100), plug in k at 427 (that's the T for which you want the 1/2 life), and solve for t. That will be the half-life because you have just 1/2 the number of atoms you started with.
ln(No/N) = kt.
You can use any convenient number (such as 100) for No, use half that for N (50 if you use 100), plug in k at 427 (that's the T for which you want the 1/2 life), and solve for t. That will be the half-life because you have just 1/2 the number of atoms you started with.
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