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

1 answer

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.
Similar Questions
  1. 1. The reaction,CO(CH2COOH)2 CO(CH3)2 + 2CO2, follows first-order kinetics and has an activation energy of 96.6 kJ mol-1. At 293
    1. answers icon 0 answers
  2. The given reaction,2N2O5 4NO2 + O2, follows first-order kinetics and has an activation energy of 103 kJ mol-1. At 298 K, k =
    1. answers icon 4 answers
  3. How do catalysts affect a chemical reaction?(1 point)Responses provide activation energy provide activation energy increase
    1. answers icon 2 answers
  4. How do catalysts affect a chemical reaction?(1 point)Responses lose activation energy lose activation energy provide activation
    1. answers icon 1 answer
more similar questions