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The rate constant; k; for a first-order reaction is equal to 3.60 10-5 $ What is the half-life for the reaction (in minutes)?2.S0 * 10 $ minutes231 minutes321 minut...

Question

The rate constant; k; for a first-order reaction is equal to 3.60 10-5 $ What is the half-life for the reaction (in minutes)?2.S0 * 10 $ minutes231 minutes321 minutes1.93 * 10* minutes

The rate constant; k; for a first-order reaction is equal to 3.60 10-5 $ What is the half-life for the reaction (in minutes)? 2.S0 * 10 $ minutes 231 minutes 321 minutes 1.93 * 10* minutes



Answers

The rate constant for a certain reaction is $1.4 \times$ $10^{-5} M^{-1} \min ^{-1}$ at $483 \mathrm{~K}$. The activation energy for the reaction is $2.11 \times 10^{3} \mathrm{~J} / \mathrm{mol}$. What is the rate constant for the reaction at $611 \mathrm{~K} ?$

So in this podcast series, we're taking a look at the model that is known as the kinetic theory of. So we assume that the molecules are very small relative to the distance between the molecules. They're in constant random motion and will often collide with one another as well as the container off the gas. So that is the kinetic theory of just to offer some contextual information. What we're taking a look at here is the half life. So the half life that is T 1/2 is Thebes. It takes for half of our materials to be consumed on. We find that our half life here 1.70 times, 10 to the minus three seconds. What?

Okay, so we will to find ah rate of the K K. Given that are half life is eight days sore. H really is eight. So let's solve the fallen for K. So let's take base kind to both sides. So we get one when it's k is equal to 10 to the power of negative 0.3 over eight Jack one On both sides, case you have negative. K is equal to 10 to hardening of seven point 3/8, minus one. Now let's multiply both sides by a negative number. We have minus plus here. And now let's put this into our calculator. Okay, so we get that Arcaro you is 0.8 to 7.

The rate constant for a certain radioactive knew Clyde to disappear by radioactive decay is 1.0 times 10 to the negative third hours. The question is, what will the half life of this particular knew Clyde be? And for that we'll need a certain equation. The half life is equal to the negative natural log of two all over the rate. Constance. Therefore, this is a simple problem of simply plugging in. When this is computed, you will find that calculator will give you a value of about 693 hours. This should then be rounded down to the proper significant figure value of 1/2 life of 690 hours.

The question here wants us to find the half life of a new client, which has a rate constant of one time to 10 to the power of negative three. So in order to do that to find the half life, we basically take the equation. The natural log of two over Lambda, where Lambda is the great constant. So in this particular case, it's going to be the natural log on to over one times 10 to the negative three, and therefore the half life is going to be a value off 6 93.15 So about 693 days.


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