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The rate constant for the second-order reaction$$2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$is $0.80 / M \cdot$ s at $10^{\circ} \mathrm{C}$. (a) Starting with a concentration of $0.086 M,$ calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when $[\mathrm{NOBr}]_{0}=0.072 M$ and $[\mathrm{NOBr}]_{0}=0.054$ $M$.

(a) $0.034 \mathrm{M}$(b) $17 s, 23 \mathrm{s}$

Chemistry 102

Chapter 6

Chemical Kinetics

Kinetics

Drexel University

University of Maryland - University College

University of Toronto

Lectures

22:42

In probability theory, the conditional probability of an event A given that another event B has occurred is defined as the probability of A given B, written as P(A|B). It is a function of the probability of B, the probability of A given B, and the probability of B.

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In chemistry, kinetics is the study of the rates of chemical reactions. The rate of a reaction is the change in concentration of a reactant over time. The rate of reaction is dependent on the concentration of the reactants, temperature, and the activation energy of the reaction.

05:22

The rate constant for the …

01:38

The rate law for the react…

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Okay, So in this question, they want us to calculate the final concentration of an OBE are after 22 seconds for part a. Right, So we know K is equal to 0.8, and we know that time is 22 seconds and they tell us that the initial concentration So a not is you go to 220.86 Okay, so since is just a second order reaction. We have to know that this is that they tell us is the second order because there's a difference between 1st 2nd and zero order in the equations that we use. So the question for the second order that we need to solve for this question is one over the concentration after time. Does he go to K T plus one over the initial? What he a not right. So we can do is we plug this values into the equation and we should get a after time is equal 2.8 22 plus one over point. Oh, a sixth. This gives us one over concentration at the time as you go to 29.23 And now we multiply both sides by concentration after time no one is able to to 9.23 Yeah, a after time. Now we divide both sides by 29.23 so we can isolate the concentration after time. And we should get the final concentration as point over three for two more left Devonport be They want us to calculate the half life we know the K is the same as above, which is 0.8 and they give us the initial consultation as so initial concentration. Does it go too important? 07 to M. So the question I went to you. Since this is a second order reaction, we need to use the equation for half life that belongs to second order. Right? And that equation is t half life is equal to one over K a. Not that was plugged the values in No. And we get 17.36 seconds as our answer. Careful, Parsi is exactly the same concept. We just have a different concentration. So we get 0.8 times by de concentration is 0.54 and we get the answer is 23.148 So the most important thing when some of these questions that you need to pick the right equation to use

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