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Consider the reaction$$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)$$Suppose that at a particular moment during the reaction nitric oxide (NO) is reacting at the rate of 0.066 M/s. (a) At what rate is $\mathrm{NO}_{2}$ being formed?(b) At what rate is molecular oxygen reacting?

(a) $0.066 \mathrm{M} / \mathrm{s}$(b) $0.033 \mathrm{M} / \mathrm{s}$

Chemistry 102

Chapter 6

Chemical Kinetics

Kinetics

Brown University

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.

04:55

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.

02:36

Consider the reaction$…

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One reaction that may occu…

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For the reaction of $\math…

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The following data were co…

01:04

Nitric oxide reacts with o…

04:40

Data for the reaction $2 \…

first stuff for this problem is to write the reaction. Great expression. Fourth, this reaction how to write the expressions for the react Inside the reaction rate expression is written as negative the change in the concentration of the reactant over the change in time and then the product side is written as the positive change and concentrations of product divided by the change in time we take this into account will start with E first term here to in no, we write this reaction rate expression as negative 1/2 times a change in the concentration of you know, divided by the change in time or a second term. 02 is written as negative a negative change in the concentration of 02 divided by the change and time, and our final term is written as 1/2 times the change of concentration of Indo two divided by so from here problem A asked us to determine the rate that in no. Two is being formed. So were given that the rate of you know over tea there is 0.66 polarity per second. We also know that these expressions are equal our rate reaction expressions here. So the negative 1/2 delta two concentration of a no over delta T is the same is in value of the negative 1/2 times the change in concentration of ino to over tea. If we substitute in this 0.66 similarity for a second into our rate expression for this, you know, we end up with negative 1/2 times 0.66 polarity per second. If we multiply through this negative 1/2 over to here to the other side, we are left with 0.66 polarity for second, as are re that in no. Two is being formed. So for be the problem, it's much the same again. Or given that the rate nitric oxide is reacting is 0.66 polarity per second. Again, we know that these two rate expressions are equivalent. So we're gonna substitute in 0.66 molar ity for second for our nitrous oxide expression there who pay attention here. We need to multiply this 0.66 by negative 1/2 and they were gonna divide by negative wine which will give us 0.33 polarity for second for the oxygen rate

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