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The rte of a certain reaction is given by the following rate law: rte =k[No] I[o2] Use this information to answer the questions below:What is the reaction order in ...

Question

The rte of a certain reaction is given by the following rate law: rte =k[No] I[o2] Use this information to answer the questions below:What is the reaction order in NO?What is the reaction order in 02?What is overall reaction order?At a certain concentration of NO and Oz the initial rate of reaction is 27.0 M { What would the initia rate of the reaction be if the concentration of NO were doubled? Be sure your answer has the correct number of significant digits.0<The rate of the reaction is mea

The rte of a certain reaction is given by the following rate law: rte =k[No] I[o2] Use this information to answer the questions below: What is the reaction order in NO? What is the reaction order in 02? What is overall reaction order? At a certain concentration of NO and Oz the initial rate of reaction is 27.0 M { What would the initia rate of the reaction be if the concentration of NO were doubled? Be sure your answer has the correct number of significant digits. 0< The rate of the reaction is measured to be 0.120 M / $ when [No] 1.6 M and [Oz] 1,7 M Calculate the value of the rate constant: Be sure your answer has the correct number of significant digits_ t =



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Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in $\mathrm{H}_{2}$ and second order in NO. Write the rate law. (b) If the rate constant for this reaction at $1000 \mathrm{~K}$ is $6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}$, what is the reaction rate when $[\mathrm{NO}]=0.035 \mathrm{M}$ and $\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathrm{c}) \mathrm{What}$ is the reaction rate at $1000 \mathrm{~K}$ when the concentration of $\mathrm{NO}$ is increased to $0.10 \mathrm{M},$ while the concentration of $\mathrm{H}_{2}$ is $0.010 \mathrm{M}$ ?

Yeah. To answer this question, we need to determine what happens to the rate when just one of the reactant concentrations change. If we look carefully at the data provided you'll notice that when only A is tripled, the rate is tripled. This is indicative of a first order reaction with respect to that reactant. If we look at experiments two and 3. When B is tripled, the rate actually increases 3 to the two or 9 fold. This is indicative of a second order reaction with respect to that reactant. So the rate law is going to be equal to rate equals K. Multiplied by A to the first power and be to the second power. Knowing the rate law allows us now to rearrange it and solve for the rate constant if we know the rate and the concentrations and we do three different times with three different experiments. So we can take the data the rate and the concentrations for any one of these three experiments and we will get a K Value of 10, 1 over Mueller one over Mueller squared seconds. Now knowing the K value, we could calculate the rate at different reactant concentrations. Rate is equal to K which we now know, multiplied by the concentrations provided .2 for a and .2 for B. Not, don't forget to square B. And we get a rate of .08, 0 molar per second

Hello. So today we're looking out of reaction that IHS first order in respects to the concentration of A and the concentration of B. And at some point the concentration of A is 1.6 times sent to the negative, too. Concentration of B is 2.4 times 10 to the negative three, and our rate is 4.1 times 10 to the negative four. So what would be our rate? Constant? Well, this is very simple. If you remember, the rate law rate is equal to concentration is equal to the rate constant times the concentration of a and since its first order in respect to a there's just a little one there, and it's also first order in respect to be so There's a little one there, so it's just a concentration off A and B this concentration of eight times the constitution of B times, the rate constant is the rate. So how about we just plug in some values and now we'll just solve four k, so we're going to have that 4.1 times 10 Still negative for leader Mullah Araji over seconds is equal to 3.8 times 10 to the negative five polarity squared times K. So then we divide and we will see that the answer is 11 universe similarity and in for seconds. And there we go. So we found the rate constant By knowing what order A and B where we could find. And when knowing the rate in the concentrations, we can easily find the right constant.

So based on the initial rate data let's compare. Let's look at N. O. From the first of the second when H. Two remains constant. Um So you have the I know concentration multiplied by a factor of three. And the initial rate also multiplied by a factor of three. So that suggests that you know is first order. So put just N. O. In the right expression. And then for H. Two let's see where I know it's constant from the second to the third. and H two is going to multiply by a factor of four. And the rate goes from this is 9 to 36. Uh Is a factor of four as well. So that means that H two is uh To the first power because it has an order of one and then we're going to fill in that the rate is equal to each concentration times its order times the rate constant. Okay um let me just put here to for completeness first order and first order. No it's to calculate the rate constant. So that was part of a this is part B. Okay. Is equal to the initial rate for any one of the experiments divided by the concentration of each species. If you rearrange the equation. So let's just use the first one and plug in the numbers since it's right there 3.0 times into the negative third. Um Mueller per second. And then we divide by five times 10 to the negative 3rd and 2.5 times 10 to the negative third. Okay. Mhm. So this moller cancels out with this moller this 10 to the negative three cancels out with this 10 to the negative three. Um Oh And we're left with three divided by five divided by 2.5. It's going to be 0.24 Divided by 10 to the negative 3rd. And then we're left with um one time second, Which is .24 Times 10 to the positive 3rd. or make this one bigger. Make that one smaller, 2.4 Times 10 to the 2nd. More clarity to the -1 seconds to the -1 as our units. And then we want the initial rate when both of the concentrations are eight times 10 to the negative three two. So we know that the rate is equal to the rate constant, which is 2.4 times time squared. Okay. And then we're gonna multiply by eight Times 10 to the -3 and we'll just do that twice all put squared. Yeah, So we have 2.4 times eight times eight is going to be 100 and 53.6 times since the negative two times 10 to the negative six, which is That's the -4. More clarity over seconds. And then we'll move this one over to to make it uh smaller, so we need to make that one bigger. So it's 1.5 3, 6 times 10 to the negative two. It's going to be our rate under those conditions.

Let's consider the reaction given for E. Called the rate lot is first order in H two second order in you know so let's raise the rate law. It is equal to k. First order with respect to H two second order risk respect. And then also there is our rate lock for B. Were used to calculate the rate we're given the rate constant here which is 6.0 to 10 to the four M -2 S -1. The each concentration is .015 Mueller. Yeah And you know is .035 more squared. The rate would be equal to 1.1 more second or B. For C The rate is equal to our constant 6.02 times 10 to the 4th And -2 S -1. Each concentration is .010 Moller and Oh is .10 moller Squared. And a rate in c works out to 6.0 more per second. And for d rates is equal to the constant 6.010 to the 4th and -2 -1. H two is 030 Moller. Yeah. and Oh is .010 boy Squared. And the d works out 2.18 more per second.


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