Trimethylamine and chlorine dioxide react in water in an electron transfer reaction to form the trimethylamine cation and chlorite ion: 1CH323 N1aq2 + ClO21aq2 + H2O1l2S 1CH323 NH+1aq2 + ClO2 -1aq2 + OH-1aq2 Initial rate data obtained at 23 °C are listed in the following table. (b) What would be the initial rate in an experiment with initial concentrations 31CH323 N4 = 4.2 * 10-2 M and 3ClO24 = 3.4 * 10-2 M?

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Hello everyone today. We have the following problem. Nitrogen dioxide is formed from the reaction of nitrogen monoxide and oxygen gas tabulated below is the initial rate of formation of our nitrogen dioxide data calculate the initial rate of formation of nitrogen dioxide when the initial concentrations of nitric oxide or nitrogen monoxide Is equal to .3 Mueller. And oxygen gas concentration is .450 Moeller. So from the given reaction present here, our rate law is going to be equal to the rate is equal to some constant K. Times. We're going to have our nitrogen monoxide raised to the power of em for a given variable times our concentration of oxygen gas raised to N. So we're taking into account our reactant because that is what the rate law is based on. And so now we need to determine what M and N. Are from our data. And so we're going to be using different experiments in comparing the numbers. So, first we're going to be solving for M. And so to do that. We're going to look at two of these experiments and we can look at experiment three and experiment one. And so why is that? So if we see here, the concentration for our oxygen is constant .150.150. However, the concentration for our nitrogen monoxide on the other hand, is going to differ and we can use this differing value here to solve for M. So what is that going to look like? Well, that's going to be the rate of three over the rate of one and that's reaction 13 and one respectively, and how we're going to set this up is we are essentially going to take our nitrogen monoxide raised to the power of M. We're going to multiply that by our concentration of oxygen gas raised to the power of And then what we're essentially going to do is the same on the bottom as well. four Reaction 3. That's going to look like 0. to the brace of the power of N. Times 0.1 50 race of the power of N over 0.1 50. Race of the power of M times 0.50 race of the power of N. And We're going to equal that to their reaction rates of formation because we also need to include that data. So we're going to put 6.75 times 10 to the negative third in the numerator. And we're gonna put 1.6875 times 10 to the negative third in the denominator. And when we simplify this out, we're going to get to race of the power of M is equal to four. And if we saw this mathematically, we can say that M is going to be equal to two. So we're gonna keep that number in the back of our pocket. We're gonna do the same process for Rates three or reaction three over the rate of reaction to. And so we're essentially going to follow the same formula with our forced formula there. However, we're going to well first we're going to keep the numerator is the same, But for the nominators instead of .150, it's going to be .3 being raised to these powers. And so we're gonna equal them to the rates as before. And when we get this, we get that 0.5 raised to the power of N is equal to 0.5. And if we saw this mathematically in is going to equal one. So since M is equal to two when N is equal to one, the rate law as we've got before is going to be equal to some constant K. Times the concentration of nitrogen monoxide raised to the power of two times some concentration of oxygen gas with no coefficient or no exponents. So from here, what can we do? Well, we can actually derive our rate or K. So the first thing on for that that we can do is we note that K. Some value K is going to be equal to the rate over the concentration of our nitrogen monoxide squared times the concentration of oxygen gas. And we derived this from the equation from before. And so what is this going to look like? Well, We're going to use experiment one To help us solve for this. So for the rate for that, we're going to get 1.6875 times 10 to the negative 3rd and divide that by 0. 50 squared times just 500.1 50. And this is going to give us 0.5 Mueller to the power of -2 seconds to the power of one. And so with this number here, we are going to be able to finally solve for our rate. So we can say that the rate is going to be equal to some constant K. Times our concentration of nitrogen monoxide. That's squared times a concentration of oxygen gas For K. We already solved. We have 0. And we're going to multiply that by our 0.300 molar squared times our concentration for our oxygen, gas. Which will be solved before was 0.150. And when we do this, we get an answer of 2.25 times 10 to the negative two molar per second. And so this is going to be the initial rate of formation. And with that we've answered the question overall, I hope that this helped. And until next time

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Textbook Question

Chlorite is reduced by bromide in acidic solution according to the following balanced equation: ClO2 -1aq2 + 4 Br-1aq2 + 4 H+1aq2S Cl-1aq2 + 2 Br21aq2 + 2 H2O1l2 (a) If Δ3Br24>Δt = 4.8 * 10-6 M>s, what is the value of Δ3ClO2 -4>Δt during the same time interval?

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Textbook Question

The reaction 2NO1g2 + 2 H21g2S N21g2 + 2 H2O1g2 is first order in H2 and second order in NO. Write the rate law, and specify the units of the rate constant.

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Textbook Question

Initial rate data at 25 °C are listed in the table for the reaction NH4 +1aq2 + NO2 -1aq2S N21g2 + 2 H2O1l2

(b) What is the value of the rate constant?

Video transcript

Hello everyone today. We have the following problem. Nitrogen dioxide is formed from the reaction of nitrogen monoxide and oxygen gas tabulated below is the initial rate of formation of our nitrogen dioxide data calculate the initial rate of formation of nitrogen dioxide when the initial concentrations of nitric oxide or nitrogen monoxide Is equal to .3 Mueller. And oxygen gas concentration is .450 Moeller. So from the given reaction present here, our rate law is going to be equal to the rate is equal to some constant K. Times. We're going to have our nitrogen monoxide raised to the power of em for a given variable times our concentration of oxygen gas raised to N. So we're taking into account our reactant because that is what the rate law is based on. And so now we need to determine what M and N. Are from our data. And so we're going to be using different experiments in comparing the numbers. So, first we're going to be solving for M. And so to do that. We're going to look at two of these experiments and we can look at experiment three and experiment one. And so why is that? So if we see here, the concentration for our oxygen is constant .150.150. However, the concentration for our nitrogen monoxide on the other hand, is going to differ and we can use this differing value here to solve for M. So what is that going to look like? Well, that's going to be the rate of three over the rate of one and that's reaction 13 and one respectively, and how we're going to set this up is we are essentially going to take our nitrogen monoxide raised to the power of M. We're going to multiply that by our concentration of oxygen gas raised to the power of And then what we're essentially going to do is the same on the bottom as well. four Reaction 3. That's going to look like 0. to the brace of the power of N. Times 0.1 50 race of the power of N over 0.1 50. Race of the power of M times 0.50 race of the power of N. And We're going to equal that to their reaction rates of formation because we also need to include that data. So we're going to put 6.75 times 10 to the negative third in the numerator. And we're gonna put 1.6875 times 10 to the negative third in the denominator. And when we simplify this out, we're going to get to race of the power of M is equal to four. And if we saw this mathematically, we can say that M is going to be equal to two. So we're gonna keep that number in the back of our pocket. We're gonna do the same process for Rates three or reaction three over the rate of reaction to. And so we're essentially going to follow the same formula with our forced formula there. However, we're going to well first we're going to keep the numerator is the same, But for the nominators instead of .150, it's going to be .3 being raised to these powers. And so we're gonna equal them to the rates as before. And when we get this, we get that 0.5 raised to the power of N is equal to 0.5. And if we saw this mathematically in is going to equal one. So since M is equal to two when N is equal to one, the rate law as we've got before is going to be equal to some constant K. Times the concentration of nitrogen monoxide raised to the power of two times some concentration of oxygen gas with no coefficient or no exponents. So from here, what can we do? Well, we can actually derive our rate or K. So the first thing on for that that we can do is we note that K. Some value K is going to be equal to the rate over the concentration of our nitrogen monoxide squared times the concentration of oxygen gas. And we derived this from the equation from before. And so what is this going to look like? Well, We're going to use experiment one To help us solve for this. So for the rate for that, we're going to get 1.6875 times 10 to the negative 3rd and divide that by 0. 50 squared times just 500.1 50. And this is going to give us 0.5 Mueller to the power of -2 seconds to the power of one. And so with this number here, we are going to be able to finally solve for our rate. So we can say that the rate is going to be equal to some constant K. Times our concentration of nitrogen monoxide. That's squared times a concentration of oxygen gas For K. We already solved. We have 0. And we're going to multiply that by our 0.300 molar squared times our concentration for our oxygen, gas. Which will be solved before was 0.150. And when we do this, we get an answer of 2.25 times 10 to the negative two molar per second. And so this is going to be the initial rate of formation. And with that we've answered the question overall, I hope that this helped. And until next time