The react ion between ethyl bromide 1C2H5Br2 and hydroxide ion in ethyl alcohol at 330 K, C2H5Br1alc2 + OH- 1alc2¡ C2H5OH1l2 + Br - 1alc2, is first order each in ethyl bromide and hydroxide ion. When 3C2H5Br4 is 0.0477 M and 3OH- 4 is 0.100 M, the rate of disappearance of ethyl bromide is 1.7 * 10-7 M>s. (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

Video transcript

Welcome back everyone. In this example, we need to consider the reaction between nitric oxide and ozone gas to produce products, nitrogen dioxide and oxygen gas. We're told that we have a first order reaction with respect to our reactants. And that the rate of disappearance of our reactant nitric oxide is equal to a value of 8.54 times 10 to the fourth power molar per second. Based on the concentration of our reactants, we need to assume that our reaction is done in a sealed vessel. And what happens to the rate of disappearance of nitric oxide when the volume of the vessel is tripled. And so, because the prompt states that we have our reaction done in a sealed vessel, what this is telling us is that because we have a closed vessel, therefore, there will be no change in our volume of the system. So the volume of our system is going to remain as the same because it's going to have a change of zero. And we would say thus, no gaseous particles will escape the system. And so we can conclude that our change in entropy of our system delta H we should recall is going to equal our change in internal energy of our system. So we should focus on the kinetics of how our reactant or rather the consumption of our reactants affects the rate of our reaction. And we want to recall that for our reaction rate, specifically for the rate law that is going to equal our rate constant K, which we are not given multiplied by the concentration of our first reactant, which given in our bounced equation. In the prompt, we have nitric oxide as our first reactant and it has a coefficient of one here. So we would raise that to one. It's then multiplied by the concentration of our second reactant, which we are given in the prompt as ozone, which also has a coefficient of one according to our prompt. And so to simplify this rate law, we would say that our rate law is equal to a rate constant K multiplied by the concentration of nitric oxide times the concentration of ozone. And so we simplify it to this rate law because we don't necessarily need to include those exponents of one because they can just be interpreted the same way here. Now, we want to recognize that in our rate law, our reactants are expressed in terms of their concentration. And so according to our prompt, we're given our concentration for each of our reactants and units of molar here. So as you can see, concentration is expressed in polarity and we want to recall that. So recall that molar is expressed in terms of moles of our reactant divided by the volume of our reactant. So if we think of this relationship with molar, in terms of our dilution formula, which we should recall is our initial concentration of our reactants times their initial volume will be equal to the final concentration of our reactants times their final volume. We can realize how tripling our volume here is going to change our rate that we express here. And so again, this equation here we should recall is for a dilution which we have here going on in our prompt for our gasses. And according to the prompt, our final volume, which would be V two, once our reactants are consumed, it's going to be equal to our initial volumes of our reactants being tripled. And so we would multiply them by three. So just to make that clear, that is three times the initial volume. And so because the prompt is asking us what will happen to the rate of disappearance of our reactant nitric oxide. This means that in terms of our dilution formula, we want to isolate for our final concentration of our reactants. And so we would say that because we're tripling the volume in our vessel, therefore, the final concentration of our reactant is going to equal in our numerator, the moles of our reactants, which according to our prompt, we have one mole of nitric oxide and one mole of ozone. So we would say one mole of our reactants divided by our volume being tripled to now three liters. And so we would have one third times our initial concentration of our reactants. M one. And as we stated before, because we understand that our rate law is expressed in terms of concentration of our reactants, which we understand is in molar, we can rewrite our rate law. So that we say that our rate is now equal to our rate constant K, where we take one third times the initial concentration of our first reactant nitric oxide multiplied by one third times the initial concentration of our second reactant ozone. And as we stated before, we have exponents of one which we simplify to just the brackets here because it's implied. And so multiplying one third by one third, we would simplify this so that we have the rate equal to 1/9 of our initial rate. And this is once our volume is tripled in our vessel. And because we recognize that 1/9 is a much smaller value than one third, we would say that therefore, for the rate of disappearance of our reactant nitric oxide, our rate is going to be decreased or we can say rather decreases by 1/9 of our original rate. And so this statement here is going to be our final answer to express why tripling the volume of our reaction vessel is going to slow down the rate of disappearance of our reactants specifically for nitric oxide. And sorry about that. So this is our final answer here. If you have any questions, please leave them down below. And I will see everyone in the next practice video.